%I
%S 0,1,1,2,2,2,3,3,4,5,3,4,5,5,5,7,4,7,5,5,9,6,7,3,6,5,5,3,5,5,6,6,8,6,
%T 5,7,9,6,8,5,5,7,7,6,9,6,7,9,8,7,4,7,8,4,7,7,7,5,7,4,6,5,6,9,6,5,8,10,
%U 11,10,11,12,12,11,10,12,12,13,13,13,9,14,10,15,17,16,16,18,17,9,5,18,20
%N a(n) = the number of integers k, 1 <= k <= n1, such that p(n)p(k) is divisible by nk (here p(n) = the nth prime.)
%e The 10th prime is 29. Checking: 292=27 is divisible by 101=9. 293=26 is not divisible by 102=8. 295=24 is not divisible by 103=7. 297=22 is not divisible by 104=6. 2911=18 is not divisible by 105=5. 2913=16 is divisible by 106=4. 2917=12 is divisible by 107=3. 2919=10 is divisible by 108=2. And 2923=6 is divisible by 109=1. There are therefore five k's where p(10)p(k) is divisible by 10k. So a(10)=5.
%p a := proc (n) local ct, k: ct := 0: for k to n1 do if `mod`(ithprime(n)ithprime(k), nk) = 0 then ct := ct+1 else end if end do: ct end proc: seq(a(n), n = 1 .. 100); # _Emeric Deutsch_, Jul 30 2009
%p A163001 := proc(n) local a,k ; a := 0 ; for k from 1 to n1 do if ( ithprime(n)ithprime(k) ) mod (nk) = 0 then a := a+1; fi; od: a ; end ; seq(A163001(n),n=1..120) ; # _R. J. Mathar_, Jul 30 2009
%K nonn
%O 1,4
%A _Leroy Quet_, Jul 20 2009
%E Extended by _Emeric Deutsch_ and _R. J. Mathar_, Jul 30 2009
