Mathesis universalis

http://oeis.org/classic.html

The Wythoff array A035513 is shown below, to the right of the broken line. It has many wonderful properties, some of which are listed after the table. It is also related to a large number of sequences in the On-Line Encyclopedia.

 0    1  |   1    2    3    5    8   13   21   34   55   89  144  1    3  |   4    7   11   18   29   47   76  123  199  322  521  2    4  |   6   10   16   26   42   68  110  178  288  466  754  3    6  |   9   15   24   39   63  102  165  267  432  699 1131  4    8  |  12   20   32   52   84  136  220  356  576  932 1508  5    9  |  14   23   37   60   97  157  254  411  665 1076 1741    6   11  |  17   28   45   73  118  191  309  500  809 1309 2118  7   12  |  19   31   50   81  131  212  343  555  898 1453 2351  8   14  |  22   36   58   94  152  246  398  644 1042 1686 2728  9   16  |  25   41   66  107  173  280  453  733 1186 1919 3105 10   17  |  27   44   71  115  186  301  487  788 1275 2063 3338 11   19  |  30   49   79 12   21  |  33   54   87 13   22  |  35   57   92 

Some properties of the Wythoff array.

(For sources see the "References" below.)

• Construction (1): the two columns to the left of the broken line consist respectively of the nonnegative integers n, and the lower Wythoff sequence A000201, whose nth term is [(n+1)tau], where tau=(1+sqrt(5))/2. The rows are then filled in by the Fibonacci rule that each term is the sum of the two previous terms. The entry n in the first column is the index of that row.

• Two definitions: The Zeckendorf expansion of n is obtained by repeatedly subtracting the largest Fibonacci number you can until nothing remains; for example 100 = 89 + 8 + 3 (see A035514, A035515, A035516, A035517).
  The Fibonacci successor to (or left shift of) n, Sn, say, is found by replacing each F_i in the Zeckendorf expansion by F_i+1; for example the successor to 100 is S100 = 144 + 13 + 5 = 162. See A022342.

• Construction (2): the two columns to the left of the broken line read n, 1+Sn; then after the broken line the sequence is
  m       Sm       SSm       SSSm       SSSSm       ...  ,

  where m = n + 1 + Sn.

• Construction (3): Let {S1, S2, S3, S4, ...} = {2,3,5,7,8,10,11,...} be the sequence of Fibonacci successors A022342. The first column of the array consists of the numbers not in that sequence: 1,4,6,9,12,... (A007067). The rest of each row is filled in by repeatedly applying S.

• Construction (4): The entry in row n and column k is
  [ (n+1) tau ] F_k+2 + n F_k+1 ,

  where {F₀, F₁, F₂, F₃, ...} = {0,1,1,2,3,5,...} are the Fibonacci numbers A000045.

• 1. The first row of the Wythoff array consists of the Fibonacci sequence 1,2,3,5,8,... A000045
  2. Every row satisfies the Fibonacci recurrence;
  3. The leading term in each row is the smallest number not found in any earlier row;
  4. Every positive integer appears exactly once in the array;
  5. The terms in any row or column are monotonically increasing;
  6. Every positive Fibonacci-type sequence (i.e. satisfying a(n)=a(n-1)+a(n-2) and eventually positive) appears as some row of the array;
  7. The terms in any two rows alternate.
  
  There are infinitely many arrays with properties 1-7, see [Kim95a].

• Another especially interesting array with properties 1-7 is the Stolarsky array: A035506,

   1   2   3    5    8   13   21   34   55   89  4   6  10   16   26   42   68  110  178  288  7  11  18   29   47   76  123  199  322  521  9  15  24   39   63  102  165  267  432  699 12  19  31   50   81  131  212  343  555  898 14  23  37   60   97  157  254  411  665 1076 17  28  45   73  118  191  309  500  809 1309 20  32  52   84  136  220  356  576  932 1508 22  36  58   94  152  246  398  644 1042 1686 25  40  65  105  170  275  445  720 1165 1885 

• The kth column of the Wythoff array consists of the numbers whose Zeckendorf expansion ends with F_k.

