

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
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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

Table of n, a(n) for n=1..65.


FORMULA

Conjecture: for k > 0 and 1 <= j <= k, a(2k^22j+1) = 4k^2+4k4j3, a(2k^22j+2) = 4k^2+4k4j, a(2k^2+2k2j+1) = 4k^2+8k4j, a(2k^2+2k2j+2) = 4k^2+8k4j+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



