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 A262976 Number of ordered ways to write n as 2^x + y^2 + pi(z^2) with x >= 0, y >= 0 and z > 0, where pi(m) denotes the number of primes not exceeding m. 4
 1, 2, 2, 3, 4, 4, 4, 6, 4, 6, 6, 7, 6, 7, 5, 6, 10, 5, 9, 10, 7, 7, 9, 9, 4, 12, 10, 9, 8, 7, 10, 9, 10, 7, 15, 10, 6, 13, 10, 9, 10, 16, 10, 10, 9, 8, 15, 9, 8, 15, 12, 12, 7, 12, 11, 14, 12, 8, 16, 6, 10, 11, 14, 8, 11, 17, 10, 16, 9, 13, 16, 15, 8, 18, 13, 10, 14, 10, 12, 16, 12, 13, 18, 11, 9, 17, 17, 9, 15, 16, 15, 9, 12, 12, 17, 12, 9, 21, 10, 11 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: (i) a(n) > 0 for all n > 0. (ii) Each positive integer can be written as 2^x + pi(y^2) + pi(z^2) with x >= 0, y > 0 and z > 0. REFERENCES Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014. EXAMPLE a(1) = 1 since 1 = 2^0 + 0^2 + pi(1^2). a(2) = 2 since 2 = 2^0 + 1^2 + pi(1^2) = 2 + 0^2 + pi(1^2). a(3) = 2 since 2 = 2^0 + 0^2 + pi(2^2) = 2 + 1^2 + pi(1^2). MATHEMATICA SQ[n_]:=IntegerQ[Sqrt[n]] f[n_]:=PrimePi[n^2] Do[r=0; Do[If[f[x]>=n, Goto[aa]]; Do[If[2^y>n-f[x], Goto[bb]]; If[SQ[n-f[x]-2^y], r=r+1], {y, 0, Log[2, n-f[x]]}]; Label[bb]; Continue, {x, 1, n}]; Label[aa]; Print[n, " ", r]; Continue, {n, 1, 100}] CROSSREFS Cf. A000079, A000290, A000720, A038107, A262746, A262887. Sequence in context: A131807 A104351 A284511 * A224709 A309965 A070172 Adjacent sequences:  A262973 A262974 A262975 * A262977 A262978 A262979 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 05 2015 STATUS approved

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Last modified October 23 07:11 EDT 2019. Contains 328336 sequences. (Running on oeis4.)