• The nth term of the vertical para-Fibonacci sequence
  0, 0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 4, 0, 5, 3, 2, 6, 1, 7, 4, 0, 8, 5, ...

  (A019586 or, for the original form, A003603) gives the index (or parameter) of the row of the Wythoff array that contains n.

  This sequence also has some nice properties.
  A. If you delete the first occurrence of each number, the sequence is unchanged. Thus if we delete the red numbers from

  0, 0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 4, 0, 5, 3, 2, 6, 1, 7, 4, 0, 8, 5, ... we get
  0, 0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 4, 0, 5, 3, 2, 6, 1, 7, 4, 0, 8, 5, ...

  again!

  B. Between any two consecutive 0's we see a permutation of the first few positive integers, and these nest, so the sequence can be rewritten as:

       0      0      0                1      0         2      1      0     3   2      1      4      0   5 3   2    6 1    7 4      0 8 5 3 9 2 10 6 1 11 7 4 12 

• The nth term of the horizontal para-Fibonacci sequence
  1, 2, 3, 1, 4, 1, 2, 5, 1, 2, 3, 1, 6, 1, 2, 3, 1, 4, 1, 2, 7, 1, 2, ...

  (A035612) gives the index (or parameter) of the column of the Wythoff array that contains n. This sequence also has a very nice property (see the entry).

References

[Con96] J. H. Conway, Unpublished notes, 1996.
[FrKi94] A. Fraenkel and C. Kimberling, Generalized Wythoff arrays, shuffles and interspersions, Discrete Mathematics 126 (1994) 137-149.
[Kim91] C. Kimberling, Problem 1615, Crux Mathematicorum, Vol. 17 (2) 44 1991, and Vol. 18, March 1992, p.82-83.
[Kim93] C. Kimberling, Orderings of the set of all positive Fibonacci sequences, in G. E. Bergum et al., editors, Applications of Fibonacci Numbers, Vol. 5 (1993), pp. 405-416.
[Kim93a] C. Kimberling, Interspersions and dispersions, Proc. Amer. Math. Soc. 117 (1993) 313-321.
[Kim94] C. Kimberling, The First Column of an Interspersion, Fibonacci Quarterly 32 (1994) 301-314.
[Kim95] C. Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
[Kim95a] C. Kimberling, Stolarsky interspersions, Ars Combinatoria 39 (1995) 129-138.
[Kim95b] C. Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly 33 (1995) 3-8.
[Kim97] C. Kimberling, Fractal Sequences and Interspersions, Ars Combinatoria, vol 45 p 157 1997.
[Mor80] D. R. Morrison, A Stolarsky array of Wythoff pairs, in A Collection of Manuscripts Related to the Fibonacci Sequence, Fibonacci Assoc., Santa Clara, CA, 1980, pp. 134-136.
[Sto76] K. B. Stolarsky, Beatty sequences, continued fractions, and certain shift operators, Canad. Math. Bull., 19 (1976), 472-482.
[Sto77] K. B. Stolarsky, A set of generalized Fibonacci sequences such that each natural number belongs to exactly one, Fib. Quart., 15 (1977), 224.

Other Links

Clark Kimberling, Fractal sequences
Clark Kimberling, Interspersions
Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.

Associated Sequences

Successive columns of the Wythoff array A035513 give sequences A000201 (just before the broken line);
A007065, A035336, A035337, A035338, A035339, A035340.
Successive rows give the Fibonacci numbers A000045, the Lucas numbers A000204, the doubled Fibonacci numbers A013588, the trebled Fibonacci numbers A022086, A022087, A000285, A022095, etc.
The main diagonal is A020941.

An analogue of Pascal's triangle that deserves to be better known.

The rule for producing these numbers is essentially the same as for Pascal's triangle: each term is the sum of the two numbers immediately above it, except that (numbering the rows by n=0,1,2,... and the entries in each row by k=0,1,2,...) if n is even and k is odd - the red entries! - we subtract C(n/2-1,(k-1)/2).

