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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. 6

%I #18 Oct 08 2015 04:25:44

%S 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,

%T 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,

%U 7,6,4,10,3,5,8,8,4,6,4,10,7,3,5,9,6,5,5,9,4,8

%N 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.

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

%C We have verified this for n up to 120000.

%C See also A262995, A263001 and A263020 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A262999/b262999.txt">Table of n, a(n) for n = 1..10000</a>

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

%e a(3) = 1 since 3 = pi(3*4/2) + pi(1^2).

%e a(5) = 1 since 5 = pi(3*4/2) + pi(2^2).

%e a(7) = 1 since 7 = pi(3*4/2) + pi(3^2).

%e a(35) = 1 since 35 = pi(13*14/2) + pi(6^2).

%e a(37) = 1 since 37 = pi(3*4/2) + pi(12^2).

%e a(217) = 1 since 217 = pi(17*18/2) + pi(33^2).

%e a(590) = 1 since 590 = 58 + 532 = pi(23*24/2) + pi(62^2).

%e a(7439) = 1 since 7439 = 3854 + 3585 = pi(269*270/2) + pi(183^2).

%e a(10381) = 1 since 10381 = 1875 + 8506 = pi(179*180/2) + pi(296^2).

%t s[n_]:=s[n]=PrimePi[n^2]

%t t[n_]:=t[n]=PrimePi[n(n+1)/2]

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

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Oct 07 2015

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