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 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 (list; graph; refs; listen; history; text; internal format)
 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 Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, On sums of primes and triangular numbers, J. Comb. and Number Theory 1(2009), no.1, 65-76. See also arXiv:0803.3737 [math.NT], 2008. 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+1-x*(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 Cf. A000040, A132399, A144590, A256071. Sequence in context: A075661 A206829 A319694 * A163495 A060709 A035222 Adjacent sequences: A335638 A335639 A335640 * A335642 A335643 A335644 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 03 2020 STATUS approved

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Last modified December 3 18:19 EST 2023. Contains 367540 sequences. (Running on oeis4.)