A139141 For n>=1, a(n) = floor((d(p(n)+1) + d(p(n)+2) + d(p(n)+3) + ... +d(p(n+1)))/(p(n+1) - p(n))), where d(m) is the number of positive divisors of m and p(n) is the n-th prime. a(0) = floor((d(1) + d(2))/2).

1, 2, 2, 3, 3, 4, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5, 7, 5, 5, 7, 5, 5, 5, 6, 5, 5, 5, 7, 6, 5, 5, 6, 5, 6, 7, 6, 6, 5, 6, 5, 10, 6, 8, 5, 7, 6, 6, 6, 7, 6, 6, 11, 5, 7, 6, 6, 9, 7, 7, 5, 6, 7, 6, 9, 6, 7, 7, 6, 7, 8, 5, 7, 7, 7, 5, 7, 8, 7, 7, 6, 13, 6, 11, 6, 8, 6, 7, 6, 9, 6, 7, 8, 6, 7, 5, 8, 7, 7, 7, 7, 6, 8

Offset: 0

Author

Leroy Quet, Apr 10 2008

Keywords

• nonn

Comments

The sequence approximates the average number of divisors over all integers between consecutive primes.

Examples

			The 9th prime is 23 and the 10th prime is 29. So a(9) = floor((d(24) + d(25) + d(26) + d(27) + d(28) + d(29))/6) = floor((8 + 3 + 4 + 4 + 6 + 2)/6) = floor(27/6) = 4.
		

Crossrefs

Cf. A001223, A139140.

Programs

Maple

A139141 := proc(n) if n = 0 then 1; else add(numtheory[tau](k),k=ithprime(n)+1..ithprime(n+1)) ; floor(%/(ithprime(n+1)-ithprime(n))) ; fi; end proc: seq(A139141(n),n=0..120) ; # R. J. Mathar, Oct 24 2009

Formula

For n>= 1, a(n) = floor(A139140(n)/A001223(n)).

Extensions

Extended beyond a(11) by R. J. Mathar, Oct 24 2009