A014551 -id:A014551 - cp's OEIS frontend

A320933 a(n) = 2^n - floor((n+3)/2).

0, 0, 2, 5, 13, 28, 60, 123, 251, 506, 1018, 2041, 4089, 8184, 16376, 32759, 65527, 131062, 262134, 524277, 1048565, 2097140, 4194292, 8388595, 16777203, 33554418, 67108850, 134217713, 268435441, 536870896

Offset: 0

Paul Curtz, Oct 28 2018

The sequence 0, 0, a(n) is an autosequence of the second kind. The difference table is:
   0,   0,   0,   0,   2,   5,  13, ...
   0,   0,   0,   2,   3,   8,  15, ...
   0,   0,   2,   1,   5,   7,  17, ...
   0,   2,  -1,   4,   2,  10,  14, ...
   2,  -3,   5,  -2,   8,   4,  20, ...
  -5,   8,  -7,  10,  -4,  16,   8, ...
  13, -15,  17, -14,  20,  -8,  32, ...
etc.

Colin Barker, Table of n, a(n) for n = 0..1000
OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,2).

Cf. A000079, A004526, A011377, A014551, A060182, A166920.

List([0..40],n->2^n-Int((n+3)/2)); # Muniru A Asiru, Oct 28 2018

[((-1)^n+2^(n+2)-2*n-5)/4: n in [0..40]]; // G. C. Greubel, Jun 04 2019

seq(2^n-floor((n+3)/2),n=0..40); # Muniru A Asiru, Oct 28 2018

a[n_]:=2^n - Floor[(n+3)/2]; Array[a, 40, 0] (* or *) CoefficientList[ Series[x^2*(2-x)/((1-x)^2*(1-x-2*x^2)), {x, 0, 40}], x] (* Stefano Spezia, Oct 28 2018 *)

concat([0,0], Vec(x^2*(2-x)/((1-x)^2*(1+x)*(1-2*x)) + O(x^40))) \\ Colin Barker, Oct 28 2018

[((-1)^n+2^(n+2)-2*n-5)/4 for n in (0..40)] # G. C. Greubel, Jun 04 2019

a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + a(n-4).
a(n+1) = a(n) + A166920(n).
a(n+4) - a(n) = 13, 28, 58, 118, ... = 15*2^n - 2 = A060182(n+2).
With b(n) = 0, 0, 0, A011377(n) = 0, 0, 0, 1, 3, 8, 18, ..., then a(n) = 2*b(n+1) - b(n).
a(n+2) - 2*a(n+1) + a(n) = A014551(n).
G.f.: x^2*(2 - x)/((1-x)^2*(1 - x - 2*x^2)). - Stefano Spezia, Oct 28 2018
a(n) = ((-1)^n + 2^(n+2) - 2*n - 5) / 4. - Colin Barker, Oct 28 2018

Three terms corrected by Colin Barker, Oct 28 2018

A087452 G.f.: (2-x)/((1+3x)(1-4x)); e.g.f.: exp(4x) + exp(-3x); a(n) = 4^n + (-3)^n.

Original entry on oeis.org

2, 1, 25, 37, 337, 781, 4825, 14197, 72097, 242461, 1107625, 4017157, 17308657, 65514541, 273218425, 1059392917, 4338014017, 17050729021, 69106897225, 273715645477, 1102998412177, 4387586157901, 17623567104025, 70274600998837, 281757406247137

Offset: 0

Paul Barry, Sep 06 2003

Generalized Lucas-Jacobsthal numbers.

Index entries for linear recurrences with constant coefficients, signature (1,12).

Cf. A014551, A053404, A087451.

CoefficientList[Series[(2 - z)/((1 + 3 z) (1 - 4 z)), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 11 2011 *)
LinearRecurrence[{1,12},{2,1},30] (* Harvey P. Dale, Nov 06 2022 *)

polsym(x^2-x-12,50) \\ Charles R Greathouse IV, Jun 11 2011

a(n) = (-3)^n+4^n.
a(n) = a(n-1) + 12*a(n-2) for n > 1; a(0)=2, a(1)=1. - Philippe Deléham, Sep 19 2009
a(n) = 2*A053404(n) - A053404(n-1), n > 0. - Ralf Stephan, Jul 21 2013

A093835 n*Jacobsthal(n).