Formally,

Reference: S. M. Losanitsch, Die Isomerie-Arten ... Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.

The sequence formed by reading the triangle by rows is A034851, and the successive diagonals are A000012, A004526, A002620, A005993, A005994, A005995, A018210, A018211, A018212, A018213, A018214. The central columns yield A034872, A032123, A005654. The row sums form A005418. The difference between Pascal's triangle and the Losanitsch triangle gives the triangle shown in A034852.

The even-numbered diagonals are the partial sums of the previous diagonals. A generating function for the (2m)-th diagonal is

and that for the (2m+1)st diagonal is obtained by dividing that by 1-x.

For example, the 5th diagonal 1,3,12,28,66,126,... has generating function

How many partially ordered sets are there with n elements? (Sequence A001035.)
If the points are distinguishable, i.e. labeled, then for n = 1, 2, 3, ... points the numbers are:

1, 3, 19, 219, 4231, 130023, 6129859, ...

At present these numbers are known up through 18 points. Click to see the full entry.

Some related sequences are:

• A000112 (unlabeled posets)
• A000798 (labeled topologies)
• A001930 (unlabeled topologies)

A selection of references:

• K. K.-H. Butler, A Moore-Penrose inverse for Boolean relation matrices, pp. 18-28 of Combinatorial Mathematics (Proceedings 2nd Australian Conf.), Lect. Notes Math. 403, 1974.
• K. K.-H. Butler and G. Markowsky, Enumeration of finite topologies, Proc. 4th S-E Conf. Combin., Graph Theory, Computing, Congress. Numer. 8 (1973), 169-184.
• C. Chaunier and N. Lygeros, Progres dans l'enumeration des posets, C. R. Acad. Sci. Paris 314 serie I (1992) 691-694.
• C. Chaunier and N. Lygeros, The Number of Orders with Thirteen Elements, Order 9:3 (1992) 203-204.
• C. Chaunier and N. Lygeros, Le nombre de posets a isomorphie pres ayant 12 elements. Theoretical Computer Science, 123 (1994), 89-94.
• J. C. Culberson and G. J. E. Rawlins, New Results from an Algorithm for Counting Posets, Order 7 (90/91), no 4, pp. 361-374.
• M. Erne, The Number of Posets with More Points Than Incomparable Pairs, Disc Math 105 (1992) 49-60.
• M. Erne, On the cardinalities of finite topologies and the number of antichains in partially ordered sets, Discr. Math. 35 (1981) 119-133.
• M. Erne and K. Stege, Counting finite posets and topologies, Order, vol. 8, pp. 247-265, 1991.
• J. W. Evans, F. Harary and M. S. Lynn; On the computer enumeration of finite topologies; Comm. Assoc. Computing Mach. 10 (1967), 295--298.
• R. Fraisse and N. Lygeros, Petits posets : denombrement, representabilite par cercles et compenseurs. C. R. Acad. Sci. Paris, 313 (1991), 417-420.
• D. Kleitman & B. L. Rothschild, Asymptotic enumeration of partial orders on a finite set, Trans. Amer. Math. Soc., 205 (1975) 205-220.
• Y. Koda (ykoda@rst.fujixerox.co.jp), The numbers of finite lattices and finite topologies, Bull. Institute Combinatorics and its Applications, Jan. 1984.
• N. Lygeros, Calculs exhaustifs sur les posets d'au plus 7 elements. SINGULARITE, vol.2 n4 p.10-24, April 1991.
• N. Lygeros and P. Zimmermann, Calculation of a(14)
• P. Renteln, On the enumeration of finite topologies, J. Combin., Inform & System Sci., vol 19 pp 201-206 1994.
• P. Renteln, Geometrical approaches to the enumeration of finite posets ..., Nieuw Archiv Wisk., vol 14 pp 349-371 1996.
• V. I. Rodionov, MR 83k:05010 T(12) and T0(12) calculated (in Russian).

What is the largest determinant of any n x n matrix with entries that are 0 and 1 ? (Sequence A003432.)
Here is the sequence (for n = 1, 2, ...):

1, 1, 2, 3, 5, 9, 32, 56, 144, 320, 1458, 3645, 9477, 25515, 131072, 327680, 1114112

The next term is believed to be 3411968, although this has not been formally proved.

