

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

ZhiWei 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 ChinaJapan Seminar (Fukuoka, Oct. 28  Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169187.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei 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>nf[x], Goto[bb]]; If[SQ[nf[x]2^y], r=r+1], {y, 0, Log[2, nf[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 A070172 A273353
Adjacent sequences: A262973 A262974 A262975 * A262977 A262978 A262979


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 05 2015


STATUS

approved