Original entry on oeis.org

0, 1, 2, 9, 20, 55, 126, 301, 680, 1539, 3410, 7513, 16380, 35503, 76454, 163845, 349520, 742747, 1572858, 3320497, 6990500, 14680071, 30758222, 64312669, 134217720, 279620275, 581610146, 1207959561, 2505397580, 5189752159

Offset: 0

Paul Barry, Apr 20 2004

Convolution of Jacobsthal numbers A001045 and modified Jacobsthal-Lucas numbers (in A014551, change the leading 2 to a 1).

Index entries for linear recurrences with constant coefficients, signature (2,3,-4,-4).

Cf. A023607.

LinearRecurrence[{2,3,-4,-4},{0,1,2,9},30] (* Harvey P. Dale, Jun 17 2017 *)

G.f.: x(1+2x^2)/(1-x-2x^2)^2; a(n)=n2^n/3-n(-1)^n/3.

a(n) = n*A001045(n). - R. J. Mathar, May 07 2019

A094360 Pair reversal of Jacobsthal-Lucas numbers.

Original entry on oeis.org

1, 2, 7, 5, 31, 17, 127, 65, 511, 257, 2047, 1025, 8191, 4097, 32767, 16385, 131071, 65537, 524287, 262145, 2097151, 1048577, 8388607, 4194305, 33554431, 16777217, 134217727, 67108865, 536870911, 268435457, 2147483647, 1073741825

Offset: 0

Paul Barry, Apr 26 2004

Pair reversal of A014551

Index entries for linear recurrences with constant coefficients, signature (-1,4,4).

Cf. A084183, A094359.

LinearRecurrence[{-1,4,4},{1,2,7},40] (* Harvey P. Dale, May 22 2020 *)

G.f. : (1+3x+5x^2)/((1+x)(1-4x^2)); a(n)=2^n(5+3(-1)^n)/4-(-1)^n.

A105225 a(n+3) = 2a(n+2) - 3a(n+1) + 2a(n); a(0) = 1, a(1) = -1, a(2) = -2.

Original entry on oeis.org

1, -1, -2, 1, 6, 5, -6, -15, -2, 29, 34, -23, -90, -43, 138, 225, -50, -499, -398, 601, 1398, 197, -2598, -2991, 2206, 8189, 3778, -12599, -20154, 5045, 45354, 35265, -55442, -125971, -15086, 236857, 267030, -206683, -740742, -327375, 1154110, 1808861, -499358, -4117079, -3118362, 5115797

Offset: 0

Creighton Dement, Apr 14 2005

Floretion Algebra Multiplication Program, FAMP Code: 2tesseq[.5'j + .5'k + .5j' + .5k' + .5'ii' + .5e]

Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (2, -3, 2).

Cf. A002249, A014551.

LinearRecurrence[{2,-3,2},{1,-1,-2},50] (* or *) CoefficientList[ Series[ (-3*x^2+3*x-1)/(2*x^3-3*x^2+2*x-1),{x,0,50}],x] (* Harvey P. Dale, Jul 23 2012 *)

a(n) - a(n+1) = A002249(n).
a(n) = (A002249(n+1) + 1)/2.
From Harvey P. Dale, Jul 23 2012: (Start)
G.f.: -(3*x^2-3*x+1)/((x-1)*(2*x^2-x+1)).
a(n)=1/2*(1+(1/2*(1-I*Sqrt[7]))^n+(1/2*(1+I*Sqrt[7]))^n). (End)

A123265 Fibonacci-Lucas triangle read by rows.

Original entry on oeis.org

2, 1, 2, 3, 3, 2, 4, 8, 5, 2, 7, 15, 15, 7, 2, 11, 30, 35, 24, 9, 2, 18, 56, 80, 66, 35, 11, 2, 29, 104, 171, 170, 110, 48, 13, 2, 47, 189, 355, 407, 315, 169, 63, 15, 2, 76, 340, 715, 932, 832, 532, 245, 80, 17, 2

Offset: 0

Philippe Deléham, Nov 07 2006

As a Riordan array, this is ((2-x)/(1-x-x^2),x/(1-x-x^2)) . Row sums form A078343, diagonals sums form A014551.

