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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 (list; graph; refs; listen; history; text; internal format)
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) > n-1 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

Zhi-Wei 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]>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, 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

Zhi-Wei Sun, Oct 07 2015

STATUS

approved

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Last modified November 19 08:44 EST 2019. Contains 329318 sequences. (Running on oeis4.)