

A262995


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


6



1, 1, 1, 3, 3, 3, 3, 5, 3, 6, 3, 6, 4, 6, 4, 7, 4, 6, 6, 6, 4, 8, 5, 6, 6, 7, 5, 8, 4, 9, 5, 7, 9, 5, 7, 8, 6, 9, 5, 9, 7, 7, 8, 8, 6, 8, 8, 8, 6, 7, 10, 8, 4, 12, 6, 8, 7, 9, 6, 10, 6, 8, 10, 8, 6, 12, 4, 12, 6, 11, 6, 11, 6, 9, 10, 8, 7, 11, 7, 10, 8, 9, 7, 10, 7, 13, 5, 7, 11, 9, 6, 8, 12, 8, 7, 11, 7, 12, 5
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OFFSET

1,4


COMMENTS

Conjecture: a(n) > 0 for all n > 0.
This has been verified for n up to 10^5. See also A262999 for a similar conjecture.
By Chebyshev's inequality, pi(n*(n+1)/2) > n1 for all n > 1.
In A262403 and A262439, the author conjectured that the sequences pi(n*(n+1)/2) (n = 1,2,3,...) and pi(1+n*(n+1)/2) (n = 1,2,3,...) are both strictly increasing.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000


EXAMPLE

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


MATHEMATICA

s[k_]:=s[k]=PrimePi[k(k+1)/2+1]
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, A000720, A111208, A262403, A262439, A262446, A262999.
Sequence in context: A068048 A176994 A264050 * A125713 A162220 A118911
Adjacent sequences: A262992 A262993 A262994 * A262996 A262997 A262998


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 07 2015


STATUS

approved