			Triangle begins:
2;
1, 2;
3, 3, 2;
4, 8, 5, 2;
7, 15, 15, 7, 2;
11, 30, 35, 24, 9, 2;
18, 56, 80, 66, 35, 11, 2;
29, 104, 171, 170, 110, 48, 13, 2;

Cf. A000032, A099920 ; A007395, A005408, A005563.

T(n,k)=T(n-1,k-1)+T(n-1,k)+T(n-2,k), T(0,0)=2, T(1,0)=1, T(n,k)=0 if k<0 or if k>n . T(n,0)=L(n)=A000032(n),Lucas numbers.

A141775 Binomial transform of (1, 2, 0, 1, 2, 0, 1, 2, 0, ...).

Original entry on oeis.org

1, 3, 5, 8, 15, 31, 64, 129, 257, 512, 1023, 2047, 4096, 8193, 16385, 32768, 65535, 131071, 262144, 524289, 1048577, 2097152, 4194303, 8388607, 16777216, 33554433, 67108865, 134217728, 268435455, 536870911, 1073741824, 2147483649, 4294967297, 8589934592, 17179869183

Offset: 0

Gary W. Adamson, Jul 03 2008

From Paul Curtz, Jun 15 2011: (Start)
A square array of a(n) and its higher order differences is defined by T(0,k) = a(k) and T(n,k) = T(n-1,k+1)-T(n-1,k):
1, 3, 5, 8, 15, 31,
2, 2, 3, 7, 16, 33,
0, 1, 4, 9, 17, 32, see A130785(n).
1, 3, 5, 8, 15, 31,
2, 2, 3, 7, 16, 33,
a(n) is identical to its third differences: T(n+3,k) = T(n,k).
The main diagonal is T(n,n) = 2^n. Subdiagonals are T(n,n-1) = A014551(n) and T(n,n-2) = A062510(n).
(End)

			a(4) = 8 = (1, 2, 0, 1) dot (1, 3, 3, 1) = (1 + 6 + 0 + 1).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Cf. A000079, A057079.

I:=[1,3,5]; [n le 3 select I[n] else 3*Self(n-1) - 3*Self(n-2) + 2*Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 15 2018

LinearRecurrence[{3,-3,2},{1,3,5},40] (* Harvey P. Dale, May 29 2012 *)

x='x+O('x^30); Vec((x-1)*(1+x)/((2*x-1)*(x^2-x+1))) \\ G. C. Greubel, Jan 15 2018

From Paul Curtz, Jun 15 2011: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3).
a(n) = 2^n - A128834(n).
a(n) - 2a(n-1)= A057079(n+1).
a(n) + a(n+3) = 9*2^n.
a(n+6) - a(n) = 63*2^n.
a(n) = A130785(n) - A130785(n-1). (End)
G.f.: (x-1)*(1+x) / ( (2*x-1)*(x^2-x+1) ). - R. J. Mathar, Jun 22 2011
a(n) = 2^n + (2*sin((Pi*n)/3))/sqrt(3). - Colin Barker, Feb 10 2017

A171590 a(n) = 1+4^(n+1)-4*(-2)^n.

Original entry on oeis.org

1, 25, 49, 289, 961, 4225, 16129, 66049, 261121, 1050625, 4190209, 16785409, 67092481, 268468225, 1073676289, 4295098369, 17179607041, 68720001025, 274876858369, 1099513724929, 4398042316801, 17592194433025, 70368727400449

Offset: 0

R. J. Mathar, Dec 12 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,6,-8).

I:=[1, 25, 49]; [n le 3 select I[n] else 3*Self(n-1) + 6*Self(n-2) - 8*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 19 2012

CoefficientList[Series[(1 + 22*x - 32*x^2)/((x - 1)*(2*x + 1)*(4*x-1)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 19 2012 *)

a(n) = A105951(2n+1) = (A014551(n+1))^2.
G.f.: (1+22*x-32*x^2)/((x-1)*(2*x+1)*(4*x-1)).
a(n) = 3*a(n-1)+6*a(n-2)-8*a(n-3).

G.f. Adapted by Vincenzo Librandi, Dec 19 2012

A245962 Triangle read by rows: T(n,k) is the number of induced subgraphs of the Lucas cube Lambda(n) that are isomorphic to the hypercube Q(k).

