A014334 -id:A014334 - cp's OEIS frontend

A103435 a(n) = 2^n * Fibonacci(n).

0, 2, 4, 16, 48, 160, 512, 1664, 5376, 17408, 56320, 182272, 589824, 1908736, 6176768, 19988480, 64684032, 209321984, 677380096, 2192048128, 7093616640, 22955425792, 74285318144, 240392339456, 777925951488, 2517421260800

Offset: 0

Ralf Stephan, Feb 08 2005

Cardinality of set of bracelets of size at most n that are tiled with two types of colored squares and four types of colored dominoes.
a(n) is also the diagonal element of the matrix A(i,j) whose first row (i=1) and first column (j=1) are the Fibonacci numbers: A(1,k)=A(k,1)=fib(k) and whose generic element is the sum of element in adjacent (preceding) row and column minus the absolute value of their difference. So a(n) = A(n,n) = A(i-1,j)+A(i,j-1)-abs(A(i-1,j)-A(i,j-1)). - Carmine Suriano, May 13 2010
a(n) is the coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) given for d=sqrt(x+1) by p(n,x)=((x+d)^n-(x-d)^n)/(2d), for n>=1.  The constant terms under this reduction are the absolute values of terms of A086344.  See A192232 for a discussion of reduction. - Clark Kimberling, Jun 29 2011
The exponential convolution of A000032 and A000045. - Vladimir Reshetnikov, Oct 06 2016

			a(5)=160=A(5,5)=A(4,5)+A(5,4)-abs[A(4,5)+A(5,4)]=80+80-0. - _Carmine Suriano_, May 13 2010
G.f. = 2*x + 4*x^2 + 16*x^3 + 48*x^4 + 160*x^5 + 512*x^6 + 1664*x^7 + ...

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 236, p. 131.

Tom Edgar, Extending Some Fibonacci-Lucas Relations, The Fibonacci Quarterly, Vol. 54, No. 1 (2016), p. 79.
Harris Kwong, An Alternate Proof of Sury's Fibonacci-Lucas Relation, The American Mathematical Monthly, Vol. 121, No. 6 (2014), p. 514.
Diego Marques, A new Fibonacci-Lucas relation, Amer. Math. Monthly, Vol. 122, No. 7 (2015), p. 683.
Ivica Martinjak and Helmut Prodinger, Complementary Families of the Fibonacci-Lucas Relations, arXiv:1508.04949 [math.CO], 2015-2016.
B. Sury, A polynomial parent to a Fibonacci-Lucas relations, Amer. Math. Monthly, Vol. 121, No. 3 (2014), p. 236.
Charles R. Wall, Problem B-607, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 4 (1987), p. 370; Product of Exponential Generating Functions, Solution to Problem B-607 by Bob Prielipp, ibid., Vol. 26, No. 4 (1988), pp. 374-375.
Index entries for linear recurrences with constant coefficients, signature (2,4).

First differences of A014334.
Partial sums of A087131.
Cf. A000032, A000045, A006483, A063727, A085449, A269991.

[2^n *Fibonacci(n): n in [0..50]]; // Vincenzo Librandi, Apr 04 2011

Expand[Table[((1 + Sqrt[5])^n - (1 - Sqrt[5])^n)5/(5 Sqrt[5]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
Table[2^n Fibonacci[n],{n,0,40}] (* or *) LinearRecurrence[{2,4},{0,2},40] (* Harvey P. Dale, Oct 14 2020 *)

a(n)=fibonacci(n)<Charles R Greathouse IV, Feb 03 2014

concat(0, Vec(2*x/(1-2*x-4*x^2) + O(x^99))) \\ Altug Alkan, May 11 2016

a(n) = A006483(n) + 1 = 2*A085449(n) = 2*A063727(n-1), n>0.
G.f.: 2*x / (1 - 2*x - 4*x^2).
a(n) = Sum_{i=0..n-1}( 2^i * Lucas(i) ).
a(n) = 2*a(n-1) + 4*a(n-2). - Carmine Suriano, May 13 2010
a(n) = a(-n) * -(-4)^n for all n in Z. - Michael Somos, Sep 20 2014
E.g.f.: 2*sinh(sqrt(5)*x)*exp(x)/sqrt(5). - Ilya Gutkovskiy, May 10 2016
Sum_{n>=1} 1/a(n) = (1/2) * A269991. - Amiram Eldar, Nov 17 2020
a(n) == 2*n (mod 10). - Amiram Eldar, Jan 15 2022
a(n) = Sum_{k=0..n} binomial(n,k) * Fibonacci(k) * Lucas(n-k) (Wall, 1987). - Amiram Eldar, Jan 27 2022