Click to see the full entry. Quite a lot is known about the above problem. See for example the survey article by J. Brenner in the Amer. Math. Monthly, June/July 1972, p. 626, and further comments in the issues of Dec. 1973, Dec. 1975 and Dec. 1977.
If n+1 is divisible by 4, and a Hadamard matrix of order n exists, then f(n) = (n+1)^{(n+1)/2}/2^n.
There are 4 equivalent versions of the problem: find the max determinant of a matrix with entries that are:
o 0 or 1, or
o in the range 0 <= x <= 1, or
o -1 or 1, or
o in the range -1 <= x <= 1.

For the most up-to-date information, see the web site The Hadamard Maximal Determinant Problem maintained by W. P. Orrick and B. Solomon. (Note that their indexing differs from that used in the OEIS.)

Expand exp(e^x - 1) in powers of x, SUM B_n xⁿ / n!. The coefficients B_n are the Bell numbers (A000110):

1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, ...

Click to see the full entry.

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, ... (A001006).
Like the Catalan numbers, the Motzkin numbers have many interpretations. For example:

• the number of ways to join n points on a circle by nonintersecting chords
• the number of paths from (0,0) to (n,0) that do not go below the horizontal axis and are made up of steps (1,1) (i. e. NE), (1,-1) (i. e. SE) and (1,0) (i.e. E).
• the number of sequences (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 0 = s(n).

A selection of references:

• T. Motzkin, Relations between hypersurface cross ratios... Bull. Amer. Math. Soc., 54, 352-360, 1948.
• R. Donaghey, Restricted plane tree representations of four Motzkin-Catalan equations, J. Combin. Theory Ser. B, 22, 114-121, 1977.
• R. Donaghey and L. W. Shapiro, Motzkin numbers, J. Combin. Theory Ser. A, 23, 291-301, 1977.
• E. Barcucci, R. Pinzani, and R. Sprugnoli, The Motzkin family, PU. M. A. Ser. A, 2, No. 3-4, 249-279, 1991.
• A. Kuznetsov, I. Pak, and A. Postnikov, Trees associated with the Motzkin numbers, J. Combin. Theory Ser. A, 76, 145-147, 1996.
• F. Bergeron et al., Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 267.
• Richard Stanley's home page, under Enumerative Combinatorics, Vol II (to be published), has a list of manifestations of Motzkin numbers.

Formulas:

• G.f.: (1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2).
• G.f. satisfies A(x) = 1 + xA(x) + x^2 A(x)^2.
• Recurrence: a(n) = (-1/2) SUM (-3)^a C(1/2,a) C(1/2, b); a+b=n+2, a>=0, b>=0.
• In Maple: seriestolist(series((1-x-(1-2*x-3*x^2)^(1/2))/(2*x^2),x,40));
• In Mathematica: a[0]=1;a[n_Integer]:=a[n]=a[n-1]+Sum[a[k]*a[n-2-k],{k,0,n-2}]; Array[ a[#]&, 30 ]

Numbers that are equal to the sum of every (smaller) number that divides them (A000396).
For example 6 is perfect because it is divisible by 1, 2 and 3, and 1 + 2 + 3 = 6.
The sequence of perfect numbers begins:

6, 28, 496, 8128, 33550336, 8589869056, 137438691328,
2305843008139952128, 2658455991569831744654692615953842176, ...

Only some thirty or so perfect numbers are known. These are some of the largest numbers that have ever been computed. Click to see the full entry.

For a sequel, see that paper that Benoit Cloitre, Matthew Vandermast and I wrote: Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.

In the early 1990's my colleague Ken Thompson computed the number of possible chess games with n moves, for n up through 7, specially for the OEIS - see A006494.

There are now several versions of this sequence, depending on exactly what is being counted. The preferred version (now known for n <= 10) is A048987:

For other versions see the entry for chess games in the Index to the OEIS.