Original entry on oeis.org

1, 1, 3, 2, 4, 3, 7, 8, 2, 11, 15, 5, 18, 30, 15, 2, 29, 56, 35, 7, 47, 104, 80, 24, 2, 76, 189, 171, 66, 9, 123, 340, 355, 170, 35, 2, 199, 605, 715, 407, 110, 11, 322, 1068, 1410, 932, 315, 48, 2, 521, 1872, 2730, 2054, 832, 169, 13, 843, 3262, 5208, 4396, 2079, 532, 63, 2

Offset: 0

Emeric Deutsch, Aug 14 2014

Number of entries in row n is 1 + floor(n/2).
The entries in row n are the coefficients of the cube polynomial of the Lucas cube Lambda(n).
For n >= 1, sum of entries in row n = A014551(n) = 2^n + (-1)^n (the Jacobsthal-Lucas numbers).
Sum_{k >= 0} k*T(n,k) = A099429(n).
T(n,0) = A000032(n) (n >= 1; the Lucas numbers); T(n,1) = A099920(n-1); T(n,2) = A245961(n).
As communicated by the authors, Theorem 5.2 and Corollary 5.3 of the Klavzar et al. paper contains a typo: 2nd binomial should be binomial(n - a - 1, a) resp. binomial(n - i - 1, i).

			Row 4 is 7, 8, 2. Indeed, the Lucas cube Lambda(4) is the graph <><> obtained by identifying a vertex of a square with a vertex of another square; it has 7 vertices (i.e., Q(0)s), 8 edges (i.e., Q(1)s), and 2 squares (i.e., Q(2)s).
Triangle starts:
   1;
   1;
   3,  2;
   4,  3;
   7,  8,  2;
  11, 15,  5;

Paolo Xausa, Table of n, a(n) for n = 0..10200 (rows 0..200 of the triangle, flattened).
Sandi Klavzar and Michel Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
Sandi Klavzar and Michel Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105.
Jun Wan and Zuo-Ru Zhang, A proof of the only mode of a unimodal sequence, arXiv:2402.12858 [math.CO], 2024.

Cf. A000032, A014551, A099429, A099920, A245961.

T := proc (n, k) options operator, arrow: add((2*binomial(n-i, i)-binomial(n-i-1, i))*binomial(i, k), i = k .. floor((1/2)*n)) end proc: for n from 0 to 20 do seq(T(n, k), k = 0 .. (1/2)*n) end do; # yields sequence in triangular form

A245962[n_, k_] := Sum[(2*Binomial[n-i, i]-Binomial[n-i-1, i])*Binomial[i, k], {i, k, n/2}]; Table[A245962[n, k], {n, 0, 15}, {k, 0, n/2}] (* Paolo Xausa, Feb 29 2024 *)

T(n,k) = Sum_{i = k..floor(n/2)} (2*binomial(n - i, i) - binomial(n - i - 1, i))*binomial(i, k).
G.f.: (1+(1+t)*z^2)/(1-z-(1+t)*z^2).
The generating polynomial of row n (i.e., the cube polynomial of Lambda(n)) is Sum_{i = 0..floor(n/2)} (2*binomial(n - i, i) - binomial(n - i - 1))(1+x)^i.
The generating polynomial of row n (i.e., the cube polynomial of Lambda(n)) is ((1+w)/2)^n + ((1-w)/2)^n, where w = sqrt(5 + 4x).
The generating function of column k (k >= 1) is z^(2k)(2-z)/(1-z-z^2)^(k+1).

A264038 Convolution of Lucas and Jacobsthal numbers.

Original entry on oeis.org

0, 2, 3, 10, 20, 47, 98, 210, 435, 902, 1848, 3775, 7670, 15542, 31403, 63330, 127500, 256367, 514938, 1033450, 2072675, 4154702, 8324528, 16673535, 33386670, 66837422, 133778523, 267724810, 535721060, 1071881327, 2144473298, 4290096450, 8582053395, 17167117142, 34339105128, 68686091455, 137384934950, 274790503142, 549614391563, 1099282801650

