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%I #13 Apr 09 2015 09:35:40
%S 1,1,1,1,5,1,4,1,1,2,0,1,1,1,29,1,0,0,29,2,29,0,29,29,1,1,1,2,0,2,1,0,
%T 1,1,0,1,1,1,0,1,0,29,1,0,5,0,9,1,0,0,0,2,0,1,0,29,0,2,0,0,0,14,1,0,9,
%U 0,1,1,0,0,29,1,0,1,0,0,0,1,0,14,0,1,9,2
%N Smallest k>=1 such that n^4 + (n+1)^4 + ... + (n+k)^4 is prime or a(n)=0 if there is no such k.
%C Every term is 0, 1, 2, 4, 5, 9, 14, or 29.
%C a(n)=0 if and only if n is in A256546.
%C From _Vladimir Shevelev_, Apr 09 2015: (Start)
%C Indeed, denote by S_k(n) = n^4 + (n+1)^4 + ... + (n+k)^4. If n=1, k=m-1, then, as is known,
%C s(m) = S_(m-1)(1) = 1^4 + 2^4 + ... + m^4 = (6*m^5 + 15*m^4 + 10*m^3 - m)/30 (1)
%C such that
%C S_k(n) = s(n+k) - s(n-1). (2)
%C Since S_(-1)(n) = 0, then S_k(n) as a polynomial is divisible by k+1. Put
%C S*_k(n) = S_k(n)/(k+1). So we have
%C S_k(n) = S*_k(n)*(k+1) = T_k(n)/30*(k+1), (3)
%C where T_k(n) = 30*S_k(n) is (by (1)) a polynomial with integer coefficients.
%C For k>=1, it is clear that (3) could be prime for some n only if k+1 is a divisor of 30, i.e., k = 1,2,4,5,9,14 or 29. The smallest n when all these values of a(n) appeared is n=62. If for some n all numbers n^4 + (n+1)^4 + ... + (n+k)^4 are composite for k = 1,2,4,5,9,14 and 29, then a(n)=0. (End)
%H Peter J. C. Moses, <a href="/A256547/b256547.txt">Table of n, a(n) for n = 1..1000</a>
%F 1) If P_1(n) is prime, then a(n)=1;
%F 2) if P_1(n) is composite, but P_2(n) is prime, then a(n)=2;
%F 3) if P_1(n) and P_2(n) are composite, but P_3(n) is prime, then a(n)=4;
%F 4) if P_1(n), P_2(n), and P_3(n) are composite, but P_4(n) is prime, then a(n)=5;
%F 5) if P_1(n), P_2(n), P_3(n), and P_4(n) are composite, but P_5(n) is prime, then a(n)=9;
%F 6) if P_1(n), P_2(n), P_3(n), P_4(n), and P_5(n) are composite, but P_6(n) is prime, then a(n)=14;
%F 7) if P_1(n), P_2(n), P_3(n), P_4(n), P_5(n), and P_6(n) are composite, but P_7(n) is prime, then a(n)=29;
%F 8) otherwise a(n)=0.
%F Here P_i(n), i=1,...,7, are defined in comment in A256546.
%Y Cf. A000583, A256385, A256503, A256546.
%K nonn
%O 1,5
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Apr 01 2015
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