An old problem proposed in 1941, solved in 2007

In 1941, Royal Vale Heath proposed this short problem in The American Mathematical Monthly [4], when H. S. M. Coxeter was in charge of the “Problems and Solutions” column:
                What is the smallest value of n for which the n² triangular numbers 0, 1, 3, 6, 10, …, n²(n²-1)/2 can be arranged to form a magic square?
This problem remained unsolved. Here is the solution found in April 2007… 66 years later:
                n = 6.

The samples were not easy to find, but are easy for the reader to check, as a factorization problem. Contents of this expanded solution:

• Triangular and polygonal numbers
• Relationship to bimagic squares
• Orders n=3, 4, 5, impossible
• Order n=6, possible! Solution of the problem.
• Order n=7, also possible
• Numbers from 1 instead of 0
• Squares of polygonal numbers, p ≤ 10
• An unsolved problem: the smallest magic square of distinct triangular numbers
• Another unsolved problem: the smallest magic square of distinct pentagonal numbers
• Acknowledgements
  • to Sebastião A. da Silva
  • to Doug Hensley and Douglas B. West
  • to George P. H. Styan and Evelyn Matheson Styan
• References

Triangular and polygonal numbers

This figure will help us to understand what triangular numbers are, and more generally polygonal numbers.

First polygonal numbers

Relationship to bimagic squares

In his partial solution [5] published in the Monthly 1942, R. V. Heath remarked that a bimagic square (magic square which is still magic after the original entries are all squared [2a]) can be directly used to construct a magic square of triangular numbers:

“Clearly, the magic property will still be retained if each of the original numbers is subtracted from its square. The resulting numbers are all even, and their halves are the triangular numbers”

Using a bimagic square of order 8 found in the well-known book [1, p. 212] by W.W. Rouse Ball and initially constructed by H. Schots in 1931 [8, p.357], Heath [5] built a magic square of 64 triangular numbers and with it showed that n£8, but the smallest possible n remained unknown, as he said:

“But it remains possible that a smaller set of triangular numbers might form a magic square without the corresponding natural numbers forming a magic square. Moreover, it has never been satisfactorily proved that there is no doubly-magic (=bimagic) square of order 7”

By an exhaustive search, we proved with Walter Trump in 2002 [2b] that a bimagic square of order smaller or equal to 7 does not exist: this means that Heath’s trick in using bimagic squares cannot be used for orders n<8. Here is a study of the problem for the smallest orders.

Order n=3, impossible

Is it possible to construct a magic square using the 9 triangular numbers 0, 1, 3, 6, 10, 15, 21, 28, 36?

No! The total sum of these numbers being 120, such a square would have a magic sum 120/3=40. The central number of any 3x3 magic square being one third of its magic sum, and 40 being not divisible by 3, since n=3, it is impossible.

Order n=4, impossible

A magic square using the 16 triangular numbers 0, 1, 3, …, 120 would have a magic sum equal to 170. There are only 10 series of 4 triangular numbers giving this sum:

        0         1         78       91
        0         10       55       105
        1         21       28       120
        1         28       36       105
        1         36       55       78
        3         10       66       91
        3         21       55       91
        6         28       45       91
        15       28       36       91
        21       28       55       66

In a magic square, each number needs to use at least two or three series: one for the row, one for the column, and one more if the number is located on a diagonal. Because the number 6 -for example- is present in only one series, a magic square of order n=4 is impossible.

Order n=5, impossible

A magic square using the 25 triangular numbers 0, 1, 3, …, 300 would have a magic sum equal to 520. There are 118 series giving this sum. Combining the series, there are 148 possible ways to get 5 series using the 25 triangular numbers. This means that it is possible to have 5 magic rows. An exhaustive search, however, shows that it is impossible to arrange these rows and make all the columns magic. The best possible arrangements are 5 magic rows and 3 magic columns, for example:

Order n=5. This is not a magic square S=520, the two last columns have sums≠S.

0	3	105	276	136
66	1	253	190	10
210	45	6	28	231
91	300	36	15	78
153	171	120	21	55

Order n=6, possible! Solution of the problem.