Offset: 0

Russell Jay Hendel, Nov 01 2015

The main theorem of the Griffith-Bramham paper found in the LINKS section is the equivalence of the following alternate definitions for a(n). (I) a(n) equals the convolution of the Lucas numbers (A000032) and the Jacobsthal numbers (A001045), where, as usual, the m-th term of the convolution of sequences {b(n)}{n>=0} and {c(n)}{n>-0} equals Sum_{t+s=m} b(t)* c(s). (II) a(n) = A014551(n+1)-A000032(n+1), the difference of the Lucas-Jacobsthal numbers and the Lucas numbers with a shift of 1. The authors prove the equivalence of (I) and (II) using the generating function method.
Referring to the simplicity of definition (II), the authors formulate the following open question: "Since the convolution takes such a simple form, we ask whether it is possible to obtain a purely combinatorial proof of this result."
I would suggest another open question: Are there convolutions of other linear homogeneous recurrences with constant coefficients which are equivalent to very simple forms?

			Let L(n)=A000032(n), j(n)=A014551(n), and J(n)=A001045(n). Then using the convolution definition (I), a(3)=10 because a(3) = L(0)J(3) + L(1)J(2) + L(2)J(1) + L(3)J(0) = 2*3 + 1*1 + 3*1 + 4*0 = 10; similarly, using definition (II) we have a(3) = j(4) - L(4) = 17 - 7 = 10.

Colin Barker, Table of n, a(n) for n = 0..1000
Martin Griffiths and Alex Bramham, The Jacobsthal Numbers: Two Results and Two Questions, Fibonacci Quarterly, Vol. 53, No. 2, May 2015, pp. 147-151.
Tamás Szakács, Convolution of second order linear recursive sequences I., Annales Mathematicae et Informaticae 46 (2016) pp. 205-216.
Index entries for linear recurrences with constant coefficients, signature (2,2,-3,-2).

Equals convolution of Lucas numbers (A000032) and Jacobsthal numbers (A001045); also equals difference of Lucas-Jacobsthal numbers (A014551) minus Lucas numbers (A000032) with a shift of 1.

LinearRecurrence[{1,2},{1,5},40]-LinearRecurrence[{1,1},{1,3},40]
LinearRecurrence[{2,2,-3,-2},{0,2,3,10},50] (* Harvey P. Dale, Dec 11 2016 *)

/* Prints first 40 terms of sequence a(n) */
Lucas(n)={if(n==0,2,if(n==1,1,Lucas(n-1)+Lucas(n-2)));}
j(n)={if(n==0,2,if(n==1,1,j(n-1)+2*j(n-2)));} /*Lucas-Jacobsthal*/
a(n)=j(n+1)-Lucas(n+1);
for(n=0,40,print(a(n)));

concat(0, Vec(-x*(x-2)/((x+1)*(2*x-1)*(x^2+x-1)) + O(x^100))) \\ Colin Barker, Nov 02 2015

a(n) = A014551(n+1)- A000032(n+1).
G.f.: 2/(1-2x)-1/(1+x)-alpha/(1-alpha*x)-beta/(1-beta*x) with alpha=(1+sqrt(5))/2 and beta=-1/alpha.
From Colin Barker, Nov 02 2015: (Start)
a(n) = 2*a(n-1)+2*a(n-2)-3*a(n-3)-2*a(n-4) for n > 3.
G.f.: 2/(1-2x)-1/(1+x)-alpha/(1-alpha*x)-beta/(1-beta*x)=-x*(x-2) / ((x+1)*(2*x-1)*(x^2+x-1)), with alpha = (sqrt(5)+1)/2, and beta=-1/alpha.(End)

cp's OEIS Frontend

A320933 a(n) = 2^n - floor((n+3)/2).

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A087452 G.f.: (2-x)/((1+3x)(1-4x)); e.g.f.: exp(4x) + exp(-3x); a(n) = 4^n + (-3)^n.

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A093835 n*Jacobsthal(n).

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A094360 Pair reversal of Jacobsthal-Lucas numbers.

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A105225 a(n+3) = 2a(n+2) - 3a(n+1) + 2a(n); a(0) = 1, a(1) = -1, a(2) = -2.

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A123265 Fibonacci-Lucas triangle read by rows.

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A141775 Binomial transform of (1, 2, 0, 1, 2, 0, 1, 2, 0, ...).

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A171590 a(n) = 1+4^(n+1)-4*(-2)^n.