A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

Original entry on oeis.org

2, 2, 12, 32, 112, 352, 1152, 3712, 12032, 38912, 125952, 407552, 1318912, 4268032, 13811712, 44695552, 144637952, 468058112, 1514668032, 4901568512, 15861809152, 51329892352, 166107021312, 537533612032, 1739495309312

Offset: 0

Paul Barry, Aug 16 2003

Number of ways to tile an n-bracelet with two types of colored squares and four types of colored dominoes.
Inverse binomial transform of even Lucas numbers (A014448).
From L. Edson Jeffery, Apr 25 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,4)=
(0 0 0 0 1)
(0 0 0 2 0)
(0 0 2 0 1)
(0 2 0 2 0)
(2 0 2 0 1).
Then a(n)=(Trace(A^n)-1)/2. Also a(n)=Trace((2*A_(5,1))^n), where A_(5,1)=[(0,1); (1,1)] is also a unit-primitive matrix. (End)
Also the number of connected dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, May 02 2017
Also the number of total dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, Apr 27 2018

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 237, p. 132.

L. Edson Jeffery, Unit-primitive matrices
Eric Weisstein's World of Mathematics, Connected Dominating Set.
Eric Weisstein's World of Mathematics, Sun Graph.
Eric Weisstein's World of Mathematics, Total Dominating Set.
Index entries for linear recurrences with constant coefficients, signature (2,4).

First differences of A006483 and A103435.

Cf. A000032, A014334, A014335, A084057, A269992.

[2] cat [2^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017

Table[Tr[MatrixPower[{{2, 2}, {2, 0}}, x]], {x, 1, 20}] (* Artur Jasinski, Jan 09 2007 *)
Join[{2}, Table[2^n LucasL[n], {n, 20}]] (* Eric W. Weisstein, May 02 2017 *)
Join[{2}, 2^# LucasL[#] & [Range[20]]] (* Eric W. Weisstein, May 02 2017 *)
LinearRecurrence[{2, 4}, {2, 12}, {0, 20}] (* Eric W. Weisstein, Apr 27 2018 *)
CoefficientList[Series[(2 (-1 + x))/(-1 + 2 x + 4 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 27 2018 *)

for(n=0,30, print1(if(n==0, 2, 2^n*(fibonacci(n+1) + fibonacci(n-1))), ", ")) \\ G. C. Greubel, Dec 18 2017

first(n) = Vec(2*(1-x)/(1-2*x-4*x^2) + O(x^n)) \\ Iain Fox, Dec 19 2017

[lucas_number2(n,2,-4) for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009

a(n) = 2*A084057(n).
Recurrence: a(n) = 2a(n-1) + 4a(n-2), a(0)=2, a(1)=2.
G.f.: 2*(1-x)/(1-2*x-4*x^2).
a(n) = (1+sqrt(5))^n + (1-sqrt(5))^n.
For n>=2, a(n) = Trace of matrix [({2,2},{2,0})^n]. - Artur Jasinski, Jan 09 2007
a(n) = 2*[A063727(n)-A063727(n-1)]. - R. J. Mathar, Nov 16 2007
a(n) = (5*A052899(n)-1)/2. - L. Edson Jeffery, Apr 25 2011
a(n) = [x^n] ( 1 + x + sqrt(1 + 2*x + 5*x^2) )^n for n >= 1. - Peter Bala, Jun 23 2015
Sum_{n>=1} 1/a(n) = (1/2) * A269992. - Amiram Eldar, Nov 17 2020
From Amiram Eldar, Jan 15 2022: (Start)
a(n) == 2 (mod 10).
a(n) = 5 * A014334(n) + 2.
a(n) = 10 * A014335(n) + 2. (End)

Edited by Ralf Stephan, Feb 08 2005

A014335 Exponential convolution of Fibonacci numbers with themselves (divided by 2).

Original entry on oeis.org

0, 0, 1, 3, 11, 35, 115, 371, 1203, 3891, 12595, 40755, 131891, 426803, 1381171, 4469555, 14463795, 46805811, 151466803, 490156851, 1586180915, 5132989235, 16610702131, 53753361203, 173949530931, 562912506675, 1821623137075, 5894896300851, 19076285150003

Offset: 0

N. J. A. Sloane

It can be noticed that A014335/A011782 is an "autosequence", that is a sequence which is identical to its inverse binomial transform, except for alternating signs. - Jean-François Alcover, Jun 15 2016

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,2,-4).

