

A141829


a(n) = the number of positive divisors of (p(n)1) that are each <= p(n+1)p(n), where p(n) is the nth prime.


3



1, 2, 2, 3, 2, 4, 2, 3, 2, 2, 5, 4, 2, 3, 2, 3, 2, 6, 3, 2, 5, 3, 2, 4, 4, 2, 3, 2, 4, 6, 3, 3, 2, 4, 2, 5, 5, 3, 2, 3, 2, 8, 2, 4, 2, 6, 7, 3, 2, 4, 3, 2, 8, 3, 3, 2, 2, 5, 4, 2, 4, 3, 3, 2, 4, 3, 5, 7, 2, 4, 3, 2, 4, 5, 3, 2, 3, 4, 5, 6, 2, 8, 2, 5, 3, 2, 5, 4, 2, 3, 2, 2, 3, 4, 3, 2, 3, 2, 6, 6, 5, 3, 2, 2, 5
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OFFSET

1,2


COMMENTS

a(n) also equals the number of positive integers k, k <= p(n+1)p(k), that divide (p(n)+k1).


LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..10000


EXAMPLE

The 16th prime is 53 and the 17th prime is 59. So the divisors of 531=52 that are <= 5953=6 are 1,2,4. There are three such divisors.
Also, 53 is divisible by 1. 54 is divisible by 2. 55 is not divisible by 3. 56 is divisible by 4. 57 is not divisible by 5. And 58 is not divisible 6. So in the span of integers p(16)=53 to p(17)1=58, there are 3 integers k where k divides (p(16)+k1). So a(16) = 3.


MAPLE

A141829 := proc(n) local p, q, a, d ; p := ithprime(n) ; q := nextprime(p) ; a := 0 ; for d in numtheory[divisors](p1) do if d <= qp then a :=a+1 ; fi; od: RETURN(a) ; end: for n from 1 to 200 do printf("%a, ", A141829(n)) ; od: # R. J. Mathar, Aug 08 2008


MATHEMATICA

Table[Function[{p, q}, DivisorSum[p  1, 1 &, # <= q  p &]] @@ {Prime@ n, Prime[n + 1]}, {n, 105}] (* Michael De Vlieger, Oct 25 2017 *)


PROG

(PARI) a(n) = #select(x>(x <= prime(n+1)prime(n)), divisors(prime(n)1)); \\ Michel Marcus, Oct 26 2017


CROSSREFS

Cf. A141830, A141831.
Sequence in context: A083901 A274517 A038148 * A111336 A083902 A205562
Adjacent sequences: A141826 A141827 A141828 * A141830 A141831 A141832


KEYWORD

nonn


AUTHOR

Leroy Quet, Jul 09 2008


EXTENSIONS

Extended beyond a(17) by R. J. Mathar, Aug 08 2008


STATUS

approved



