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Smallest k such that n * sum(i=0..k, binomial(k,i) mod (n-1) ) <= 2^n.
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%I #16 Dec 14 2014 15:09:49

%S 1,4,3,6,5,8,7,8,9,10,10,9,10,10,10,11,11,11,12,11,12,11,11,12,13,12,

%T 11,13,12,13,13,13,12,13,13,14,13,14,13,14,14,14,14,14,14,15,15,15,14,

%U 14,15,14,15,15,15,15,16,15

%N Smallest k such that n * sum(i=0..k, binomial(k,i) mod (n-1) ) <= 2^n.

%C Aside from the third value, the sequence is the same as A251738.

%e For n = 3,

%e 3 * sum(i=0..1, binomial(1,i) mod 2) = 3 * (1 + 1) = 6 > 2^1,

%e 3 * sum(i=0..2, binomial(2,i) mod 2) = 3 * (1 + 0 + 1) = 6 > 2^2,

%e 3 * sum(i=0..3, binomial(3,i) mod 2) = 3 * (1 + 1 + 1 + 1) = 12 > 2^3,

%e 3 * sum(i=0..4, binomial(4,i) mod 2) = 3 * (1 + 0 + 0 + 0 + 1) = 6 <= 2^4,

%e so A251739(3) = 4.

%Y Cf. A251738.

%K nonn

%O 2,2

%A _Jens Voß_, Dec 07 2014