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A256558 Number of ways to write n = p + floor(k*(k+1)/4), where p is a prime and k is a positive integer.

%I A256558 #10 Jun 20 2024 08:24:50
%S A256558 0,1,2,1,2,2,2,3,1,3,1,4,2,3,1,3,3,3,2,4,3,2,3,4,3,2,3,1,5,3,3,3,3,3,
%T A256558 3,3,3,4,2,3,5,3,2,6,2,5,4,4,1,6,3,4,3,3,3,5,3,4,4,2,3,6,4,5,4,2,3,5,
%U A256558 3,5,6,2,4,6,4,5,3,3,5,5,6,3,6,2,3,6,4,4,7,3,3,5,5,3,3,2,6,6,4,5
%N A256558 Number of ways to write n = p + floor(k*(k+1)/4), where p is a prime and k is a positive integer.
%C A256558 Conjecture: (i) a(n) > 0 for all n > 1.
%C A256558 (ii) For any integer m > 4 not equal to 12, each integer n > 1 can be written as p + floor((k^2-1)/m), where p is a prime and k is a positive integer.
%C A256558 We also have some other conjectures on representations n = p + floor(k*(k+1)/m) with m > 4.
%H A256558 Zhi-Wei Sun, <a href="/A256558/b256558.txt">Table of n, a(n) for n = 1..10000</a>
%H A256558 Zhi-Wei Sun, <a href="http://arxiv.org/abs/0803.3737">On sums of primes and triangular numbers</a>, arXiv:0803.3737 [math.NT], 2008-2009.
%H A256558 Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/109p.pdf">On sums of primes and triangular numbers</a>, J. Comb. Number Theory 1(2009), 65-76.
%e A256558  a(15) = 1 since 15 = 5 + floor(6*7/4) with 5 prime.
%e A256558 a(420) = 1 since 420 = 419 + floor(2*3/4) with 419 prime.
%e A256558 a(945) = 1 since 945 = 877 + floor(16*17/4) with 877 prime.
%t A256558 Do[r=0;Do[If[PrimeQ[n-Floor[k(k+1)/4]],r=r+1],{k,1,(Sqrt[16n+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]
%Y A256558 Cf. A000040, A000217, A132399, A256544, A256555.
%K A256558 nonn
%O A256558 1,3
%A A256558 _Zhi-Wei Sun_, Apr 01 2015