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%I #29 Oct 13 2017 21:38:56
%S 1,1,1,2,4,6,9,14,22,35,55,89,142,230,373,609,996,1637,2698,4461,7398,
%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 A256969 except for n = 2 and n = 3. The following argument shows that they are equal for n > 3. First note that b(k+1) > b(k). Next, Product_{i=1..k} p_i is 2 times an odd number, i.e., it is not divisible by 4. Similarly since p_i - 1 is even for i > 1, Product_{i=1..k} (p_i - 1) is divisible by 2^(k-1), i.e., it is divisible by 4 for k >= 3. Thus b(k) is not an integer for k >= 3. Since b(3) = 15/4 > 3, this means that A256969(n) = A256968(n) for n > 3 - _Chai Wah Wu_, Apr 17 2015
%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) = 2.
%o (Python)
%o from sympy import prime
%o A256968_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 ........A256968_list.append(k)
%o ........count += 1 # _Chai Wah Wu_, Apr 17 2015
%Y Cf. A060753, A038110, A256969, A001611.
%K nonn,more
%O 0,4
%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