

A256071


Number of ordered ways to write n = p + x*(3*x1)/2, where p is prime or zero, and x is an integer.


4



1, 1, 2, 2, 2, 3, 1, 4, 2, 2, 2, 1, 4, 2, 3, 3, 1, 3, 4, 3, 3, 1, 3, 2, 4, 3, 3, 1, 3, 4, 2, 4, 2, 3, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 3, 4, 3, 2, 4, 3, 1, 3, 3, 5, 4, 3, 2, 3, 4, 5, 3, 2, 4, 4, 4, 2, 3, 2, 5, 4, 3, 3, 4, 5, 5, 3, 4, 3, 3, 4, 5, 4, 4, 5, 3, 3, 3, 3, 6, 3, 3, 2, 2, 4, 7, 3, 3, 3, 4, 5, 3
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OFFSET

0,3


COMMENTS

Conjecture: a(n) > 0 for all n. Moreover, each nonnegative integer n is either an odd prime, or a generalized pentagonal number, or the sum of an odd prime and a generalized pentagonal number.
This is similar to the author's earlier conjecture on sums of primes and triangular numbers (see the reference and also A132399).
The conjecture has been verified for all n = 0..10^9.


REFERENCES

ZhiWei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), 6576.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, On sums of primes and triangular numbers, arXiv:0803.3737 [math.NT], 20082009.
ZhiWei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 20092015.


EXAMPLE

a(11) = 1 since 11 = 11 + 0*(3*01)/2 with 11 prime.
a(15) = 1 since 15 = 0 + (3)*(3*(3)1)/2.
a(50) = 1 since 50 = 43 + (2)*(3*(2)1)/2 with 43 prime.


MATHEMATICA

P[n_]:=(n==0)PrimeQ[n]
Do[r=0; Do[If[P[nx(3x1)/2], r=r+1], {x, Floor[(Sqrt[24n+1]1)/6], Floor[(Sqrt[24n+1]+1)/6]}]; Print[n, " ", r]; Label[aa]; Continue, {n, 0, 100}]


PROG

(PARI) a(n)=if(n==0, return(1)); sum(x=1, (1+sqrt(24*n+1))\6, isprime(nx*(3*x1)/2))+sum(x=0, (sqrt(24*n+1)1)\6, isprime(nx*(3*x+1)/2))+ispolygonal(n, 5)+(x>3*x^2+x==2*n)(round((sqrt(24*n+1)1)/6)) \\ Charles R Greathouse IV, Apr 07 2015


CROSSREFS

Cf. A000040, A001318, A132399, A256119.
Sequence in context: A135151 A256855 A273943 * A248808 A233206 A014843
Adjacent sequences: A256068 A256069 A256070 * A256072 A256073 A256074


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 13 2015


STATUS

approved



