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

Author

Alex Ratushnyak, Aug 05 2012

Keywords

• nonn
• base

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

Crossrefs

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

Programs

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 *)

Python

a = 1
for n in range(1,13):
    print(a, end=', ')
    a = a*(4**n) + 2**n - 1

Formula

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