A238442 Triangle read by rows demonstrating Euler's pentagonal theorem for the sum of divisors.

1, 1, 2, 3, 1, 4, 3, 7, 4, -5, 6, 7, -1, 12, 6, -3, -7, 8, 12, -4, -1, 15, 8, -7, -3, 13, 15, -6, -4, 18, 13, -12, -7, 12, 18, -8, -6, 12, 28, 12, -15, -12, 1, 14, 28, -13, -8, 3, 24, 14, -18, -15, 4, 15, 24, 24, -12, -13, 7, 1, 31, 24, -28, -18, 6, 3

Offset

1,3

Comments

The law found by Leonhard Euler for the sum of divisors of n is that S(n) = S(n - 1) + S(n - 2) - S(n - 5) - S(n - 7) + S(n - 12) + S(n - 15) - S(n - 22) - S(n - 26) + S(n - 35) + S(n - 40) + ..., where the constants are the positive generalized pentagonal numbers, and S(0) = n, which is also a positive member of A001318.

Therefore column k lists A001318(k) together with the elements of A000203, starting at row A001318(k), but with all elements of column k multiplied by A057077(k-1).

The first element of column k is A057077(k-1)*A001318(k)which is also the last term of row A001318(k).

For Euler's pentagonal theorem for the partition numbers see A175003.

Note that both of Euler's pentagonal theorems refer to generalized pentagonal numbers (A001318), not to pentagonal numbers (A000326).

Links

Table of n, a(n) for n=1..66.

L. Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, p. 8.

L. Euler, De mirabilibus proprietatibus numerorum pentagonalium

L. Euler, Découverte d'une loi tout extraordinaire des nombres par rapport à la somme de leurs diviseurs

L. Euler, Discovery of a most extraordinary law of numbers, relating to the sum of their divisors

L. Euler, Observatio de summis divisorum, p. 8.

L. Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.

L. Euler, J. Bell, A demonstration of a theorem on the order observed in the sums of divisors, arXiv:math/0507201 [math.HO], 2005-2009.

Index entries for sequences related to sigma(n)

Formula

T(n,k) = A057077(k-1)*A001318(k), if n = A001318(k) and k = A235963(n). Otherwise T(n,k) = A057077(k-1)*A000203(n - A001318(k)), n >= 1, 1 <= k <= A235963(n).

Example

Triangle begins:
   1;
   1,   2;
   3,   1;
   4,   3;
   7,   4,  -5;
   6,   7,  -1;
  12,   6,  -3,  -7;
   8,  12,  -4,  -1;
  15,   8,  -7,  -3;
  13,  15,  -6,  -4;
  18,  13, -12,  -7;
  12,  18,  -8,  -6,  12;
  28,  12, -15, -12,   1;
  14,  28, -13,  -8,   3;
  24,  14, -18, -15,   4,  15;
  24,  24, -12, -13,   7,   1;
  31,  24, -28, -18,   6,   3;
  18,  31, -14, -12,  12,   4;
  39,  18, -24, -28,   8,   7;
  20,  39, -24, -14,  15,   6;
  42,  20, -31, -24,  13,  12;
  32,  42, -18, -24,  18,   8, -22;
  36,  32, -39, -31,  12,  15,  -1;
  24,  36, -20, -18,  28,  13,  -3;
  60,  24, -42, -39,  14,  18,  -4;
  31,  60, -32, -20,  24,  12,  -7, -26;
  ...
For n = 21 the sum of divisors of 21 is 1 + 3 + 7 + 21 = 32. On the other hand, from Euler's Pentagonal Number Theorem we have that the sum of divisors of 21 is S_21 = S_20 + S_19 - S_16 - S_14 + S_9 + S_6, the same as the sum of the 21st row of triangle: 42 + 20 - 31 - 24 + 13 + 12 = 32, equaling the sum of divisors of 21.
For n = 22 the sum of divisors of 22 is 1 + 2 + 11 + 22 = 36. On the other hand, from Euler's Pentagonal Number Theorem we have that the sum of divisors of 22 is S_22 = S_21 + S_20 - S_17 - S_15 + S_10 + S_7 - S_0, the same as the sum of the 22nd row of triangle is 32 + 42 - 18 - 24 + 18 + 8 - 22 = 36, equaling the sum of divisors of 22. Note that S_0 = n, hence in this case S_0 = 22.

Mathematica

rows = m = 18;
a057077[n_] := {1, 1, -1, -1}[[Mod[n, 4] + 1]];
a001318[n_] := (1/8)((2n + 1) Mod[n, 2] + 3n^2 + 2n);
a235963[n_] := Flatten[Table[k, {k, 0, m}, {(k+1)/(Mod[k, 2]+1)}]][[n+1]];
T[n_, k_] := If[n == a001318[k] && k == a235963[n], a001318[k] a057077[k - 1], a057077[k - 1] DivisorSigma[1, n - a001318[k]]];
Table[T[n, k], {n, 1, m}, {k, 1, a235963[n]}] // Flatten (* Jean-François Alcover, Nov 29 2018 *)

Prog

10 '(GWbasic) A program with four A-numbers.
20 Dim A000203(30), A001318(10), A057077(30), A235963(30), T(30, 10)
30 For n = 1 to 26
40   For k = 1 to A235963(n)
50     If n = A001318(k) and k = A235963(n) then T(n, k) = A057077(k-1)*A001318(k) else T(n, k) = A057077(k-1)*A000203(n - A001318(k))
60   print T(n, k);
70   next k
80 print
90 next n
100 End
# Omar E. Pol, Feb 26 2014

Crossrefs

Row sums give A000203, the sum of divisors of n.

Row n has length A235963(n).

Cf. A001318, A027750, A057077, A175003, A196020, A237270, A237273.

Sequence in context: A258254 A100035 A201927 * A090244 A210976 A258263

Adjacent sequences: A238439 A238440 A238441 * A238443 A238444 A238445

Keyword

sign,tabf

Author

Omar E. Pol, Feb 26 2014

Status

approved