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%I #30 Sep 25 2015 05:47:23
%S 0,0,1,0,1,0,4,0,4,1,0,9,1,0,9,1,0,16,4,0,16,4,1,0,25,4,1,0,25,9,1,0,
%T 36,9,1,0,36,9,4,0,49,16,4,1,0,49,16,4,1,0,64,16,4,1,0,64,25,9,1,0,81,
%U 25,9,1,0,81,25,9,4,0,100,36,9,4,1,0,100,36,16,4,1,0,121,36,16,4,1,0,121,49,16,4,1,0
%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k+1 copies of the squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.
%C Gives an identity for the sum of remainders of n mod k, for k = 1,2,3,...,n. Alternating sum of row n equals A004125(n), i.e., sum_{k=1..A003056(n))} (-1)^(k-1)*T(n,k) = A004125(n).
%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%e Triangle begins:
%e 0;
%e 0;
%e 1, 0;
%e 1, 0;
%e 4, 0;
%e 4, 1, 0;
%e 9, 1, 0;
%e 9, 1, 0;
%e 16, 4, 0;
%e 16, 4, 1, 0;
%e 25, 4, 1, 0;
%e 25, 9, 1, 0;
%e 36, 9, 1, 0;
%e 36, 9, 4, 0;
%e 49, 16, 4, 1, 0;
%e 49, 16, 4, 1, 0;
%e 64, 16, 4, 1, 0;
%e 64, 25, 9, 1, 0;
%e 81, 25, 9, 1, 0;
%e 81, 25, 9, 4, 0;
%e 100, 36, 9, 4, 1, 0;
%e 100, 36, 16, 4, 1, 0;
%e 121, 36, 16, 4, 1, 0;
%e 121, 49, 16, 4, 1, 0;
%e ...
%e For n = 24 the 24th row of triangle is 121, 49, 16, 4, 1, 0 therefore the alternating row sum is 121 - 49 + 16 - 4 + 1 - 0 = 85 equaling A004125(24).
%Y Cf. A000203, A000217, A000290, A003056, A004125, A120444, A196020, A211343, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A237048, A237591, A237593, A261699.
%K nonn,tabf
%O 1,7
%A _Omar E. Pol_, Jan 23 2014
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