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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.
1
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
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]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jul 02 2019
STATUS
approved