A215203 a(0) = 0, a(n) = a(n - 1)*2^(n + 1) + 2^n - 1. That is, add one 0 and n 1's to the binary representation of previous term.

0, 1, 11, 183, 5871, 375775, 48099263, 12313411455, 6304466665215, 6455773865180671, 13221424875890015231, 54154956291645502388223, 443637401941159955564326911, 7268555193403964711965932118015, 238176016577461115681699663643131903

Offset: 0

Author

Alex Ratushnyak, Aug 05 2012

Keywords

• nonn
• base
• easy

Examples

			Binary representations:
a(0): 0;
a(1): 1;
a(2): 1011;
a(3): 10110111;
a(4): 1011011101111;
a(5): 1011011101111011111;
a(6): 10110111011110111110111111;
a(7): 1011011101111011111011111101111111;
a(8): 1011011101111011111011111101111111011111111, etc.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..80

Crossrefs

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

Cf. A215172: add n 0's and n 1's to the binary representation of previous term.

Programs

Mathematica

nxt[{n_,a_}]:={n+1,FromDigits[Join[IntegerDigits[a,2],PadRight[{0},n+2,1]],2]}; NestList[nxt,{0,0},15][[All,2]] (* Harvey P. Dale, Feb 11 2023 *)

Python

a = 0
for n in range(1, 10):
    print(a, end=', ')
    a = a*(2**(n+1)) + 2**n - 1

Formula

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

a(n)*2 + A076131(n+1) + 1 = 2^A000217(n+1).