A147537 Numbers whose binary representation is the concatenation of 2n-1 digits 1 and n digits 0.

2, 28, 248, 2032, 16352, 131008, 1048448, 8388352, 67108352, 536869888, 4294965248, 34359734272, 274877898752, 2199023239168, 17592186011648, 140737488289792, 1125899906711552, 9007199254478848, 72057594037403648, 576460752302374912, 4611686018425290752

Offset: 1

Author

Omar E. Pol, Nov 06 2008

Keywords

• base
• easy
• nonn

Comments

a(n) is the number whose binary representation is A138118(n).

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (10,-16).

Crossrefs

Cf. A138118.

Programs

GAP

List([1..20], n-> 2^n*(2^(2*n-1)-1)); # G. C. Greubel, Jan 12 2020

Magma

[2^n*(2^(2*n-1)-1): n in [1..20]] // G. C. Greubel, Jan 12 2020

Maple

seq(2^n*(2^(2*n-1)-1), n = 1..20); # G. C. Greubel, Jan 12 2020

Mathematica

Table[FromDigits[Join[Table[1, {2n - 1}], Table[0, {n}]], 2], {n, 1, 20}] (* Stefan Steinerberger, Nov 11 2008 *)

PARI

vector(20, n, 2^n*(2^(2*n-1)-1)) \\ G. C. Greubel, Jan 12 2020

Python

def a(n): return ((1 << (2*n-1)) - 1) << n
print([a(n) for n in range(1, 22)]) # Michael S. Branicky, Feb 28 2021

Sage

[2^n*(2^(2*n-1)-1) for n in (1..20)] # G. C. Greubel, Jan 12 2020

Formula

From Colin Barker, Nov 04 2012: (Start)

a(n) = 2^(n-1)*(4^n - 2) = 2*A147590(n).

a(n) = 10*a(n-1) - 16*a(n-2).

G.f.: 2*x*(1+4*x)/((1-2*x)*(1-8*x)). (End)