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A344463
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a(n) is the least prime p such that (p^2-2*n)/(2*n-1) and (p^2+2*n)/(2*n+1) are both prime, or 0 if such p does not exist.
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1
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2, 0, 239, 251, 2069, 131, 521, 0, 2243, 379, 643, 40849, 89101, 11717, 81839, 18787, 659, 0, 2887, 41081, 199261, 13931, 4231, 223439, 17443, 953, 3499, 6689, 122131, 298777, 9883, 0, 93131, 12329, 48989, 67307, 9199, 44549, 13903, 294353, 605071, 42331, 7309, 167677, 41651, 46747, 129581
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OFFSET
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1,1
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COMMENTS
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a(n) = 0 if 2*n is a square.
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LINKS
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EXAMPLE
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a(3) = 239 because 239, (239^2-2*3)/(2*3-1) = 11423, and (239^2+2*3)/(2*3+1) = 8161 are primes, and no smaller prime works.
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MAPLE
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f:= proc(n) local M, p, i, j;
if issqr(2*n) then return 0 fi;
M:= sort(map(t -> rhs(op(t)), [msolve(p^2=1, 4*n^2-1)]));
for i from 0 to 10^6 do
for j in M do
p:= i*(4*n^2-1)+j;
if isprime(p) and isprime((p^2-2*n)/(2*n-1)) and isprime((p^2+2*n)/(2*n+1)) then return p fi
od od;
FAIL
end proc:
map(f, [$1..100]);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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