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A253178 revision #49

A253178

Least k>=1 such that 2*A007494(n)^k+1 is prime.

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 47, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 1, 2729, 1, 1, 2, 1, 2, 175, 1, 1, 1, 1, 1, 1, 3, 3, 3, 43, 1, 1, 2, 1, 1, 3, 2, 1, 1, 3, 1, 11, 1, 1, 4, 1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 192275, 2, 1233, 1, 3, 5, 51, 1, 1, 1, 1, 286, 1, 1, 755, 2, 1, 4, 1, 6, 1, 2

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OFFSET

1,8

COMMENTS

If n=3j+1, then all 2*n^k+1 are divisible by 3 and cannot be prime, so we limited the n in A007494 (n which is not in the form 3j+1).

Conjecture: a(n) is defined for all n.

a(145) > 200000, a(146) - a(156) = {1, 1, 66, 1, 4, 3, 1, 1, 1, 1, 6}, a(157) > 100000, a(158) - a(180) = {2, 1, 2, 11, 1, 1, 3, 321, 1, 1, 3, 1, 2, 12183, 5, 1, 1, 957, 2, 3, 16, 3, 1}.

a(n) = 1 if and only if n is in A144769.

LINKS

Eric Chen, Table of n, a(n) for n = 1..144

FORMULA

a(n) = A119624(A007494(n)).

MATHEMATICA

A007494[n_] := 2n - Floor[n/2];

Table[k=1; While[!PrimeQ[2*A007494[n]^k+1], k++]; k, {n, 1, 144}]

PROG

(PARI) a007494(n) = n+(n+1)>>1

a(n) = for(k=1, 2^24, if(ispseudoprime(2*a007494(n)^k+1), return(k)))

CROSSREFS