

A274876


The number of ways (2n)^2 is expressible as (p+1)(q+1) where p and q are distinct primes.


2



0, 0, 1, 0, 0, 3, 0, 1, 2, 0, 0, 4, 0, 1, 1, 1, 0, 4, 0, 2, 3, 1, 0, 4, 0, 1, 1, 2, 0, 5, 0, 1, 2, 1, 1, 4, 0, 1, 3, 1, 0, 7, 0, 2, 4, 0, 0, 4, 0, 2, 3, 1, 0, 2, 0, 2, 0, 0, 0, 9, 0, 0, 2, 0, 0, 5, 0, 3, 0, 2, 0, 8, 0, 0, 2, 2, 2, 6, 0, 2, 2, 1, 0, 6, 0, 1, 1, 2, 0, 8, 1, 1, 1, 0, 0, 2, 0, 1, 0, 4, 0, 4, 0, 2, 5
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,6


COMMENTS

No odd number squared is expressible as (p+1)(q+1) where p and q are distinct primes, since q must be odd and therefore (q+1) is even.


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000


EXAMPLE

a(1) = 0 since 2 is not expressible as (p+1)(q+1); same for a(2); a(3) = 1 since 6^2 = 36 = (2+1)(11+1); a(6) = 3 since 12^2 = 144 = (2+1)(47+1) = (5+1)(23+1) = (7+1)(17+1); a(9) = 2 since 18^2 = 324 = (2+1)(107+1) = (5+1)(53+1); etc.


MATHEMATICA

f[n_] := Block[{c = 0, p = 2}, While[p < 2n 1, If[ PrimeQ[(2n)^2/(p +1) 1], c++]; p = NextPrime@ p]; c]; Array[f, 105]


PROG

(PARI) a(n)=sumdiv(4*n^2, d, d<2*n && isprime(d1) && isprime(4*n^2/d1)) \\ Charles R Greathouse IV, Jul 10 2016


CROSSREFS

Cf. A274848, A274877.
Sequence in context: A303301 A160499 A329272 * A065718 A025428 A199176
Adjacent sequences: A274873 A274874 A274875 * A274877 A274878 A274879


KEYWORD

nonn


AUTHOR

Zak Seidov and Robert G. Wilson v, Jul 10 2016


STATUS

approved