A magic square using the 36 triangular numbers 0, 1, 3, …, 630 would have a magic sum equal to 1295. There are 1921 series giving this sum.

Good news: it is possible to arrange these series to form magic squares! Here is an example, solution of our problem.

A solution of Heath’s Problem E 496.
Order n=6. Magic square with consecutive triangular numbers from 0 to 630. S=1295

0	406	120	528	105	136
1	300	435	378	171	10
66	276	496	15	91	351
595	78	153	28	210	231
3	190	55	21	465	561
630	45	36	325	253	6

Order n=7, also possible

The order n=7 allows also magic squares of the first triangular numbers.

Order n=7. Magic square with consecutive triangular numbers from 0 to 1176.   S=2800.

0	378	1176	210	595	6	435
3	351	45	465	703	1128	105
946	171	561	820	190	21	91
741	528	36	325	120	15	1035
1081	300	55	496	780	10	78
28	666	66	231	276	630	903
1	406	861	253	136	990	153

Numbers from 1 instead of 0

If we prefer to use consecutive polygonal numbers starting from 1 instead of 0 (see below the da Silva’s challenge), a similar reasoning shows that the minimum order is again 6. The 6 series of order 4 and the 91 series of order 5 are not sufficient to construct a magic square. Here are examples of order 6 and 7 starting from 1.

Order n=6. Magic square with consecutive triangular numbers from 1 to 666. S=1406.

28	666	78	1	528	105
45	276	351	3	406	325
66	378	136	171	190	465
496	21	153	630	15	91
210	10	435	595	36	120
561	55	253	6	231	300




Order n=7. Magic square with consecutive triangular numbers from 1 to 1225.  S=2975.

36	406	276	3	528	946	780
45	903	351	6	1225	10	435
561	861	496	741	105	21	190
990	120	630	1	66	703	465
253	136	666	1081	153	91	595
55	378	231	15	820	1176	300
1035	171	325	1128	78	28	210

Squares of polygonal numbers, p ≤ 10

We can generalize Heath’s Problem E496 to other polygonal numbers.

Reminder: the i-th p-gonal number is equal to

((p - 2)i² - (p - 4)i)/2

With p=3, we get triangular numbers. With p=4, we get square numbers. With p=5, we get pentagonal numbers. And so on…

Any bimagic square can be used to construct magic squares of k₂i²+k₁i+k₀ numbers: using the same bimagic square of order n=8 as R.V. Heath, Charles W. Trigg published squares of polygonal numbers for p=3, 5, 6, 7, 8 in [9], and for p=9, 10 in [10]. But is it possible to construct squares of orders n<8? Yes!

• The case p=3, triangular numbers. The smallest solution is n=6, as analyzed above.
• The case p=4, square numbers. The smallest solution is slightly larger: n=7. This question was solved in 2005: I constructed a magic square of squares of order 7 using the first squares 0², 1², …, 48². The magic square is on my web site [2c] and is also published in the MAA MathTrek column of Ivars Peterson [6].
• The cases p=5, 6, 7, 8, 9, 10. The smallest solution is again n=7 for all these p. It might be boring to give all my samples, but here is one with p=5, a magic square of pentagonal numbers of order 7.
   

  Order n=7. Magic square with consecutive pentagonal numbers from 0 to 3432.  S=8064.

  1617	3015	35	0	1162	715	1520
  2882	330	12	5	210	3290	1335
  1926	2752	2501	247	117	376	145
  70	176	2380	1	3432	925	1080
  51	532	2262	2625	1717	287	590
  1426	782	22	2035	1001	651	2147
  92	477	852	3151	425	1820	1247

An interesing remark: a magic square of polygonal numbers can be turned into a magic square of squares by multiplying each term by 8(p – 2) then adding (p - 4)² to each term, because:

8(p – 2) [((p - 2)i² - (p - 4)i)/2] + (p - 4)²   =   [2(p - 2)i² - (p - 4)]²

An unsolved problem: the smallest magic square of distinct triangular numbers

All the above examples use the first consecutive polygonal numbers. But what is the smallest order n if we allow any polygonal numbers, consecutive or not, but distinct?

