A215172 a(0)=1, a(n) = a(n-1)*4^n + 2^n - 1. That is, add n 0's and n 1's to the binary representation of previous term.

1, 5, 83, 5319, 1361679, 1394359327, 5711295803455, 93573870443806847, 6132457173405325525247, 1607586853265165654490350079, 1685676992249374341322873324438527, 7070241751299519797307892876185811552255

Offset

0,2

Comments

Binary representations:

a(0) 1

a(1) 101

a(2) 1010011

a(3) 1010011000111

a(4) 101001100011100001111

a(5) 1010011000111000011110000011111

a(6) 1010011000111000011110000011111000000111111

a(7) 101001100011100001111000001111100000011111100000001111111

a(8) 1010011000111000011110000011111000000111111000000011111110000000011111111

Links

Harvey P. Dale, Table of n, a(n) for n = 0..57

Formula

a(0)=1, a(n) = a(n-1)*4^n + 2^n - 1.

Mathematica

nxt[{n_, a_}]:={n+1, FromDigits[Join[IntegerDigits[a], PadRight[{}, n, 0], PadRight[ {}, n, 1]]]}; FromDigits[IntegerDigits[#], 2]&/@NestList[nxt, {1, 1}, 12][[All, 2]] (* Harvey P. Dale, Apr 30 2019 *)

Prog

(Python)
a = 1
for n in range(1, 13):
    #print 'a('+str(n-1)+')', bin(a)[2:], a
    print a,
    a = a*(4**n) + 2**n - 1

Crossrefs

Cf. A076131: add n 0's and one 1 to the binary representation of previous term.

Sequence in context: A250549 A035512 A054953 * A301811 A216146 A111834

Adjacent sequences: A215169 A215170 A215171 * A215173 A215174 A215175

Keyword

nonn,base

Author

Alex Ratushnyak, Aug 05 2012

Status

approved