%I #52 Aug 05 2019 11:29:37

%S 1,2,11,27,61,809,13945,268027,565447,2358365,73551683,2734683311,

%T 112599773191,4860900544813,9968041656757,40762420985117,

%U 83151858555707,5085105491885327,341472595155548909,24295409051193284539

%N a(n) is the numerator of Product_{i=1..n} (1/(1-1/prime(i))) - 1.

%C The denominators are A038110(n+1).

%C a(n)/A038110(n+1) = Sum_{k >=2} 1/k where k is a positive integer whose prime factors are among the first n primes. In particular, for n=1,2,3,4,5, a(n)/A038110(n+1) is the sum of the reciprocals of the terms (excepting the first, 1) in A000079, A003586, A051037, A002473, A051038.

%C Appears to be a duplicate of A161527. - _Michel Marcus_, Aug 05 2019

%H Robert Israel, <a href="/A254196/b254196.txt">Table of n, a(n) for n = 1..422</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SmoothNumber.html">Smooth Number</a>

%F a(n) = A038111(n+1)/prime(n+1)-A038110(n+1). - _Robert Israel_, Jan 28 2015, corrected Jul 07 2019.

%e a(1)=1 because 1/2 + 1/4 + 1/8 + 1/16 + ... = 1/1.

%e a(2)=2 because 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + 1/12 + ... = 2/1.

%e a(3)=11 because 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/12 + 1/15 + ... = 11/4.

%e a(4)=27 because Sum_{n>=2} 1/A002473(n) = 27/8.

%e a(5)=61 because Sum_{n>=2} 1/A051038(n) = 61/16.

%p seq(numer(mul(1/(1-1/ithprime(i)),i=1..n)-1),n=1..20); # _Robert Israel_, Jan 28 2015

%t Numerator[Table[Product[1/(1 - 1/p), {p, Prime[Range[n]]}] - 1, {n,1,20}]]

%t b[0] := 0; b[n_] := b[n - 1] + (1 - b[n - 1]) / Prime[n]

%t Numerator@ Table[b[n], {n, 1, 20}] (* _Fred Daniel Kline_, Jun 27 2017 *)

%o (PARI) a(n) = numerator(prod(i=1, n, (1/(1-1/prime(i)))) - 1); \\ _Michel Marcus_, Jun 29 2017

%Y Cf. A038110, A000079, A003586, A051037, A002473, A051038, A161527.

%K nonn

%O 1,2

%A _Geoffrey Critzer_, Jan 26 2015