A263020 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(m*(3*m-1)/2), where pi(x) denotes the number of primes not exceeding x.

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

Offset: 1

Author

Zhi-Wei Sun, Oct 07 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 1.

We have verified this for n up to 3*10^5. It seems that a(n) = 1 only for n = 2, 4, 10, 20, 31, 68, 147, 252, 580, 600, 772, 1326, 1381, 2779, 3136, 3422, 3729, 7151, 9518, 13481, 18070, 18673, 36965, 48181, 69250, 91130, 93580, 99868.

Note that n*(n+1)/2 (n = 0,1,2,...) are the triangular numbers while n*(3n-1)/2 (n = 0,1,2,...) are the pentagonal numbers.

Examples

			a(2) = 1 since 2 = 2 + 0 = pi(2*3/2) + pi(1*(3*1-1)/2).
a(4) = 1 since 4 = 4 + 0 = pi(4*5/2) + pi(1*(3*1-1)/2).
a(10) = 1 since 10 = 2 + 8 = pi(2*3/2) + pi(4*(3*4-1)/2).
a(20) = 1 since 20 = 9 + 11 = pi(7*8/2) + pi(5*(3*5-1)/2).
a(31) = 1 since 31 = 16 + 15 = pi(10*11/2) + pi(6*(3*6-1)/2).
a(68) = 1 since 68 = 2 + 66 = pi(2*3/2) + pi(15*(3*15-1)/2).
a(147) = 1 since 147 = pi(31*32/2) + pi(13*(3*13-1)/2).
a(252) = 1 since 252 = pi(29*30/2) + pi(26*(3*26-1)/2).
a(580) = 1 since 580 = pi(5*6/2) + pi(53*(3*53-1)/2).
a(600) = 1 since 600 = pi(42*43/2) + pi(46*(3*46-1)/2).
a(772) = 1 since 772 = pi(107*108/2) + pi(6*(3*6-1)/2).
a(1326) = 1 since 1326 = pi(139*140/2) + pi(22*(3*22-1)/2).
a(1381) = 1 since 1381 = pi(145*146/2) + pi(18*(3*18-1)/2).
a(2779) = 1 since 2779 = pi(212*213/2) + pi(33*(3*33-1)/2).
a(3136) = 1 since 3136 = pi(147*148/2) + pi(102*(3*102-1)/2).
a(3422) = 1 since 3422 = pi(151*152/2) + pi(109*(3*109-1)/2).
a(3729) = 1 since 3729 = pi(29*30/2) + pi(151*(3*151-1)/2).
a(7151) = 1 since 7151 = pi(100*101/2) + pi(208*(3*208-1)/2).
a(9518) = 1 since 9518 = pi(82*83/2) + pi(250*(3*250-1)/2).
a(13481) = 1 since 13481 = pi(539*540/2) + pi(6*(3*6-1)/2).
a(18070) = 1 since 18070 = pi(632*633/2) + pi(17*(3*17-1)/2).
a(18673) = 1 since 18673 = 14493 + 4180 = pi(561*562/2) + pi(163*(3*163-1)/2).
a(36965) = 1 since 36965 = 3780 + 33185 = pi(266*267/2) + pi(511*(3*511-1)/2).
a(48181) = 1 since 48181 = 30755 + 17426 = pi(848*849/2) + pi(359*(3*359-1)/2).
a(69250) = 1 since 69250 = 20669 + 48581 = pi(682*683/2) + pi(629*(3*629-1)/2).
a(91130) = 1 since 91130 = 81433 + 9697 = pi(1442*1443/2) + pi(260*(3*260-1)/2).
a(93580) = 1 since 93580 = 91865 + 1715 = pi(1539*1540/2) + pi(99*(3*99-1)/2).
a(99868) = 1 since 99868 = 66079 + 33789 = pi(1287*1288/2) + pi(516*(3*516-1)/2).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Conjectures on the prime-counting function, a message to Number Theory Mailing List, Oct. 19, 2015.

Crossrefs

Cf. A000217, A000326, A000720, A111208, A262403, A262995, A262999, A263001, A263107.

Programs

Mathematica

s[n_]:=s[n]=PrimePi[n(3n-1)/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}]