OFFSET
1,2
COMMENTS
Conjecture: gives an identity for the sum of all divisors of all positive integers <= n. Alternating sum of row n equals A024916(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A024916(n).
Column 1 is A210843.
Column k lists the partial sums of the k-th column of triangle A252117 which gives an identity for sigma.
The first element of column k is A000330(k).
The second element of column k is A002492(k).
EXAMPLE
Triangle begins:
1;
4;
13, 5;
35, 20;
86, 65;
194, 175, 14;
415, 430, 56;
844, 970, 182;
1654, 2075, 490;
3133, 4220, 1204, 30;
5773, 8270, 2716, 120;
10372, 15665, 5810, 390;
18240, 28865, 11816, 1050;
31449, 51860, 23156, 2580;
53292, 91200, 43862, 5820, 55;
88873, 157245, 80822, 12450, 220;
146095, 266460, 145208, 25320, 715;
236977, 444365, 255360, 49620, 1925;
379746, 730475, 440286, 93990, 4730;
601656, 1184885, 746088, 173190, 10670;
943305, 1898730, 1244222, 311160, 22825, 91;
...
For n = 6 the sum of all divisors of all positive integers <= 6 is [1] + [1+2] + [1+3] + [1+2+4] + [1+5] + [1+2+3+6] = 1 + 3 + 4 + 7 + 6 + 12 = 33. On the other hand the 6th row of triangle is 194, 175, 14, so the alternating row sum is 194 - 175 + 14 = 33, equaling the sum of all divisors of all positive integers <= 6.
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Dec 14 2014
STATUS
approved