%I #15 May 07 2018 03:36:12
%S 0,1,1,3,3,2,2,3,3,4,3,5,4,4,3,4,5,7,4,7,4,8,7,6,7,6,5,5,5,7,5,8,5,5,
%T 8,6,9,9,6,8,6,6,7,8,4,7,8,7,3,10,6,7,8,7,7,9,5,8,7,6,5,5,6,3,11,7,9,
%U 12,8,12,10,11,11,9,7,9,7,8,8,11,7,11,8,9,15,11,8,9,8,9
%N Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m squarefree, where k and m are nonnegative integers.
%C Conjecture: a(n) > 0 for all n > 1.
%C This has been verified for all n = 2..10^10.
%C Note that a(n) <= A303821(n).
%H Zhi-Wei Sun, <a href="/A303934/b303934.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/116f.pdf">Mixed sums of primes and other terms</a>, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
%H Zhi-Wei Sun, <a href="https://doi.org/10.1007/978-3-319-68032-3_20">Conjectures on representations involving primes</a>, 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 <a href="http://arxiv.org/abs/1211.1588">arXiv:1211.1588 [math.NT]</a>, 2012-2017.)
%e a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime and 2^0 + 5^0 squarefree.
%e a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime and 2^1 + 5^0 squarefree.
%t tab={};Do[r=0;Do[If[SquareFreeQ[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]
%Y Cf. A000040, A000079, A000351, A005117, A118955, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303821, A303932, A304034, A304081, A304122.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, May 03 2018