A126277 -id:A126277 - cp's OEIS frontend

A124671 Row sums of A126277 = binomial transform of (1, 2, 2, 3, 4, 4, 4, ...)

1, 3, 7, 16, 37, 85, 191, 418, 893, 1871, 3863, 7892, 16005, 32297, 64959, 130374, 261309, 523299, 1047415, 2095800, 4192741, 8386813, 16775167, 33552106, 67106237, 134214775, 268432151, 536867228, 1073737733, 2147479121, 4294962303, 8589929102, 17179863165

Offset: 1

Gary W. Adamson, Dec 23 2006

			a(4) = 16 = sums of 4th row terms of A126277: (1 + 4 + 7 + 4).
a(4) = 16 = 1*1 + 3*2 + 3*2 + 1*3.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-14,16,-9,2).

Cf. A126277.

I:=[1,3,7,16,37]; [n le 5 select I[n] else 6*Self(n-1)-14*Self(n-2)+16*Self(n-3)-9*Self(n-4)+2*Self(n-5): n in [1..40]]; // Vincenzo Librandi, Mar 15 2014

CoefficientList[Series[(1 - 3 x + 3 x^2)/((1 - 2 x) (x - 1)^4), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *)

Vec(x*(1-3*x+3*x^2)/((1-2*x)*(x-1)^4) + O(x^100)) \\ Colin Barker, Mar 13 2014

G.f.: x*(1-3*x+3*x^2)/((1-2*x)*(x-1)^4). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 14 2009; corrected by R. J. Mathar, Sep 16 2009
a(n) = 6*a(n-1) - 14*a(n-2) + 16*a(n-3) - 9*a(n-4) + 2*a(n-5) for n > 4. - Vincenzo Librandi, Mar 15 2014
a(n) = -2 + 2^(1+n) - (5*n)/6 - n^3/6. - Colin Barker, Jul 21 2017

More terms from Colin Barker, Mar 13 2014

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

Original entry on oeis.org

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

James Sellers, Dec 26 1999

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

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).

A050487(2^m-1).
Equals (1/2) A051667.
Cf. A000079, A000225, A000295, A005408, A008292, A033484, A054852, A060724, A066810, A125165, A126277, A142963, A156919, A156920, A156925.

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

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

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

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

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

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

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

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

A095264 a(n) = 2^(n+2) - 3*n - 4.

Original entry on oeis.org

1, 6, 19, 48, 109, 234, 487, 996, 2017, 4062, 8155, 16344, 32725, 65490, 131023, 262092, 524233, 1048518, 2097091, 4194240, 8388541, 16777146, 33554359, 67108788, 134217649, 268435374, 536870827, 1073741736, 2147483557, 4294967202, 8589934495, 17179869084, 34359738265

Offset: 1

Gary W. Adamson, May 31 2004

A sequence derived from a 3rd-order matrix generator.
The number of positive 3-strand braids of degree at most n. - R. J. Mathar, May 04 2006
Define a triangle T by T(n,n) = n*(n+1)/2, T(n,1) = n*(n-1) + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c). Its rows are 1; 3,3; 7,6,6; 13,13,12,10; 21,26,25,22,15; etc. The sum of the terms in the n-th row is a(n). - J. M. Bergot, May 03 2013

			a(5) = 109 = 2^7 - 3*5 - 4.
a(5) = 109 since M^5 * [1 0 0] = [1 5 109].
a(7) = 487 = 4*234 - 5*109 + 2*48.

P. Dehornoy, Combinatorics of normal sequences of braids, arXiv:math/0511114 [math.CO], 2005.
Shishuo Fu, Zhicong Lin, and Yaling Wang, Refined Wilf-equivalences by Comtet statistics, arXiv:2009.04269 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A036563, A125232, A126277.

a[n_] := (MatrixPower[{{1, 0, 0}, {1, 1, 0}, {1, 3, 2}}, n].{{1}, {0}, {0}})[[3, 1]]; Table[ a[n], {n, 30}] (* Robert G. Wilson v, Jun 05 2004 *)
Table[2^(n+2)-3n-4,{n,40}] (* or *) LinearRecurrence[{4,-5,2},{1,6,19},40] (* Harvey P. Dale, Sep 24 2021 *)

Let M = [1 0 0 / 1 1 0 / 1 3 2], then M^n * [1 0 0] = [1 n a(n)]. The characteristic polynomial of M is x^3 - 4*x^2 + 5*x - 2.
a(n+3) = 4*a(n+2) - 5*a(n+1) + 2*a(n).
a(n) = Sum_{i=2..n+1} A036563(i) [A036563 is 2^i-3]. - Gerald McGarvey, Jun 28 2004
Row sums of A125232; 5th diagonal from the right of A126277; binomial transform of [1, 5, 8, 8, 8, ...]. - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (3n-2). - Gary W. Adamson, Sep 30 2007
G.f.: -x*(1+2*x)/((2*x-1)*(x-1)^2). - R. J. Mathar, Nov 18 2007
E.g.f.: exp(x)*(4*exp(x) - 3*x - 4). - Elmo R. Oliveira, Apr 01 2025

Edited, corrected and extended by Robert G. Wilson v, Jun 05 2004

More terms from Elmo R. Oliveira, Apr 01 2025

A126284 a(n) = 52^n - 4n - 5.

Original entry on oeis.org

1, 7, 23, 59, 135, 291, 607, 1243, 2519, 5075, 10191, 20427, 40903, 81859, 163775, 327611, 655287, 1310643, 2621359, 5242795, 10485671, 20971427, 41942943, 83885979, 167772055, 335544211, 671088527, 1342177163, 2684354439

Offset: 1

Gary W. Adamson, Dec 24 2006

Row sums of A125233.
A triangle with left and right borders being the odd numbers 1,3,5,7,... will give the same partial sums for the sum of its rows. - J. M. Bergot, Sep 29 2012
The triangle in the above comment is constructed the same way as Pascal's triangle, i.e., C(n, k) = C(n-1, k) + C(n-1, k-1). - Michael B. Porter, Oct 03 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000384, A125233, A126277.

List([1..30],n->5*2^n-4*n-5); # Muniru A Asiru, Oct 24 2018

[5*2^n - 4*n - 5: n in [1..30]]; // G. C. Greubel, Oct 23 2018

A126284:=n->5*2^n-4*n-5; seq(A126284(n), n=1..50); # Wesley Ivan Hurt, Mar 27 2014

CoefficientList[Series[(1 + 3 x)/(1 - 4 x + 5 x^2 - 2 x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 28 2014 *)

a(n)=5<Charles R Greathouse IV, Oct 03 2012

a(1) = 1; a(2) = 7; a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3), n > 2.
The 6th diagonal from the right of A126277.
G.f.: x*(1+3*x)/(1-4*x+5*x^2-2*x^3). - Colin Barker, Feb 12 2012
E.g.f.: 5*exp(2*x) - (5+4*x)*exp(x). - G. C. Greubel, Oct 23 2018

More terms from Vladimir Joseph Stephan Orlovsky, Oct 18 2008

New definition from R. J. Mathar, Sep 29 2012

cp's OEIS Frontend

A124671 Row sums of A126277 = binomial transform of (1, 2, 2, 3, 4, 4, 4, ...)

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

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A095264 a(n) = 2^(n+2) - 3*n - 4.

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A126284 a(n) = 5*2^n - 4*n - 5.

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

A126284 a(n) = 52^n - 4n - 5.