

A252117


Irregular triangle read by row: T(n,k), n>=1, k>=1, in which column k lists the numbers of A000716 multiplied by A000330(k), and the first element of column k is in row A000217(k).


3



1, 3, 9, 5, 22, 15, 51, 45, 108, 110, 14, 221, 255, 42, 429, 540, 126, 810, 1105, 308, 1479, 2145, 714, 30, 2640, 4050, 1512, 90, 4599, 7395, 3094, 270, 7868, 13200, 6006, 660, 13209, 22995, 11340, 1530, 21843, 39340, 20706, 3240, 55, 35581, 66045, 36960, 6630, 165, 57222, 109215, 64386, 12870, 495
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OFFSET

1,2


COMMENTS

Gives an identity for sigma(n). Alternating sum of row n equals A000203(n), the sum of the divisors of n.
Column 1 is A000716, but here the offset is 1 not 0.
The 1st element of column k is A000330(k).
The 2nd element of column k is A059270(k).
The 3rd element of column k is A220443(k).
The partial sums of column k give the kth column of A249120.
In the HondaYoda paper see "3. String theory and Riemann hypothesis". The coefficients that are mentioned in 3.11 are the first 16 terms of A000716, the coefficients that are mentioned in 3.12 are the first 5 terms of A000330, and the coefficients that are mentioned in 3.13 are the first 16 terms of A000203.
Another triangle with the same row lengths and whose alternating row sums give A000203 is A196020. (End)


LINKS



FORMULA



EXAMPLE

Triangle begins:
1;
3;
9, 5;
22, 15;
51, 45;
108, 110, 14;
221, 255, 42;
429, 540, 126;
810, 1105, 308;
1479, 2145, 714, 30;
2640, 4050, 1512, 90;
4599, 7395, 3094, 270;
7868, 13200, 6006, 660;
13209, 22995, 11340, 1530;
21843, 39340, 20706, 3240, 55;
35581, 66045, 36960, 6630, 165;
57222, 109215, 64386, 12870, 495;
90882, 177905, 110152, 24300, 1210;
142769, 286110, 184926, 44370, 2805;
221910, 454410, 305802, 79200, 5940;
341649, 713845, 498134, 137970, 12155, 91;
...
For n = 6 the divisors of 6 are 1, 2, 3, 6, so the sum of the divisors of 6 is 1 + 2 + 3 + 6 = 12. On the other hand, the 6th row of the triangle is 108, 110, 14, so the alternating row sum is 108  110 + 14 = 12, equaling the sum of the divisors of 6.
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of the divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 21843, 39340, 20706, 3240, 55, so the alternating row sum is 21843  39340 + 20706  3240 + 55 = 24, equaling the sum of the divisors of 15.


PROG

(PARI) A003056(n) = (sqrtint(8*n+1)1)\2;
A000716(n) = polcoef(1/eta('x+O('x^(n+1)))^3, n, x);


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



