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 A308416 Values of m for which 2*p + m cannot be a square when p is a prime. 0
 1, 4, 8, 9, 13, 16, 17, 20, 24, 25, 28, 29, 33, 36, 37, 40, 41, 44, 48, 49, 52, 53, 56, 57, 61, 64, 65, 68, 69, 72, 73, 76, 80, 81, 84, 85, 88, 89, 92, 93, 97, 100, 101, 104, 105, 108, 109, 112, 113, 116, 120, 121, 124, 125, 128, 129, 132, 133, 136, 137, 141, 144, 145, 148, 149 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. 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). LINKS FORMULA 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 PROG (Python) a=[] a.append(0) #Offset starts at 1 iMax=15 #Example value for i in range(1, iMax+1):   for j in range(0, i):    m=i*i+j*4    a.append(m) a.sort() CROSSREFS Cf. A042948, A028347. Sequence in context: A235054 A292364 A071835 * A010429 A140282 A161757 Adjacent sequences:  A308413 A308414 A308415 * A308417 A308418 A308419 KEYWORD nonn AUTHOR Bob Andriesse, May 25 2019 STATUS approved

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Last modified December 2 03:22 EST 2020. Contains 338865 sequences. (Running on oeis4.)