%I #32 Nov 06 2024 04:11:22
%S 0,1,3,1,6,1,10,4,15,4,1,21,9,1,28,9,1,36,16,4,45,16,4,1,55,25,4,1,66,
%T 25,9,1,78,36,9,1,91,36,9,4,105,49,16,4,1,120,49,16,4,1,136,64,16,4,1,
%U 153,64,25,9,1,171,81,25,9,1,190,81,25,9,4,210,100,36,9,4,1
%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k copies of the positive squares in nondecreasing order, except the first column which lists the triangular numbers, and the first element of column k is in row k(k+1)/2.
%C Gives an identity for the sum of all aliquot divisors of all positive integers <= n.
%C Alternating sum of row n equals A153485(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A153485(n).
%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%C Column 1 is A000217. Columns 2-3 are A008794, A211547, but without the zeros.
%C Column k lists the partial sums of the k-th column of triangle A231347 which gives an identity for the sum of aliquot divisors of n. - _Omar E. Pol_, Nov 11 2014
%e Triangle begins:
%e 0;
%e 1;
%e 3, 1;
%e 6, 1;
%e 10, 4;
%e 15, 4, 1;
%e 21, 9, 1;
%e 28, 9, 1;
%e 36, 16, 4;
%e 45, 16, 4, 1;
%e 55, 25, 4, 1;
%e 66, 25, 9, 1;
%e 78, 36, 9, 1;
%e 91, 36, 9, 4;
%e 105, 49, 16, 4, 1;
%e 120, 49, 16, 4, 1;
%e 136, 64, 16, 4, 1;
%e 153, 64, 25, 9, 1;
%e 171, 81, 25, 9, 1;
%e 190, 81, 25, 9, 4;
%e 210, 100, 36, 9, 4, 1;
%e 231, 100, 36, 16, 4, 1;
%e 253, 121, 36, 16, 4, 1;
%e 276, 121, 49, 16, 4, 1;
%e ...
%e For n = 6 the divisors of all positive integers <= 6 are [1], [1, 2], [1, 3], [1, 2, 4], [1, 5], [1, 2, 3, 6] hence the sum of all aliquot divisors is [0] + [1] + [1] + [1+2] + [1] + [1+2+3] = 0 + 1 + 1 + 3 + 1 + 6 = 12. On the other hand the 6th row of triangle is 15, 4, 1, therefore the alternating row sum is 15 - 4 + 1 = 12, equaling the sum of all aliquot divisors of all positive integers <= 6.
%Y Cf. A000203, A000217, A001065, A008794, A003056, A153485, A196020, A211547, A211343, A228813, A231345, A231347, A235791, A235794, A235799, A236104, A236106, A236112, A237591, A237593, A286001.
%K nonn,tabf
%O 1,3
%A _Omar E. Pol_, Jan 28 2014