Cf. A000032, A000045, A014334, A081057, A203579.
Cf. (partial sums of) A063727.
Column k=2 of A346415.

[(2^n*Lucas(n)-2)/10: n in [0..40]]; // G. C. Greubel, Jan 06 2023

a[0]:=0:a[1]:=0:for n from 2 to 50 do a[n]:=2*a[n-1]+4*a[n-2]+1 od: seq(a[n], n=0..29); # Zerinvary Lajos, Dec 14 2008
# second Maple program:
a:= n-> (<<0|1|0>, <0|0|1>, <-4|2|3>>^n)[1,3]:
seq(a(n), n=0..30);  # Alois P. Heinz, Oct 04 2016

LinearRecurrence[{3,2,-4}, {0,0,1}, 41] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
Table[(2^n LucasL[n] - 2)/10, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 06 2016 *)

[(2^n*lucas_number2(n,1,-1) -2)/10 for n in range(41)] # G. C. Greubel, Jan 06 2023

a(n) = A014334(n)/2.
G.f.: x^2/((1-x)*(1-2*x-4*x^2)). - Vladeta Jovovic, Mar 05 2003
E.g.f.: exp(x)*(cosh(sqrt(5)*x)-1)/5. - Vladeta Jovovic, Sep 01 2004
From Benoit Cloitre, Sep 25 2004: (Start)
a(n+1) = Sum_{i=0..n} A000045(i)*2^(i-1).
a(n) = (1/5)*(2^(n-1)*A000032(n) - 1). (End)
a(n) = 2*a(n-1) + 4*a(n-2) + 1, a(0)=0; a(1)=0. - Zerinvary Lajos, Dec 14 2008
G.f.: G(0)*x^2/(2*(1-x)^2), where G(k)= 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = (A203579(n) - 2)/5. - Vladimir Reshetnikov, Oct 06 2016

A081057 E.g.f.: Sum_{n>=0} a(n)x^n/n! = {Sum_{n>=0} F(n+1)x^n/n!}^2, where F(n) is the n-th Fibonacci number.

Original entry on oeis.org

1, 2, 6, 18, 58, 186, 602, 1946, 6298, 20378, 65946, 213402, 690586, 2234778, 7231898, 23402906, 75733402, 245078426, 793090458, 2566494618, 8305351066, 26876680602, 86974765466, 281456253338, 910811568538, 2947448150426, 9538142575002, 30866077751706

Offset: 0

Paul D. Hanna, Mar 03 2003

nonn

a(n) ~ c*(sqrt(5)+1)^n, where c = (sqrt(5)+3)/10.
The inverse binomial transform is 1,1,3,5,... (1 followed by A056487). Partial sum of 1,1,4,12,..., i.e., 1 plus n-th partial sum of A087206. [R. J. Mathar, Oct 04 2010]
From R. J. Mathar, Oct 12 2010: (Start)
Apparently the row n=4 of an array which counts walks with k steps on an n X n board, starting at a corner, each step to one of the <= 4 adjacent squares:
1,2,4,8,16,32,64,128,256,512,1024,2048,4096,
1,2,6,16,48,128,384,1024,3072,8192,24576,65536,196608,
1,2,6,18,58,186,602,1946,6298,20378,65946,213402,690586,
1,2,6,18,60,198,684,2322,8100,27702,96876,331938,1161540,
1,2,6,18,60,200,698,2432,8658,30762,110374,395428,1422916,
1,2,6,18,60,200,700,2448,8800,31552,115104,418176,1537536,
1,2,6,18,60,200,700,2450,8818,31730,116182,425172,1573416,
1,2,6,18,60,200,700,2450,8820,31750,116400,426600,1583400,
1,2,6,18,60,200,700,2450,8820,31752,116422,426862,1585246,
1,2,6,18,60,200,700,2450,8820,31752,116424,426886,1585556,
1,2,6,18,60,200,700,2450,8820,31752,116424,426888,1585582,
(End)
Decomposing rook walks of length=n on a 4 X 4 board into combinations of independent vertical and horizontal walks in 4-wide corridors leads to an exponential convolution of the Fibonacci numbers, cf. A052899. [David Scambler, Oct 17 2010]

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

a(n) = A052899(n-1) + A052899(n). a(n) - 2*a(n-1) = A014334(n).
Cf. A000032, A000045, A000204.
Row sums of A109906.