The first 4x4 magic square of squares, using 16 distinct squares, was constructed by Euler, in a letter sent to Lagrange in 1770 [3]. I found the first 5x5 magic square of squares in 2004 [2c] [3] [6]. Now I am please to give the first 4x4 and 5x5 magic square of triangular numbers:

Order n=4. Magic square with distinct triangular numbers. S=1402.

66	465	780	91
1	630	105	666
300	171	496	435
1035	136	21	210




Order n=5. Magic square with distinct triangular numbers.  S=823.

351	0	210	91	171
36	136	153	378	120
105	406	15	231	66
325	253	10	45	190
6	28	435	78	276

It is still unknown if a 3x3 magic square of squares is possible [2d] [3] [6] [7], but what about a 3x3 magic square of triangular numbers? As remarked by John P. Robertson (author of [7]), in a private communication of April 2007:

“If there is a 3x3 magic square of squares, then all the entries are odd, and so congruent to 1 modulo 8.  Because if T is a triangular number then 8T + 1 is a square, and if S is an odd square then (S - 1)/8 is a triangular number, the question of whether there is a 3x3 magic square of squares is equivalent to the question of whether there is a 3x3 magic square of triangular numbers.”

Open problem. Who will construct a 3x3 magic square of distinct triangular numbers, or its equivalent 3x3 magic square of squares? Or who will prove that it is impossible?

Another unsolved problem: the smallest magic square of distinct pentagonal numbers

We have seen that we do not have the answer to the problem of the smallest magic square of triangular or of square numbers: there are 4x4 magic squares, but we still don’t know if 3x3 squares are possible.

But after triangular numbers (p=3) and square numbers (p=4), what about pentagonal numbers (p=5)? We find that 6x6 squares are possible, as shown in the figure below, but it should be possible to construct 5x5 squares or smaller ones.

Order n=6. Magic square with distinct pentagonal numbers. S=4578.

1426	1520	1080	176	0	376
1335	5	782	2147	22	287
1	1820	651	1926	145	35
92	51	925	247	2262	1001
1247	852	715	70	532	1162
477	330	425	12	1617	1717

Open problem. What is the smallest possible magic square of distinct pentagonal numbers: 3x3, 4x4, 5x5 or 6x6?

In November 2007, Lee Morgenstern worked on this very difficult problem. He constructed the first known 4x4 and 5x5 magic squares of distinct pentagonal numbers. Congratulations!

Order n=5. Magic squares with distinct pentagonal numbers.
Smallest possible magic sums: S=7555 on the left, S=6333 (if 0 is allowed) on the right.

4030	1001	145	2262	117		1426	1247	376	3290	0
70	176	2501	2882	1926		4187	35	715	477	925
782	3151	1162	425	2035		5	145	3876	51	2262
1426	2147	22	1335	2625		70	1335	782	1001	3151
1247	1080	3725	651	852		651	3577	590	1520	1

Order n=4. Magic squares with distinct pentagonal numbers.
Same smallest possible magic sum: S=2591234.

3725	1908012	659022	20475		12650	1969401	578151	31032
760060	115787	500837	1214550		722107	83426	455126	1330575
300832	543305	1431305	315792		247051	495650	1557032	291501
1526617	24130	70	1040417		1609426	42757	925	938126

Lee constructed also this 3x3 semi-magic square:

Order n=3. Semi-magic square
with distinct pentagonal numbers.
Smallest possible magic sum: S=412249.

356972	651	54626
19780	275847	116622
35497	135751	241001

All this means that the above open problem becomes now:

Open problem. Who will construct a 3x3 magic square of distinct pentagonal numbers? Or who will prove that it is impossible?

As seen above, a magic square of polygonal numbers can be turned into a magic square of squares: an example of a 3x3 magic square of pentagonal numbers would also solve the 3x3 magic square of squares problem.

