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Let b(n) = Product_{i=1..n} p_i/(p_i - 1), p_i = i-th prime; a(n) = minimum k such that b(k) > n.
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%I #24 Oct 10 2018 10:23:50

%S 1,1,2,3,4,6,9,14,22,35,55,89,142,230,373,609,996,1637,2698,4461,7398,

%T 12301,20503,34253,57348,96198,161659,272124,458789,774616,1309627,

%U 2216968,3757384,6375166

%N Let b(n) = Product_{i=1..n} p_i/(p_i - 1), p_i = i-th prime; a(n) = minimum k such that b(k) > n.

%C A001611 is similar but strictly different.

%C Equal to A256968 except for n = 2 and n = 3. See comment in A256968. - _Chai Wah Wu_, Apr 17 2015

%C a(n) appears to be the same as A005579(n) for n > 2. - _Georg Fischer_, Oct 09 2018

%H Popular Computing (Calabasas, CA), <a href="/A256968/a256968.png">Problem 182 (Suggested by Victor Meally)</a>, Annotated and scanned copy of page 10 of Vol. 5 (No. 53, Aug 1977).

%e The b(n) sequence for n >= 0 begins 1, 2, 3, 15/4, 35/8, 77/16, 1001/192, 17017/3072, 323323/55296, 676039/110592, 2800733/442368, 86822723/13271040, 3212440751/477757440, 131710070791/19110297600, 5663533044013/802632499200, ... = A060753/A038110. So a(3) = 3.

%o (Python)

%o from sympy import prime

%o A256969_list, count, bn, bd = [], 0, 1, 1

%o for k in range(1,10**4):

%o ....p = prime(k)

%o ....bn *= p

%o ....bd *= p-1

%o ....while bn > count*bd:

%o ........A256969_list.append(k)

%o ........count += 1 # _Chai Wah Wu_, Apr 17 2015

%Y Cf. A001611, A005579, A060753, A038110, A256968.

%K nonn,more

%O 0,3

%A _N. J. A. Sloane_, Apr 17 2015

%E More terms from _Chai Wah Wu_, Apr 17 2015

%E a(32)-a(33) from _Chai Wah Wu_, Apr 19 2015