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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A303932 Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic residue modulo p, and k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 1, 3, 4, 2, 3, 3, 1, 3, 5, 2, 1, 4, 2, 1, 4, 3, 4, 4, 2, 3, 7, 4, 2, 6, 3, 2, 4, 4, 3, 3, 2, 4, 6, 2, 1, 6, 2, 2, 6, 5, 6, 5, 5, 6, 8, 3, 5, 8, 5, 3, 7, 6, 5, 7, 6, 9, 7, 5, 7, 7, 3, 5, 9, 5, 7, 9, 6, 11, 10, 5, 11, 10, 4, 5, 13, 3, 5
Offset: 1

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Author

Zhi-Wei Sun, May 02 2018

Keywords

Comments

Conjecture 1. a(n) > 0 for all n > 1, i.e., any even number greater than two can be written as the sum of a power 2, a power of 3 and a prime p with 11 a quadratic residue modulo p.
Conjecture 2. For any integer n > 2, we can write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic nonresidue modulo p, and k and m are nonnegative integers.
We have verified Conjectures 1 and 2 for n up to 5*10^8.

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 3^0 with 11 a quadratic residue modulo the prime 2.
a(3) = 1 since 2*3 = 2 + 2^0 + 3^1 with 11 a quadratic residue modulo the prime 2.
a(10) = 1 since 2*10 = 7 + 2^2 + 3^2 with 11 a quadratic residue modulo the prime 7.
a(14) = 1 since 2*14 = 19 + 2^3 + 3^0 with 11 a quadratic residue modulo the prime 19.
a(17) = 1 since 2*17 = 5 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 5.
a(38) = 1 since 2*38 = 37 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 37.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=PQ[n]=n==2||(n>2&&PrimeQ[n]&&JacobiSymbol[11,n]==1);
    tab={};Do[r=0;Do[If[PQ[2n-2^k-3^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[3,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]