%I #31 Aug 01 2019 11:15:01
%S 1,4,8,9,13,16,17,20,24,25,28,29,33,36,37,40,41,44,48,49,52,53,56,57,
%T 61,64,65,68,69,72,73,76,80,81,84,85,88,89,92,93,97,100,101,104,105,
%U 108,109,112,113,116,120,121,124,125,128,129,132,133,136,137,141,144,145,148,149
%N Values of m for which 2*p + m cannot be a square when p is a prime.
%C m = i^2 + 4*j is a term for i > 0, 0 <= j < i. Proof: If p = 2, then i^2 < 2*p + m < (i+2)^2. Therefore (i+1)^2 = 4 + i^2 + 4*j, which leads to a contradiction. If p > 2 is such that 2*p + i^2 + 4*j = k^2, then k + i and k - i are both even numbers. Therefore 4 | 2*p + 4*j, which is also a contradiction.
%C The terms of this sequence can be obtained by starting with A042948 (numbers congruent to 0 or 1 mod 4) and deleting the terms of A028347 (n^2 - 4).
%F Conjecture: for k > 0 and 1 <= j <= k, a(2k^2-2j+1) = 4k^2+4k-4j-3, a(2k^2-2j+2) = 4k^2+4k-4j, a(2k^2+2k-2j+1) = 4k^2+8k-4j, a(2k^2+2k-2j+2) = 4k^2+8k-4j+1. - _Jinyuan Wang_, Jul 23 2019
%o (Python)
%o a=[]
%o a.append(0) #Offset starts at 1
%o iMax=15 #Example value
%o for i in range(1,iMax+1):
%o for j in range(0,i):
%o m=i*i+j*4
%o a.append(m)
%o a.sort()
%Y Cf. A042948, A028347.
%K nonn
%O 1,2
%A _Bob Andriesse_, May 25 2019
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