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a(n) is the number of ways n can be written in the form n=2^k1*p1^k2+2^k3*p2^k4, where p1 and p2 are odd prime numbers, and k1, k2, k3, k4 are nonnegative integers.
4

%I #14 Aug 11 2015 09:50:29

%S 1,1,2,2,3,3,4,4,5,5,6,6,7,7,7,7,8,8,9,9,9,9,10,10,11,11,12,12,13,12,

%T 13,13,13,13,14,13,14,14,14,14,15,14,15,16,15,15,17,16,18,18,17,17,20,

%U 18,19,19,18,18,21,18,19,21,19,20,21,18,20,20,20,18,23

%N a(n) is the number of ways n can be written in the form n=2^k1*p1^k2+2^k3*p2^k4, where p1 and p2 are odd prime numbers, and k1, k2, k3, k4 are nonnegative integers.

%C Sequence is defined for n >= 2.

%C When ki=0, corresponding terms 2^k1, p1^k2, 2^k3, or p2^k4 are 1.

%C All terms of this sequence are greater than zero.

%H Lei Zhou, <a href="/A238263/b238263.txt">Table of n, a(n) for n = 2..10000</a>

%e n=2, 2=1*1+1*1. This is the only choice that matches the definition. So a(2)=1;

%e ...

%e n=37, 37 = 1*1+2^2*3^2 = 1*3+2*17*1 = 1*5+2^5*1 = 2*3*1+1*31 = 2^3*1+1*29 = 1*3^2+2^2*7 = 2*5+1*3^3 = 1*11+2*13 = 2^2*3+1*5^2 = 1*13+2^3*3 = 2*7+1*23 = 1*17+2^2*5 = 2*3^2+1*19, 13 ways matching the definition. So a(37)=13.

%t Table[ct = 0; Do[If[f1 = FactorInteger[i]; l1 = Length[f1]; If[f1[[1, 1]] == 2, l1--]; f2 = FactorInteger[n - i]; l2 = Length[f2]; If[f2[[1, 1]] == 2, l2--]; (l1 <= 1) && (l2 <= 1), ct++], {i, 1, Floor[n/2]}]; ct, {n, 2, 72}]

%Y Cf. A000961.

%K nonn,easy

%O 2,3

%A _Lei Zhou_, Feb 21 2014