Acknowledgements

Particular thanks to Sebastião A. da Silva, Brazil, who challenged me to find solutions of order n < 8 in March 2007. Without knowing the references [4] and [5] to Problem E496 by Heath and [9] and [10] to articles by Charles W. Trigg, Sebastião had independently found the relationship with bimagic squares and sent me this solution of order 8, constructed using the G. Pfeffermann’s first bimagic square built in 1890 [2a].

Order n=8. Magic square with consecutive triangular numbers from 1 to 2080.  S=5720.
Constructed by Sebastião A. da Silva using the Pfeffermann’s first bimagic square.

1596	595	36	1653	171	1128	45	496
561	210	1485	1176	28	435	1770	55
351	946	91	276	2080	741	10	1225
190	15	630	465	1431	78	1081	1830
120	325	2016	3	861	300	1275	820
21	1540	153	66	666	1711	528	1035
1891	136	903	1378	378	1	780	253
990	1953	406	703	105	1326	231	6

But Sebastião was unsuccessful in finding examples of smaller orders. He offered a bottle of Brazilian Curaçao if I succeeded in answering his question, asking in French:

               “Est-il possible de construire un carré triangulaire quand il n´existe pas un bimagique du même ordre ?”
               (“Is it possible to construct a triangular square when there is no bimagic of the same order?”)

A bottle? Very interesting! It’s because I worked on his challenge that I looked for mathematical references and found that the same problem had been proposed already a long time ago in the Monthly –without the reward of a bottle- and which had remained unsolved. The only difference is the starting triangular number: Sebastião started from 1, while Heath started from 0. Both cases are solved now. Because I won his challenge, and because it seems unfortunately difficult to send a bottle through the airmail post, Sebastião sent me in May 2007 this nice gift instead of a bottle. Thanks Sebastião for your interesting challenge and for your gift. We will drink together another bottle when you come to Paris, or when I go to Rio!

Pão de Açucar, Rio de Janaeiro, received from Sebastião A. da Silva

Thanks to Doug Hensley, Dept. of Mathematics, Texas A&M University, and Douglas B. West, Dept. of Mathematics, University of Illinois at Urbana-Champaign, both in charge of the “Problems and Solutions” column of The American Mathematical Monthly. They were the first to read the solution n=6, after Sebastião, and immediately accepted to publish it, 66 years after the problem was proposed in the same column.

Thanks to George P. H. Styan, McGill University, Montreal, Canada, who sent me a copy of the paper [9], wich I was unable to find it in the main mathematical libraries of Paris. And thanks also to him and to Evelyn Matheson Styan, his wife, for correcting my English in this expanded solution of Problem E 496!

References

• [1] W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13^th edition, Dover Publications, 1987
• [2] C. Boyer, Multimagic squares and cubes web site, www.multimagie.com/indexengl.htm
  • [2a] www.multimagie.com/English/Bimagic.htm
  • [2b] www.multimagie.com/English/Smallestbi.htm
  • [2c] www.multimagie.com/English/SquaresOfSquares4_7.htm
  • [2d] www.multimagie.com/English/SquaresOfSquares.htm
• [3] C. Boyer, Some notes on the magic squares of squares problem, The Mathematical Intelligencer 27(2005), n°2, 52-64
  
• [4] R. V. Heath, Problem E 496, The American Mathematical Monthly 48(1941), 699
• [5] R. V. Heath, A magic square of triangular numbers - Solution E 496, The American Mathematical Monthly 49(1942), 476
• [6] I. Peterson, Magic squares of squares, MAA Online, MathTrek column, June 27, 2005, http://www.maa.org/mathland/mathtrek_06_27_05.html
• [7] J. P. Robertson, Magic squares of squares, Mathematics Magazine 69(1996), n°4, 289-293
• [8] H. Schots, Helsch vierkant, Bulletins de la Classe des Sciences – Académie Royale de Belgique, 5^ème Série 17(1931), 339-361
• [9] C. W. Trigg, Magic squares with polygonal elements, School Science and Mathematics 71(1971), 195-197
• [10] C. W. Trigg, Magic squares with nonagonal and decagonal elements, Journal of Recreational Mathematics 5(1972), 203-204

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