G.f.: (1-x-2x^2)/(1-3x-2x^2+4x^3). - Michael Somos, Mar 04 2003
a(n) - 2*a(n-1) = A014334(n), n > 0. - Vladeta Jovovic, Mar 05 2003
From Vladeta Jovovic, Mar 05 2003: (Start)
a(n) = 2/5 + (3/10 - 1/10*5^(1/2))*(1 - 5^(1/2))^n + (3/10 + 1/10*5^(1/2))*(1 + 5^(1/2))^n.
Recurrence: a(n) = 3*a(n-1) + 2*a(n-2) - 4*a(n-3).
G.f.: (1+x)*(1-2*x)/(1-2*x-4*x^2)/(1-x). (End)
a(n) = Sum_{k=0..n} ( F(k+1) * F(n-k+1) * C(n,k) ), where F(k) = Fibonacci(k). - David Scambler, Oct 17 2010
a(n) = (2^n*Lucas(n+2)+2)/5. - Ira M. Gessel, Mar 06 2022

Corrected and extended by Vladeta Jovovic and Michael Somos, Mar 05 2003

A082987 a(n) = Sum_{k=0..n} 3^k*F(k) where F(k) is the k-th Fibonacci number.

Original entry on oeis.org

0, 3, 12, 66, 309, 1524, 7356, 35787, 173568, 842790, 4090485, 19856568, 96384072, 467861331, 2271040644, 11023873914, 53510987541, 259747827852, 1260842371428, 6120257564955, 29708354037720, 144207380197758

Offset: 0

Benoit Cloitre, May 29 2003

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,6,-9).

Cf. A014334, A082988, A119282.

LinearRecurrence[{4,6,-9},{0,3,12},30] (* Harvey P. Dale, Feb 03 2019 *)

a(n)=if(n<0,0,sum(k=0,n,fibonacci(k)*3^k))

a(0)=0, a(1)=3, a(2)=12, a(n)=4a(n-1)+6a(n-2)-9a(n-3).

G.f.: 3*x / ((x-1)*(9*x^2+3*x-1)). - Colin Barker, Jun 26 2013

Offset changed to 0 by Seiichi Manyama, Oct 03 2023

A082988 a(n) = Sum_{k=0..n} 4^k*F(k) where F(k) is the k-th Fibonacci number.

Original entry on oeis.org

0, 4, 20, 148, 916, 6036, 38804, 251796, 1628052, 10540948, 68212628, 441505684, 2857424788, 18493790100, 119693957012, 774676469652, 5013809190804, 32450060277652, 210021188163476, 1359285717096340, 8797481879000980

Offset: 0

Benoit Cloitre, May 29 2003

More generally for any complex number z, the sequence a(n) = Sum_{k=0..n} z^k*F(k) satisfies the recurrence: a(0) = 0, a(1) = z, a(2) = z(z+1), for n > 2 a(n) = (z+1)*a(n-1)+z*(z-1)*a(n-2)-z^2*a(n-3).

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,12,-16).

Cf. A000045, A014334, A082987, A119282.

LinearRecurrence[{5, 12, -16}, {0, 4, 20}, 21] (* Amiram Eldar, Apr 29 2025 *)

a(n)=if(n<0,0,sum(k=0,n,fibonacci(k)*4^k));

a(0) = 0, a(1) = 4, a(2) = 20, a(n) = 5a(n-1)+12a(n-2)-16a(n-3).

O.g.f.: 4*x/((x-1)*(16*x^2+4*x-1)). - R. J. Mathar, Dec 05 2007

A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 3, 1, 2, 4, 5, 1, 3, 6, 9, 16, 1, 4, 9, 15, 25, 39, 1, 5, 13, 24, 40, 64, 105, 1, 6, 18, 37, 64, 104, 169, 272, 1, 7, 24, 55, 101, 168, 273, 441, 715, 1, 8, 31, 79, 156, 269, 441, 714, 1156, 1869, 1, 9, 39, 110, 235, 425, 710, 1155, 1870, 3025, 4896

Offset: 0

Johannes W. Meijer, Oct 19 2013

Triangle T(n, k) is related to the Kn1p sums of the ‘Races with Ties’ triangle A035317. See A230447 for the Kn1p sums and A180662 for the definitions of these sums.
The row sums equal ((-1)^n*3*A083581(n) + A022379(2*n+2))/15.
Note that the partial fraction expansion of the G.f. of the terms in the n-th row of the square array Tsq(n, k) = T(n+k, k) is related to A014334, the exponential convolution of the Fibonacci numbers with themselves, and to A000032, the Lucas numbers.

