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%I #21 Feb 23 2021 05:27:25
%S 3,5,7,15,11,13,29,17,19,25,23,35,53,29,31,51,103,37,191,41,43,69,47,
%T 65,101,53,81,87,59,61,311,85,67,137,71,73,149,229,79,123,83,129,173,
%U 89,181,141,283,97,197,101,103,159,107,109,121,113,229,177,709,143
%N a(n) = A061026(2n): smallest k such that 2n divides phi(k), phi = A000010.
%C A061026(n) = A061026(2n) for odd n > 1 since phi(m) is even for m >= 3. In this sequence the redundant values are omitted.
%C We have the obvious inequality A070846(n) >= A307437(n) >= a(n). For odd p = prime(k), A307437(p) = a(p), and if A341861(k) > 0 we have A070846(p) = a(p).
%C The smallest n such that A070846(n) > A307437(n) > a(n) is n = 40, where A070846(40) = 241, A307437(40) = 187 and a(40) = 123.
%H Jianing Song, <a href="/A341845/b341845.txt">Table of n, a(n) for n = 1..20000</a>
%e a(12) = 35 since phi(35) = 24 is divisible by 2*12, and there is no m < 12 such that phi(m) is divisible by 2*12.
%e a(16) = 51 since phi(51) = 32 is divisible by 2*16, and there is no m < 16 such that phi(m) is divisible by 2*16.
%o (PARI) a(n) = for(m=1, (2*n)^2, if(eulerphi(m)%(2*n)==0, return(m)))
%o (Python)
%o from sympy import totient as phi
%o def a(n):
%o k = 1
%o while phi(k)%(2*n) != 0: k += 1
%o return k
%o print([a(n) for n in range(1, 61)]) # _Michael S. Branicky_, Feb 21 2021
%Y Cf. A061026, A000010, A070846, A307437, A341861.
%K nonn
%O 1,1
%A _Jianing Song_, Feb 21 2021
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