A262999 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(m^2), where pi(x) denotes the number of primes not exceeding x.

0, 2, 1, 3, 1, 4, 1, 4, 3, 3, 4, 3, 4, 3, 5, 2, 4, 6, 2, 6, 3, 5, 3, 5, 5, 4, 6, 3, 5, 5, 4, 5, 6, 6, 1, 10, 1, 6, 7, 3, 6, 6, 6, 3, 6, 6, 4, 9, 2, 8, 4, 7, 3, 8, 5, 4, 8, 6, 2, 7, 6, 6, 4, 8, 5, 7, 3, 7, 7, 6, 4, 10, 3, 5, 8, 8, 4, 6, 4, 10, 7, 3, 5, 9, 6, 5, 5, 9, 4, 8

Offset: 1

Author

Zhi-Wei Sun, Oct 07 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 3, 5, 7, 35, 37, 217, 7439, 10381.

We have verified this for n up to 120000.

See also A262995, A263001 and A263020 for similar conjectures.

Examples

			a(2) = 2 since 2 = pi(1*2/2) + pi(2^2) = pi(2*3/2) + pi(1^2).
a(3) = 1 since 3 = pi(3*4/2) + pi(1^2).
a(5) = 1 since 5 = pi(3*4/2) + pi(2^2).
a(7) = 1 since 7 = pi(3*4/2) + pi(3^2).
a(35) = 1 since 35 = pi(13*14/2) + pi(6^2).
a(37) = 1 since 37 = pi(3*4/2) + pi(12^2).
a(217) = 1 since 217 = pi(17*18/2) + pi(33^2).
a(590) = 1 since 590 = 58 + 532 = pi(23*24/2) + pi(62^2).
a(7439) = 1 since 7439 = 3854 + 3585 = pi(269*270/2) + pi(183^2).
a(10381) = 1 since 10381 = 1875 + 8506 = pi(179*180/2) + pi(296^2).
		

Links

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

Crossrefs

Cf. A000217, A000290, A000720, A038107, A111208, A262995, A263001, A263020.

Programs

Mathematica

s[n_]:=s[n]=PrimePi[n^2]
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}]