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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Oct 07 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0.

This has been verified for n up to 10^5. See also A262999 for a similar conjecture.

By Chebyshev's inequality, pi(n*(n+1)/2) > n-1 for all n > 1.

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.

Examples

			a(1) = 1 since 1 = pi(1*2/2) + pi(1+1*2/2).
a(2) = 1 since 2 = pi(1*2/2) + pi(1+2*3/2).
a(3) = 1 since 3 = pi(2*3/2) + pi(1+1*2/2).
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).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A000720, A111208, A262403, A262439, A262446, A262999.

Programs

Mathematica

s[k_]:=s[k]=PrimePi[k(k+1)/2+1]
t[n_]:=t[n]=PrimePi[n(n+1)/2]
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}]