A308950 Number of ways to write n as (p-1)/6 + 2^a*3^b, where p is a prime, and a and b are nonnegative integers.

0, 1, 2, 3, 3, 3, 4, 4, 5, 4, 5, 4, 6, 7, 6, 4, 5, 6, 9, 6, 6, 6, 5, 6, 7, 6, 7, 7, 10, 7, 6, 5, 8, 10, 8, 7, 8, 8, 11, 5, 10, 8, 8, 7, 6, 6, 6, 9, 10, 8, 6, 5, 10, 9, 8, 7, 9, 7, 11, 7, 8, 8, 7, 13, 10, 7, 10, 5, 10, 10, 10, 8, 8, 13, 9, 8, 8, 10, 11, 9, 8, 11, 8, 10, 10, 8, 8, 10, 9, 8, 8, 8, 10, 10, 8, 5, 11, 8, 15, 7

Offset: 1

Author

Zhi-Wei Sun, Jul 02 2019

Keywords

• nonn

Comments

Conjecture: Let r be 1 or -1. Then, any integer n > 1 can be written as (p-r)/6 + 2^a*3^b, where p is a prime, and a and b are nonnegative integers; in other words, 6*n+r can be written as p + 2^k*3^m, where p is a prime, and k and m are positive integers.

We have verified this for all n = 2..10^9.

Conjecture verified up to n = 10^11. - Giovanni Resta, Jul 03 2019

Examples

			a(2) = 1 since 2 = (7-1)/6 + 2^0*3^0 with 7 prime.
a(3) = 2 since 3 = (13-1)/6 + 2^0*3^0 = (7-1)/6 + 2^1*3^0 with 13 and 7 prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A000079, A000244, A002476, A003586, A007528, A308411.

Programs

Mathematica

tab={};Do[r=0;Do[If[PrimeQ[6(n-2^a*3^b)+1],r=r+1],{a,0,Log[2,n]},{b,0,Log[3,n/2^a]}];tab=Append[tab,r],{n,1,100}];Print[tab]