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%I #13 Apr 26 2015 00:18:06
%S 1,1,2,1,2,2,2,2,2,2,2,2,2,2,3,2,3,2,3,3,3,3,5,3,3,3,5,3,3,3,5,3,4,3,
%T 3,3,6,5,1,2,5,4,2,1,2,3,4,3,4,5,3,3,3,3,3,2,2,2,4,3,3,2,3,3,1,3,4,4,
%U 5,4,4,3,4,3,3,1,5,3,5,3,2,1,3,3,3,1,2,2,4,2,4,4,5,3,4,4,5,3,3,2
%N Number of unordered ways to write n = a + b, where a and b are distinct elements of the set {floor(x/3): 3*x-1 and 3*x+1 are twin prime}, one of a and b is even, and one of a and b has the form p-1 or p-2 with p prime.
%C Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 4, 39, 44, 65, 76, 82, 86, 108, 110, 123, 154, 175, 178, 196, 205, 221, 284, 308, 621, 735, 4655.
%C This is much stronger than the Twin Prime Conjecture. Note that a(n) <= A257317(n) <= A256707(n). We have verified that a(n) > 0 for all n = 1..10^7.
%H Zhi-Wei Sun, <a href="/A257474/b257474.txt">Table of n, a(n) for n = 1..20000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1504.01608">Natural numbers represented by floor(x^2/a)+floor(y^2/b)+floor(z^2/c)</a>, arXiv:1504.01608 [math.NT], 2015.
%e a(205) = 1 since 205 = 25 + 180 = floor(76/3) + floor(540/3) with 180 even and 180 + 1 prime, and {3*76-1,3*76+1} = {227,229} and {3*540-1,3*540+1} = {1619,1621} twin prime pairs.
%e a(308) = 1 since 308 = 128 + 180 = floor(384/3) + floor(540/3) with 180 + 1 prime, and {3*128-1,3*128+1} = {1151,1153} and {3*540-1,3*540+1} = {1619,1621} twin prime pairs.
%e a(621) = 1 since 621 = 310 + 311 = floor(930/3) + floor(934/3) with 310 even and 310 + 1 prime, {3*930-1,3*930+1} ={2789,2791} and {3*934-1,3*934+1} = {2801,2803} twin prime pairs.
%e a(735) = 1 since 735 = 311 + 424 = floor(934/3) + floor(1274/3) with 424 even, 311 + 2 = 313 prime, and {3*934-1,3*934+1} = {2801,2803} and {3*1274-1,3*1274+1} = {3821,3823} twin prime pairs.
%e a(4655) = 1 since 4655 = 15 + 4640 = floor(46/3) + floor(13920/3) with 4640 even, 15 + 2 prime, and {3*46-1,3*46+1} = {137,139} and {3*13920-1,3*13920+1} = {41759,41761} twin prime pairs.
%t TQ[n_]:=PrimeQ[3n-1]&&PrimeQ[3n+1]
%t PQ[n_]:=TQ[3*n]||TQ[3*n+1]||TQ[3n+2]
%t RQ[n_]:=PrimeQ[n+1]||PrimeQ[n+2]
%t Do[r=0;Do[If[Mod[x(n-x),2]==0&&(RQ[x]||RQ[n-x])&&PQ[x]&&PQ[n-x],r=r+1],{x,0,(n-1)/2}];
%t Print[n," ",r];Continue,{n,1,100}]
%Y Cf. A000040, A014574, A256707, A257121, A257317.
%K nonn
%O 1,3
%A _Zhi-Wei Sun_, Apr 25 2015
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