

A335641


Number of ordered ways to write 2n+1 as p + x*(9x+7) with p prime and x an integer.


1



1, 2, 2, 1, 1, 2, 1, 1, 3, 2, 2, 2, 2, 3, 2, 3, 2, 1, 3, 2, 2, 3, 2, 1, 1, 4, 2, 2, 3, 3, 5, 2, 3, 4, 2, 4, 3, 2, 3, 3, 4, 1, 2, 4, 3, 2, 2, 3, 2, 3, 4, 4, 3, 4, 3, 4, 2, 2, 5, 4, 4, 3, 3, 5, 4, 5, 2, 1, 6, 1, 3, 2, 3, 4, 3, 5, 2, 4, 4, 3, 5, 2, 3, 4, 1, 5, 4, 3, 4, 4, 4, 3, 3, 5, 4, 3, 6, 4, 6, 5
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OFFSET

1,2


COMMENTS

Conjecture 1: a(n) > 0 for all n > 0. Also, a(n) = 1 only for n = 1, 4, 5, 7, 8, 18, 24, 25, 42, 68, 70, 85, 117, 118, 196, 238, 287, 497, 628, 677, 732.
We have verified a(n) > 0 for all n = 1..2*10^8.
Conjecture 2: Let f(x) be any of the polynomials x*(3x+1), x*(5x+1), 2x*(3x+1), 2x*(3x+2). Then, each odd integer greater than one can be written as p + f(x) with p prime and x an integer.


LINKS



EXAMPLE

a(68) = 1, and 2*68+1 = 137 + 0*(9*0+7) with 137 prime.
a(117) = 1, and 2*117+1 = 233 + (1)*(9*(1)+7) with 233 prime.
a(238) = 1, and 2*238+1 = 461 + 1*(9*1+7) with 461 prime.
a(287) = 1, and 2*287+1 = 293 + (6)*(9*(6)+7) with 293 prime.
a(732) = 1, and 2*732+1 = 673 + 9*(9*9+7) with 673 prime.


MATHEMATICA

tab={}; Do[r=0; Do[If[PrimeQ[2n+1x*(9*x+7)], r=r+1], {x, Floor[(Sqrt[36(2n+1)+49]+7)/18], (Sqrt[36(2n+1)+49]7)/18}];
tab=Append[tab, r], {n, 1, 100}]; Print[tab]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



