

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
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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

ZhiWei 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]>ns[k], Goto[aa]]; If[t[j]==ns[k], r=r+1]; Continue, {j, 1, ns[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

ZhiWei Sun, Oct 07 2015


STATUS

approved



