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%I #10 Mar 22 2017 03:44:07
%S 3,6,15,25,27,30,35,40,48,50,54,60,63,66,78,80,100,108,112,118,120,
%T 123,124,126,140,144,158,175,192,198,200,207,216,220,224,225,232,238,
%U 243,247,304,310,316,319,341,345,348,358,364,368,375,385,391,408,416,425,432
%N Integers n such that sigma(binomial(n,k)) = sigma(binomial(n-1,k-1)) + sigma(binomial(n-1,k)) for some k.
%C Consider the triangle formed by replacing each m in Pascal's triangle with sigma(m). Then this sequence consists of the row indices where there is a term that is 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.
%e So a(1)= 3, and its column index is 1 which will be corresponding value in A284169.
%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)));
%Y Cf. A000203, A007318, A074801, A284169.
%K nonn
%O 1,1
%A _Michel Marcus_, Mar 21 2017
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