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A304034
Number of ways to write n as p + 2^k + (1+(n mod 2))*3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))*3^m squarefree.
16
0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 2, 1, 3, 1, 4, 2, 5, 1, 3, 2, 5, 1, 7, 3, 3, 4, 4, 4, 6, 2, 3, 5, 6, 2, 7, 3, 5, 5, 6, 5, 9, 3, 4, 6, 7, 2, 12, 2, 5, 6, 7, 4, 10, 3, 3, 5, 8, 2, 8, 3, 4, 6, 8, 5, 9, 4, 2, 7, 7, 3, 13, 5, 5, 9, 7, 5, 13, 3, 6, 10, 7, 5, 10, 5, 7, 7, 9, 8, 13
OFFSET
1,10
COMMENTS
Conjecture: a(n) > 0 for all n > 11.
This has been verified for n up to 10^10.
See also A304081 for a similar conjecture.
LINKS
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)
EXAMPLE
a(8) = 1 since 8 = 3 + 2^1 + 3^1 with 3 prime and 2^1 + 3^1 = 5 squarefree.
a(13) = 1 since 13 = 3 + 2^2 + 2*3^1 with 3 prime and 2^2 + 2*3^1 = 2*5 squarefree.
a(19) = 1 since 19 = 5 + 2^3 + 2*3^1 with 5 prime and 2^3 + 2*3^1 = 2*7 squarefree.
a(23) = 1 since 23 = 13 + 2^2 + 2*3^1 with 13 prime and 2^2 + 2*3 = 2*5 squarefree.
MATHEMATICA
tab={}; Do[r=0; Do[If[SquareFreeQ[2^k+(1+Mod[n, 2])*3^m]&&PrimeQ[n-2^k-(1+Mod[n, 2])*3^m], r=r+1], {k, 1, Log[2, n]}, {m, 1, If[2^k==n, -1, Log[3, (n-2^k)/(1+Mod[n, 2])]]}]; tab=Append[tab, r], {n, 1, 90}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 06 2018
STATUS
approved