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%I #15 May 13 2013 01:49:38
%S 11,155,2164,33723,539379,8396540,136109403,2147745531,34360623100,
%T 549771505659,8797030442667,140737513521148,2251823188540923,
%U 36028801313906427,576460760876579772
%N Least k with 2n divisors such that k-1 and k+1 in binary representation have same number 2n of 0's as 1's.
%C Does a(n) exist for every n? It seems plausible at first glance; asymptotically there should be enough numbers in the range 16^n * [1/2, 1] that have 2n divisors (since 16 > e). [_Charles R Greathouse IV_, Jun 05 2011]
%C a(16) <= 9223372071079772155. - _Donovan Johnson_, Sep 25 2011
%o (PARI) a(n)=my(v=vector(4*n,i,i>2*n));for(k=1<<(4*n-1)+1<<(2*n-1)-1,1<<(4*n)-1<<(2*n),if(vecsort(binary(k-1))==v & vecsort(binary(k+1))==v & numdiv(k)==2*n, return(k))) \\ _Charles R Greathouse IV_, Jun 05 2011
%Y Cf. A000005, A191292, A191369.
%K nonn,base
%O 1,1
%A _Juri-Stepan Gerasimov_, Jun 05 2011
%E a(5)-a(11) from _Charles R Greathouse IV_, Jun 05 2011
%E a(12)-a(15) from _Donovan Johnson_, Sep 25 2011
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