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

0,3

Links

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

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).

Example

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.

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 *)

Prog

(Python)
a = 0
for n in range(1, 10):
    #print 'a('+str(n-1)+')', bin(a)[2:], a
    print a,
    a = a*(2**(n+1)) + 2**n - 1

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.

Sequence in context: A068648 A209351 A287062 * A034787 A001408 A298643

Adjacent sequences: A215200 A215201 A215202 * A215204 A215205 A215206

Keyword

nonn,base,easy

Author

Alex Ratushnyak, Aug 05 2012

Status

approved