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

1,1

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

Links

Table of n, a(n) for n=1..15.

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

Cf. A000005, A191292, A191369.

Sequence in context: A191369 A223713 A223779 * A243947 A156933 A242008

Adjacent sequences: A191531 A191532 A191533 * A191535 A191536 A191537

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