A050488 a(n) = 3*(2^n-1) - 2*n.

0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821, 25769803707

Offset: 0

Author

James Sellers, Dec 26 1999

Keywords

• nonn
• easy

Comments

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).

Number of subgroups of C_(2^n) X C_(2^n) (see A060724).

Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008

Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010

(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012

The partial sums of A033484. - J. M. Bergot, Oct 03 2012

Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012

Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013

Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015

Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.
• Tamas Lengyel, On p-adic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 73-94.
• Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Crossrefs

A050487(2^m-1).

Equals (1/2) A051667.

Cf. A000079, A000225, A000295, A005408, A008292, A033484, A054852, A060724, A066810, A125165, A126277, A142963, A156919, A156920, A156925.

Programs

GAP

List([0..30],n->3*(2^n-1)-2*n); # Muniru A Asiru, Oct 26 2018

Haskell

a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list)
-- Reinhard Zumkeller, Jul 24 2015

Magma

[3*(2^n-1) - 2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018

Maple

seq(coeff(series(x*(x+1)/((1-x)^2*(1-2*x)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018

Mathematica

Table[3(2^n-1)-2n,{n,0,30}] (* or *) LinearRecurrence[{4,-5,2}, {0,1,5}, 40] (* Harvey P. Dale, Apr 09 2018 *)

PARI

a(n)=3*(2^n-1)-2*n \\ Charles R Greathouse IV, Sep 24 2015

Python

for n in range(0, 30): print(3*(2**n-1) - 2*n, end=', ') # Stefano Spezia, Oct 27 2018

Formula

Row sums of A125165: (1, 5, 15, 37, ...). Binomial transform of [1, 4, 6, 6, 6, ...] = [1, 5, 15, 37, ...]. 4th diagonal from the right of A126777 = (1, 5, 15, ...). - Gary W. Adamson, Dec 23 2006

a(n) = 2*a(n-1) + (2n-1). - Gary W. Adamson, Sep 30 2007

From Johannes W. Meijer, Feb 20 2009: (Start)

a(n+1) = A156920(n+1,1).

a(n+1) = A156919(n+1,1)/2^n.

a(n+1) = A142963(n+2,1)/2.

a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.

G.f.: z*(1+z)/((1-z)^2*(1-2*z)).

(End)

a(n) = 2*n + 2*a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010

a(n+1) = Sum_{k=0..n} A000079(k) * A005408(n-k), convolution of the powers of 2 with the odd numbers. - Reinhard Zumkeller, Mar 08 2012

E.g.f.: exp(x)*(3*exp(x) - 2*x - 3). - Stefano Spezia, May 15 2023