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%I #10 Jun 24 2017 14:08:19
%S 1,2,5,5,9,3,17,13,21,10,18,6,28,32,33,26,20,27,28,19,12,41,34,42,54,
%T 25,5,28,84,22,40,5,54,44,56,25,29,86,81,89,88,134,10,71,148,69,87,27,
%U 91,92,18,128,23,51,52,153,108,202,112,138,58,201,141,162,176,77,4
%N a(n) is the least k such that sigma(binomial(m,k)) = sigma(binomial(m-1,k-1)) + sigma(binomial(m-1,k)) where m = A284168(n).
%C Consider the triangle formed by replacing each m in Pascal's triangle with sigma(m). Then this sequence consists of the least column indices of the terms that are equal to the sum of its NW and N neighbors, as in a Pascal triangle.
%e Here is the triangle also described in A074801.
%e 1,
%e 1, 1,
%e 1, 3, 1,
%e 1, 4, 4, 1,
%e 1, 7, 12, 7, 1,
%e 1, 6, 18, 18, 6, 1,
%e On row index 3, we have 4 which is the sum of 1 and 3 its NW and N neighbors, this is A284168(1). The corresponding column index is 1, so a(1) = 1.
%o (PARI) T(n, k) = sigma(binomial(n, k));
%o isokT(n, k) = T(n-1, k-1) + T (n-1, k) == T(n, k);
%o isokn(n) = for (k=1, n-1, if (isokT(n,k), return(1)));
%o listak(nn) = for (n=1, nn, for (k=1, n-1, if (isokT(n,k), print1(k, ", "); break)));
%Y Cf. A000203, A007318, A074801, A284168.
%K nonn
%O 1,2
%A _Michel Marcus_, Mar 21 2017
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