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A276830 Number of ways to write n as ((p-1)/2)^2 + P_2, where p is an odd prime and P_2 is a product of at most two primes.

%I A276830 #7 Sep 20 2016 08:14:02
%S A276830 0,1,1,1,2,2,2,2,1,3,3,2,2,3,3,2,1,3,2,2,1,2,3,2,1,4,3,2,2,4,2,3,1,3,
%T A276830 4,2,2,5,4,4,2,5,3,3,2,3,5,3,1,5,3,2,2,2,3,3,2,4,4,3,2,5,3,2,3,5,3,4,
%U A276830 3,4,5,2,3,5,4,2,3,5,2,3
%N A276830 Number of ways to write n as ((p-1)/2)^2 + P_2, where p is an odd prime and P_2 is a product of at most two primes.
%C A276830 Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 9, 17, 21, 25, 33, 49, 109, 169, 189, 361, 841, 961, 12769, 19321.
%C A276830 See also A276825 for a similar conjecture involving cubes, and some comments on x^2 + P_2.
%H A276830 Zhi-Wei Sun, <a href="/A276830/b276830.txt">Table of n, a(n) for n = 1..10000</a>
%e A276830 a(2) = 1 since 2 = ((3-1)/2)^2 + 1 with 3 prime.
%e A276830 a(3) = 1 since 3 = ((3-1)/2)^2 + 2 with 3 and 2 both prime.
%e A276830 a(4) = 1 since 4 = ((3-1)/2)^2 + 3 with 3 prime.
%e A276830 a(9) = 1 since 9 = ((5-1)/2)^2 + 5 with 5 prime.
%e A276830 a(17) = 1 since 17 = ((5-1)/2)^2 + 13 with 5 and 13 both prime.
%e A276830 a(21) = 1 since 21 = ((5-1)/2)^2 + 17 with 5 and 17 both prime.
%e A276830 a(25) = 1 since 25 = ((5-1)/2)^2 + 3*7 with 5, 3 and 7 all prime.
%e A276830 a(33) = 1 since 33 = ((5-1)/2)^2 + 29 with 5 and 29 both prime.
%e A276830 a(49) = 1 since 49 = ((13-1)/2)^2 + 13 with 13 prime.
%e A276830 a(109) = 1 since 109 = ((13-1)/2)^2 + 73 with 13 and 73 both prime.
%e A276830 a(169) = 1 since 169 = ((13-1)/2)^2 + 7*19 with 13, 7 and 19 all prime.
%e A276830 a(189) = 1 since 189 = ((5-1)/2)^2 + 5*37 with 5 and 37 both prime.
%e A276830 a(361) = 1 since 361 = ((37-1)/2)^2 + + 37 with 37 prime.
%e A276830 a(841) = 1 since 841 = ((37-1)/2)^2 + 11*47 with 37, 11 and 47 all prime.
%e A276830 a(961) = 1 since 961 = ((61-1)/2)^2 + 61 with 61 prime.
%e A276830 a(12769) = 1 since 12769 = ((109-1)/2)^2 + 59*167 with 109, 59 and 167 all prime.
%e A276830 a(19321) = 1 since 19321 = ((277-1)/2)^2 + 277 with 277 prime.
%t A276830 PP[n_]:=PP[n]=PrimeQ[Sqrt[n]]||(SquareFreeQ[n]&&Length[FactorInteger[n]]<=2)
%t A276830 Do[r=0;Do[If[PP[n-((Prime[k]-1)/2)^2],r=r+1;If[r>1,Goto[aa]]],{k,2,PrimePi[2*Sqrt[n]+1]}];Print[n," ",r];
%t A276830 Label[aa];If[Mod[n,50000]==0,Print[n]];Continue,{n,10^5,1000000}]
%Y A276830 Cf. A000040, A000290, A037143, A235645, A276711, A276825.
%K A276830 nonn
%O A276830 1,5
%A A276830 _Zhi-Wei Sun_, Sep 20 2016