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.

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

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 3 = 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: A104351 A366274 A284511 * A349608 A224709 A309965

Adjacent sequences: A262973 A262974 A262975 * A262977 A262978 A262979

Keyword

nonn

Author

Zhi-Wei Sun, Oct 05 2015

Status

approved