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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
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE 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). 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}] CROSSREFS Cf. A000217, A000290, A000720, A038107, A111208, A262995, A263001, A263020. Sequence in context: A029234 A102613 A097019 * A324933 A085343 A049077 Adjacent sequences:  A262996 A262997 A262998 * A263000 A263001 A263002 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 07 2015 STATUS approved

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Last modified January 16 23:43 EST 2022. Contains 350377 sequences. (Running on oeis4.)