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A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(1+m*(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

%I A262995 #12 Oct 07 2015 11:31:05
%S A262995 1,1,1,3,3,3,3,5,3,6,3,6,4,6,4,7,4,6,6,6,4,8,5,6,6,7,5,8,4,9,5,7,9,5,
%T A262995 7,8,6,9,5,9,7,7,8,8,6,8,8,8,6,7,10,8,4,12,6,8,7,9,6,10,6,8,10,8,6,12,
%U A262995 4,12,6,11,6,11,6,9,10,8,7,11,7,10,8,9,7,10,7,13,5,7,11,9,6,8,12,8,7,11,7,12,5
%N A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(1+m*(m+1)/2), where pi(x) denotes the number of primes not exceeding x.
%C A262995 Conjecture: a(n) > 0 for all n > 0.
%C A262995 This has been verified for n up to 10^5. See also A262999 for a similar conjecture.
%C A262995 By Chebyshev's inequality, pi(n*(n+1)/2) > n-1 for all n > 1.
%C A262995 In A262403 and A262439, the author conjectured that the sequences pi(n*(n+1)/2) (n = 1,2,3,...) and pi(1+n*(n+1)/2) (n = 1,2,3,...) are both strictly increasing.
%H A262995 Zhi-Wei Sun, <a href="/A262995/b262995.txt">Table of n, a(n) for n = 1..10000</a>
%e A262995 a(1) = 1 since 1 = pi(1*2/2) + pi(1+1*2/2).
%e A262995 a(2) = 1 since 2 = pi(1*2/2) + pi(1+2*3/2).
%e A262995 a(3) = 1 since 3 = pi(2*3/2) + pi(1+1*2/2).
%e A262995 a(4) = 3 since 4 = pi(1*2/2) + pi(1+3*4/2) = pi(2*3/2) + pi(1+2*3/2) = pi(3*4/2) + pi(1+1*2/2).
%t A262995 s[k_]:=s[k]=PrimePi[k(k+1)/2+1]
%t A262995 t[n_]:=t[n]=PrimePi[n(n+1)/2]
%t A262995 Do[r=0;Do[If[s[k]>n,Goto[bb]];Do[If[t[j]>n-s[k],Goto[aa]];If[t[j]==n-s[k],r=r+1];Continue,{j,1,n-s[k]+1}];Label[aa];Continue,{k,1,n}];Label[bb];Print[n," ",r];Continue,{n,1,100}]
%Y A262995 Cf. A000217, A000720, A111208, A262403, A262439, A262446, A262999.
%K A262995 nonn
%O A262995 1,4
%A A262995 _Zhi-Wei Sun_, Oct 07 2015