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%I #18 May 05 2018 02:45:16
%S 0,0,1,2,3,4,4,5,3,6,5,6,8,7,5,7,7,6,8,11,5,8,9,5,10,8,7,8,7,5,7,10,6,
%T 9,9,5,11,12,8,13,12,9,8,15,9,11,12,11,7,10,9,10,14,9,12,12,11,11,12,
%U 9,9,12,8,5,13,9,10,14,10,13,9,15,10,12,9,12,11,9,11,13
%N Number of ways to write 2*n+1 as p + 2^k + binomial(2*m,m), where p is a prime, and k and m are positive integers.
%C Conjecture: a(n) > 0 for all n > 2.
%C This has been verified for n up to 10^9.
%H Zhi-Wei Sun, <a href="/A303998/b303998.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(3) = 1 since 2*3+1 = 3 + 2^1 + binomial(2*1,1) with 3 prime.
%e a(4) = 2 since 2*4+1 = 3 + 2^2 + binomial(2*1,1) = 5 + 2^1 + binomial(2*1,1) with 3 and 5 both prime.
%t c[n_]:=c[n]=Binomial[2n,n];
%t tab={};Do[r=0;k=1;Label[bb];If[c[k]>2n,Goto[aa]];Do[If[PrimeQ[2n+1-c[k]-2^m],r=r+1],{m,1,Log[2,2n+1-c[k]]}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]
%Y Cf. A000040, A000079, A000984, 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, A303934, A303997, A304031.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, May 04 2018
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