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 A232269 Number of ways to write 2*n + 1 = x + y (x, y > 0) with x^3 + y^2 and x^2 + y^2 both prime. 2
 1, 3, 1, 2, 3, 2, 1, 6, 4, 1, 4, 6, 3, 8, 1, 1, 6, 1, 1, 9, 2, 4, 5, 3, 1, 2, 7, 4, 5, 8, 1, 12, 4, 4, 12, 3, 4, 9, 10, 1, 5, 9, 5, 11, 7, 4, 9, 2, 4, 19, 1, 1, 14, 4, 6, 16, 8, 5, 8, 7, 2, 11, 8, 1, 16, 3, 5, 9, 4, 3, 8, 8, 6, 16, 4, 3, 12, 13, 5, 11, 5, 3, 10, 10, 7, 12, 7, 4, 17, 20, 1, 17, 5, 6, 15, 4, 5, 18, 5, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: a(n) > 0 for all n > 0. Also, any odd integer greater than one can be written as x + y (0 < x < y) with x^3 + y^2 prime. The conjecture implies that there are infinitely many primes of the form x^3 + y^2 (x, y > 0) with x^2 + y^2 also prime. Note that Ming-Zhi Zhang ever asked (cf. A036468) whether any odd integer greater than one can be written as x + y (x, y > 0) with x^2 + y^2 prime. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588. EXAMPLE a(10) = 1 since 2*10 + 1 = 1 + 20 with 1^2 + 20^2 = 1^3 + 20^2 = 401 prime. a(15) = 1 since 2*15 + 1 = 25 + 6 with 25^2 + 6^2 = 661 and 25^3 + 6^2 = 15661 both prime. a(40) = 1 since 2*40 + 1 = 55 + 26 with 55^2 + 26^2 = 3701 and 55^3 + 26^2 = 167051 both prime. a(91) =1 since 2*91 + 1 = 85 + 98 with 85^2 + 98^2 = 16829 and 85^3 + 98^2 = 623729 both prime. MATHEMATICA a[n_]:=Sum[If[PrimeQ[x^3+(2n+1-x)^2]&&PrimeQ[x^2+(2n+1-x)^2], 1, 0], {x, 1, 2n}] Table[a[n], {n, 1, 100}] CROSSREFS Cf. A000040, A036468, A066649, A220413, A232174, A232186. Sequence in context: A036584 A260453 A210243 * A305391 A165084 A029279 Adjacent sequences:  A232266 A232267 A232268 * A232270 A232271 A232272 KEYWORD nonn AUTHOR Zhi-Wei Sun, Nov 22 2013 STATUS approved

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Last modified February 17 15:19 EST 2019. Contains 320220 sequences. (Running on oeis4.)