

A050488


a(n) = 3*(2^n1)  2*n.


23



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
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OFFSET

0,3


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) = 127 = 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(r1,c1) + T(r1,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(n1,2) for n > 0. Also, 4th NWSE 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
Tamas Lengyel, On padic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 7394.
Index entries for linear recurrences with constant coefficients, signature (4,5,2).


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(n1) + (2n1).  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(n1)  5a(n2) + 2a(n3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1z)^2*(12*z)).
(End)
a(n) = 2*n + 2*a(n1)  1 (with a(0)=0).  Vincenzo Librandi, Aug 06 2010
a(n+1) = Sum_{k=0..n} A000079(k) * A005408(nk), convolution of the powers of 2 with the odd numbers.  Reinhard Zumkeller, Mar 08 2012


MAPLE

seq(coeff(series(x*(x+1)/((1x)^2*(12*x)), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018


MATHEMATICA

Table[3(2^n1)2n, {n, 0, 30}] (* or *) LinearRecurrence[{4, 5, 2}, {0, 1, 5}, 40] (* Harvey P. Dale, Apr 09 2018 *)


PROG

(Haskell)
a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list)
 Reinhard Zumkeller, Jul 24 2015
(PARI) a(n)=3*(2^n1)2*n \\ Charles R Greathouse IV, Sep 24 2015
(MAGMA) [3*(2^n1)  2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018
(GAP) List([0..30], n>3*(2^n1)2*n); # Muniru A Asiru, Oct 26 2018
(Python) for n in range(0, 30): print(3*(2**n1)  2*n, end=', ') # Stefano Spezia, Oct 27 2018


CROSSREFS

A050487(2^m1).
Equals (1/2) A051667.
Cf. A000225, A054852, A126277, A125165, A156925, A000079, A005408, A008292, A000295.
Sequence in context: A213487 A005491 A348780 * A142964 A188282 A014316
Adjacent sequences: A050485 A050486 A050487 * A050489 A050490 A050491


KEYWORD

nonn,easy


AUTHOR

James A. Sellers, Dec 26 1999


STATUS

approved