			The first few rows of triangle T(n, k), n >= 0 and 0 <= k <= n.
n/k 0   1   2    3    4     5     6     7
------------------------------------------------
0|  1
1|  1,  0
2|  1,  1,  3
3|  1,  2,  4,   5
4|  1,  3,  6,   9,  16
5|  1,  4,  9,  15,  25,   39
6|  1,  5, 13,  24,  40,   64,  105
7|  1,  6, 18,  37,  64,  104,  169,   272
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
n/k 0   1   2    3    4    5      6     7
------------------------------------------------
0|  1,  0,  3,   5,  16,  39,   105,  272
1|  1,  1,  4,   9,  25,  64,   169,  441
2|  1,  2,  6,  15,  40,  104,  273,  714
3|  1,  3,  9,  24,  64,  168,  441, 1155
4|  1,  4, 13,  37, 101,  269,  710, 1865
5|  1,  5, 18,  55, 156,  425, 1135, 3000
6|  1,  6, 24,  79, 235,  660, 1795, 4795
7|  1,  7, 31, 110, 345, 1005, 2800, 7595

Cf. (Triangle columns) A000012, A001477, A152950, A226205, A007598, A001654, A064831, A080145

T := proc(n, k) option remember: if k=0 then return(1) elif k=n then return(combinat[fibonacci](n+2)*combinat[fibonacci](n-1)) else procname(n-1, k-1) + procname(n-1, k) fi: end: seq(seq(T(n, k), k=0..n), n=0..10); # End first program.
T := proc(n, k): add(A035317(n+k-p-2, p), p=0..k) end: A035317 := proc(n, k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(T(n, k), k=0..n), n=0..10); # End second program.

T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = F(n+2) * F(n-1) = A226205(n+1) with F(n) = A000045(n), the Fibonacci numbers, n >= 0 and 0 <= k <= n.
T(n, k) = sum(A035317(n+k-p-2, p), p=0..k), n >= 0 and 0 <= k <= n.
T(n+p+2, p-2) = A080239(n+2*p-1) - sum(A035317(n-k+p-1, k+p-1), k=0..floor(n/2)), n >= 0 and p >= 2.
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
Tsq(n, k) = sum(Tsq(n-1, i), i=0..k), n >= 1 and k >= 0, with Tsq(0, k) = A226205(k+1).
The two G.f.’s given below generate the terms in the n-th row of the square array Tsq(n, k). The remarkable second G.f. is the partial fraction expansion of the first G.f..
G.f.: 1/((1-x)^(n-2)*(1+x)*(x^2-3*x+1)), n >= 0.
G.f.: sum((-1)^(n+k-1)*A014334(k+2)/(2^(k+2)*(x-1)^(n-k-2)), k=0..n-3) + 1/(5*2^(n-2)*(1+x)) + (A000032(n+1) - A000032(n-1)*x)/(5*(x^2-3*x+1)), n >= 0.

A277220 Exponential convolution of Fibonacci (A000045) and Catalan (A000108) numbers.

Original entry on oeis.org

0, 1, 3, 11, 43, 180, 790, 3590, 16745, 79705, 385615, 1890747, 9375216, 46931897, 236873261, 1204089630, 6159064015, 31678706490, 163739008070, 850051218980, 4430529313065, 23175017046351, 121617754070653, 640122809255716, 3378402106118508, 17875011275340275

Offset: 0

Vladimir Reshetnikov, Oct 06 2016

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A277251, A000045, A000108, A014330, A014334, A090826.

