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A238442
Triangle read by rows demonstrating Euler's pentagonal theorem for the sum of divisors.
10
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).
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
(GW-BASIC)
10 'A program with four A-numbers. Omar E. Pol, Feb 26 2014
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
CROSSREFS
Row sums give A000203, the sum of divisors of n.
Row n has length A235963(n).
Sequence in context: A258254 A100035 A201927 * A090244 A210976 A258263
KEYWORD
sign,tabf
AUTHOR
Omar E. Pol, Feb 26 2014
STATUS
approved