OFFSET
1,1
COMMENTS
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]
a(16) <= 9223372071079772155. - Donovan Johnson, Sep 25 2011
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Juri-Stepan Gerasimov, Jun 05 2011
EXTENSIONS
a(5)-a(11) from Charles R Greathouse IV, Jun 05 2011
a(12)-a(15) from Donovan Johnson, Sep 25 2011
STATUS
approved