A191534 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.

11, 155, 2164, 33723, 539379, 8396540, 136109403, 2147745531, 34360623100, 549771505659, 8797030442667, 140737513521148, 2251823188540923, 36028801313906427, 576460760876579772

Offset: 1

Author

Juri-Stepan Gerasimov, Jun 05 2011

Keywords

• nonn
• base

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

Crossrefs

Cf. A000005, A191292, A191369.

Programs

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

Extensions

a(5)-a(11) from Charles R Greathouse IV, Jun 05 2011

a(12)-a(15) from Donovan Johnson, Sep 25 2011