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

1,3

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

Links

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

Example

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.

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]

Crossrefs

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

Sequence in context: A234016 A029119 A178042 * A193832 A293137 A087823

Adjacent sequences: A308947 A308948 A308949 * A308951 A308952 A308953

Keyword

nonn

Author

Zhi-Wei Sun, Jul 02 2019

Status

approved