A042950 Row sums of the Lucas triangle A029635.

2, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472, 6442450944

Offset: 0

Author

N. J. A. Sloane and J. H. Conway

Keywords

• nonn
• easy

Comments

Map a binary sequence b=[ b_1,...] to a binary sequence c=[ c_1,...] so that C = 1/Product((1-x^i)^c_i == 1 + Sum b_i*x^i mod 2.

This produces 2 new sequences: d={i:c_i=1} and e=[ 1,e_1,... ] where C = 1 + Sum e_i*x^i.

This sequence is d when b=[ 0,1,1,1,1,...].

Number of rises after n+1 iterations of morphism A007413.

a(n) written in base 2: a(0) = 10, a(n) for n >= 1: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, (n-1) times 0 (see A003953(n)). - Jaroslav Krizek, Aug 17 2009

Row sums of the Lucas triangle A029635. - Sergio Falcon, Mar 17 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• S. Kitaev and T. Mansour, Counting the occurrences of generalized patterns in words generated by a morphism, arXiv:math/0210170 [math.CO], 2002.
• Index entries for linear recurrences with constant coefficients, signature (2).

Crossrefs

Cf. A003945, A007283.

Programs

Magma

[2] cat [2^(n+1) - 2^(n-1): n in [1..40]]; // Vincenzo Librandi, Aug 08 2015

Mathematica

Table[ Ceiling[3*2^(n - 1)], {n, 0, 32}] (* Robert G. Wilson v, Jul 08 2006 *)
a[0] = 2; a[1] = 3; a[n_] := 2a[n - 1]; Table[a[n], {n, 0, 32}] (* Robert G. Wilson v, Jul 08 2006 *)
f[s_] := Append[s, 1 + Plus @@ s]; Nest[f, {2}, 32] (* Robert G. Wilson v, Jul 08 2006 *)
CoefficientList[Series[(2 - x)/(1 - 2x), {x, 0, 32}], x] (* Robert G. Wilson v, Jul 08 2006 *)

PARI

a(n)=ceil(3*2^(n-1))

Python

def A042950(n): return (3*2**n + int(n==0))//2 # G. C. Greubel, Jun 06 2025

Formula

G.f.: (2-x)/(1-2*x).

a(n) = 2*a(n-1), n > 1; a(0)=2, a(1)=3.

a(n) = A003945(n), for n > 0.

From Paul Barry, Dec 06 2004: (Start)

Binomial transform of 2, 1, 2, 1, 2, 1, ... = (3+(-1)^n)/2.

a(n) = (3*2^n + 0^n)/2. (End)

a(0) = 2, a(n) = 3*2^(n-1) = 2^n + 2^(n-1) for n >= 1. - Jaroslav Krizek, Aug 17 2009

a(n) = 2^(n+1) - 2^(n-1), for n > 0. - Ilya Gutkovskiy, Aug 08 2015

E.g.f.: (3*exp(2*x) + 1)/2. - G. C. Greubel, Jun 06 2025