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A357124
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a(n) is the least k >= 1 such that A000045(n) + k*A000032(n) is prime, or -1 if there is no such k.
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1
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1, 1, 2, -1, 2, 6, -1, 2, 8, -1, 4, 2, -1, 6, 2, -1, 10, 4, -1, 20, 2, -1, 4, 44, -1, 4, 56, -1, 8, 22, -1, 12, 16, -1, 10, 2, -1, 34, 8, -1, 8, 16, -1, 26, 10, -1, 10, 14, -1, 60, 4, -1, 14, 28, -1, 32, 16, -1, 8, 20, -1, 66, 44, -1, 74, 12, -1, 110, 40, -1, 48, 6, -1, 10, 4, -1, 32, 34, -1, 62
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OFFSET
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0,3
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COMMENTS
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a(n) = -1 if n > 0 is divisible by 3, as A000045(n) and A000032(n) are both even.
If n is not divisible by 3, A000032(n) and A000045(n) are coprime, so Dirichlet's theorem implies A000032(n) + k*A000045(n) is prime for infinitely many k.
a(n) is even if n > 1 is not divisible by 3, since A000045(n) and A000032(n) are both odd.
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LINKS
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EXAMPLE
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a(4) = 2 because A000032(4) = 7 and A000045(4) = 3, and 7+2*3 = 13 is prime while 7+1*3 = 10 is not prime.
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MAPLE
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F:= combinat:-fibonacci:
L:= n -> 2*F(n+1)-F(n):
f:= proc(n) local a, b, k;
if n mod 3 = 0 then return -1 fi;
a:= F(n); b:= L(n);
for k from 2 by 2 do
if isprime(a+k*b) then return k fi
od
end proc:
f(0):= 1: f(1):= 1:
map(f, [$0..100]);
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MATHEMATICA
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a={}; nmax=79; For[n=0, n<=nmax, n++, If[n>0 && Divisible[n, 3], AppendTo[a, -1], For[k=1, k>0, k++, If[PrimeQ[Fibonacci[n]+k LucasL[n]], AppendTo[a, k]; k=-1]]]]; a(* Stefano Spezia, Sep 15 2022 *)
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PROG
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(Python)
if n == 0: return 1
elif n % 3 == 0: return -1
k = 1
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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