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A236540 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.

%I A236540 #32 Nov 06 2024 04:11:22
%S A236540 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 A236540 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 A236540 153,64,25,9,1,171,81,25,9,1,190,81,25,9,4,210,100,36,9,4,1
%N A236540 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 A236540 Gives an identity for the sum of all aliquot divisors of all positive integers <= n.
%C A236540 Alternating sum of row n equals A153485(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A153485(n).
%C A236540 Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%C A236540 Column 1 is A000217. Columns 2-3 are A008794, A211547, but without the zeros.
%C A236540 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 A236540 Triangle begins:
%e A236540     0;
%e A236540     1;
%e A236540     3,   1;
%e A236540     6,   1;
%e A236540    10,   4;
%e A236540    15,   4,   1;
%e A236540    21,   9,   1;
%e A236540    28,   9,   1;
%e A236540    36,  16,   4;
%e A236540    45,  16,   4,   1;
%e A236540    55,  25,   4,   1;
%e A236540    66,  25,   9,   1;
%e A236540    78,  36,   9,   1;
%e A236540    91,  36,   9,   4;
%e A236540   105,  49,  16,   4,  1;
%e A236540   120,  49,  16,   4,  1;
%e A236540   136,  64,  16,   4,  1;
%e A236540   153,  64,  25,   9,  1;
%e A236540   171,  81,  25,   9,  1;
%e A236540   190,  81,  25,   9,  4;
%e A236540   210, 100,  36,   9,  4,  1;
%e A236540   231, 100,  36,  16,  4,  1;
%e A236540   253, 121,  36,  16,  4,  1;
%e A236540   276, 121,  49,  16,  4,  1;
%e A236540   ...
%e A236540 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 A236540 Cf. A000203, A000217, A001065, A008794, A003056, A153485, A196020, A211547, A211343, A228813, A231345, A231347, A235791, A235794, A235799, A236104, A236106, A236112, A237591, A237593, A286001.
%K A236540 nonn,tabf
%O A236540 1,3
%A A236540 _Omar E. Pol_, Jan 28 2014