[(&+[Binomial(n,k)*Fibonacci(k)*Catalan(n-k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Oct 22 2018

Table[Sum[Binomial[n, k] Fibonacci[k] CatalanNumber[n - k], {k, 0, n}], {n, 0, 30}] (* or *)
Round@Table[(GoldenRatio^n Hypergeometric2F1[1/2, -n, 2, -4/GoldenRatio] - (-GoldenRatio)^(-n) Hypergeometric2F1[1/2, -n, 2, 4 GoldenRatio])/Sqrt[5], {n, 0, 30}] (* Round is equivalent to FullSimplify here, but is much faster *)

for(n=0, 30, print1(sum(k=0,n, binomial(n,k)*fibonacci(k)* binomial(2*n-2*k,n-k)/(n-k+1)), ", ")) \\ G. C. Greubel, Oct 22 2018

a(n) = Sum_{k=0..n} binomial(n,k) * A000045(k) * A000108(n-k).
a(n) = (phi^n * hypergeom([1/2, -n], [2], -4/phi) - (-phi)^(-n) * hypergeom([1/2, -n], [2], 4*phi))/sqrt(5), where phi = (1+sqrt(5))/2 = A001622.
Recurrence: 19*(n+1)*(n+2)*(11*n+13)*a(n) + 2*(55*n^3+208*n^2+311*n+230)*a(n+1) + 2*(55*n^3+373*n^2+674*n+206)*a(n+3) = (n+2)*(297*n^2+1022*n+617)*a(n+2) + (n+3)*(n+5)*(11*n+2)*a(n+4).
E.g.f.: 2*exp(5*x/2)*sinh(x*sqrt(5)/2)*(BesselI_0(2*x) - BesselI_1(2*x))/sqrt(5) (the product of e.g.f. for Fibonacci and Catalan numbers).
a(n) ~ (phi + 4)^(n + 3/2) / (8 * sqrt(5*Pi) * n^(3/2)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 10 2018

A277251 Exponential convolution of Lucas (A000032) and Catalan (A000108) numbers.

Original entry on oeis.org

2, 3, 9, 29, 107, 430, 1840, 8230, 38015, 179873, 867079, 4242111, 21006358, 105072063, 530058079, 2693632580, 13775807415, 70847283680, 366167521240, 1900884870494, 9907318315587, 51822028122623, 271949090063769, 1431369293422604, 7554372307564282

Offset: 0

Vladimir Reshetnikov, Oct 06 2016

nonn

Cf. A277220, A000032, A000108, A014330, A014334, A090826.

Table[Sum[Binomial[n, k] LucasL[k] CatalanNumber[n - k], {k, 0, n}], {n, 0,
   30}] (* or *)
Round@Table[GoldenRatio^n Hypergeometric2F1[1/2, -n, 2, -4/GoldenRatio] + (-GoldenRatio)^(-n) Hypergeometric2F1[1/2, -n, 2, 4 GoldenRatio], {n, 0, 30}] (* Round is equivalent to FullSimplify here, but is much faster *)

a(n) = Sum_{k=0..n} binomial(n,k) * A000032(k) * A000108(n-k).
a(n) = phi^n * hypergeom([1/2, -n], [2], -4/phi) + (-phi)^(-n) * hypergeom([1/2, -n], [2], 4*phi), where phi = (1+sqrt(5))/2 = A001622.
Recurrence: 19*(n+1)*(n+2)*(11*n+13)*a(n) + 2*(55*n^3+208*n^2+311*n+230)*a(n+1) + 2*(55*n^3+373*n^2+674*n+206)*a(n+3) = (n+2)*(297*n^2+1022*n+617)*a(n+2) + (n+3)*(n+5)*(11*n+2)*a(n+4).
E.g.f.: 2*exp(5*x/2)*cosh(x*sqrt(5)/2)*(BesselI_0(2*x) - BesselI_1(2*x)) (the product of e.g.f. for Lucas and Catalan numbers).
a(n) ~ (phi + 4)^(n + 3/2) / (8 * sqrt(Pi) * n^(3/2)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Mar 10 2018

cp's OEIS Frontend

A103435 a(n) = 2^n * Fibonacci(n).

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A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

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A014335 Exponential convolution of Fibonacci numbers with themselves (divided by 2).

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A081057 E.g.f.: Sum_{n>=0} a(n)*x^n/n! = {Sum_{n>=0} F(n+1)*x^n/n!}^2, where F(n) is the n-th Fibonacci number.

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A082987 a(n) = Sum_{k=0..n} 3^k*F(k) where F(k) is the k-th Fibonacci number.

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A082988 a(n) = Sum_{k=0..n} 4^k*F(k) where F(k) is the k-th Fibonacci number.

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A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

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A081057 E.g.f.: Sum_{n>=0} a(n)x^n/n! = {Sum_{n>=0} F(n+1)x^n/n!}^2, where F(n) is the n-th Fibonacci number.