A014335 -id:A014335 - cp's OEIS frontend

A204451 2*A014335 - A203578. Difference of the exponential convolution of A000045 (Fibonacci) with itself and the corresponding exponential half-convolution.

0, 0, 0, 3, 8, 35, 75, 371, 888, 3891, 9445, 40755, 102323, 426803, 1091167, 4469555, 11625960, 46805811, 123364443, 490156851, 1306737465, 5132989235, 13816838695, 53753361203, 145912841523, 562912506675, 1539304050375, 5894896300851, 16225419029303, 61732155503411, 170909837010835

Offset: 0

Wolfdieter Lang, Jan 16 2012

nonn

See A203578 for the exponential (or binomial) half-convolution of A000045 (Fibonacci). The present sequence has to be added to this sequence in order to obtain the (full) exponential convolution 2*A014335.

Cf. A000045, A014335, A203578.

a(n) = sum(binomial(n,k)*F(k)*F(n-k), k=floor(n/2)+1..n), n>=0, with the Fibonacci numbers A000045(n).
E.g.f.: exp(x)*(cosh((2*phi-1)*x)-1)/5 - (BesselI(0,2*phi*x) + BesselI(0,2*(phi-1)*x) - 2*BesselI(0,2*I*x))/10. See the e.g.f. of A203578, also for phi and BesselI.
Bisection: a(2*k) =((2^(2*k) - binomial(2*k,k))*L(2*k)/2 - (1 - (-1)^k*binomial(2*k,k)))/5 and a(2*k+1) = (2^(2*k)*L(2*k+1) - 1)/5 = A203578(2*k), k>=0, with the Lucas numbers L(n)= A000032(n). Compare with A203578. See A000346(k-1), with A000346(-1)=0, for (2^(2*k) - binomial(2*k,k))/2, k>=0.

A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 6, 16, 56, 176, 576, 1856, 6016, 19456, 62976, 203776, 659456, 2134016, 6905856, 22347776, 72318976, 234029056, 757334016, 2450784256, 7930904576, 25664946176, 83053510656, 268766806016, 869747654656, 2814562533376, 9108115685376, 29474481504256

Offset: 0

Paul Barry, May 10 2003

Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)*x) (1,0,5,0,25,...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
Numerators of fractions in the approximation of the square root of 5 satisfying: a(n) = (a(n-1)+c)/(a(n-1)+1), with c=5 and a(1)=1. For denominators see A063727. - Mark Dols, Jul 24 2009
Equals right border of triangle A143969. (1, 6, 16, 56, ...) = row sums of triangle A143969 and INVERT transform of (1, 5, 5, 5, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 5 types of other natural numbers. - Milan Janjic, Aug 13 2010
From Gary W. Adamson, Jul 30 2016: (Start)
The sequence is case N=1 in an infinite set obtained by taking powers of the 2 X 2 matrix M = [(1,5); (1,N)], then extracting the upper left terms. The infinite set begins:
N=1 (A084057):  1, 6, 16,  56,  176,   576,   1856, ...
N=2 (A108306):  1, 6, 21,  81,  306,  1161,   4401, ...
N=3 (A164549):  1, 6, 26, 116,  516,  2296,  10216, ...
N=4 (A015449):  1, 6, 31, 161,  836,  4341,  22541, ...
N=5 (A000400):  1, 6, 36, 216, 1296,  7776,  46656, ...
N=6 (A049685):  1, 6, 41, 281, 1926, 13201,  90481, ...
N=7 (.......):  1, 6, 46, 356, 2756, 21336, 222712, ...
...
Sequences in the above set can be obtained by taking INVERT transforms of the following:
N=1 INVERT transform of (1, 5,  5,  5,   5,    5, ...
N=2 ..."......"......". (1, 5, 10, 20,  40,   80, ...
N=3 ..."......"......". (1, 5, 15, 45, 135,  405, ...
N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...
...
with the pattern (1, 5, N*5, (N^2)*5, (N^3)*5, ...
It appears that the sequence generated from powers (n>0) of the matrix P = [(1,a); (1,b)], (a,b > 0), then extracting the upper left terms, is equal to the INVERT transform of the sequence starting: (1, a, b*a, (b^2)*a, (b^3)*a, ...). (End)

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A046717, A002533, A098158, A143969, A269992.
a(n) = A087131(n)/2.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A108306, A164549, A015449, A000400, A049685

I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 31 2016

f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or *)
LinearRecurrence[{2, 4}, {1, 1}, 28] (* Robert G. Wilson v, Sep 18 2013 *)
RecurrenceTable[{a[1] == 1, a[2] == 1, a[n] == 2 a[n-1] + 4 a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Jul 31 2016 *)
Table[2^(n-1) LucasL[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 19 2016 *)

lucas(n)=fibonacci(n-1)+fibonacci(n+1)
a(n)=lucas(n)/2*2^n \\ Charles R Greathouse IV, Sep 18 2013

from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,1,2,4, lambda n: 0); [next(it) for i in range(1,26)] # Zerinvary Lajos, Jul 09 2008

[lucas_number2(n,2,-4)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.
G.f.: (1-x) / (1-2*x-4*x^2).
E.g.f.: exp(x) * cosh(sqrt(5)*x).
a(2n+1) = 2*a(n)*a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*5^k . - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} A098158(n,k)*5^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = 2^(n-1)*A000032(n). - Mark Dols, Jul 24 2009
If p(1)=1, and p(i)=5 for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for i<=j, A(i,j):=-1, (i=j+1), and A(i,j):=0 otherwise, then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/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) = A063727(n) - A063272(n-1). - R. J. Mathar, Jun 06 2019
a(n) = 1 + 5*A014335(n). - R. J. Mathar, Jun 06 2019
Sum_{n>=1} 1/a(n) = A269992. - Amiram Eldar, Feb 01 2021

A179596 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2x - 11x^2 - 6*x^3).

Original entry on oeis.org

1, 3, 17, 73, 351, 1607, 7513, 34809, 161903, 751783, 3493353, 16227737, 75393055, 350251335, 1627192697, 7559508409, 35119644495, 163157037671, 757987215241, 3521419711833, 16359641017343, 76002822156295, 353090213774361

Offset: 0

Johannes W. Meijer, Jul 28 2010; edited Jun 21 2013

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 or 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the center square the king goes crazy and turns into a red king.
On a 3 X 3 chessboard there are 2^9 = 512 ways to go crazy on the center square (off the center the piece behaves like a normal king). The red king is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program and A180140. For the corner squares the 512 red kings lead to 47 different red king sequences, see the overview of the red king sequences.
The sequence above corresponds to four A[5] vectors with decimal [binary] values 367 [101 101 111], 463 [111 001 111], 487 [111 100 111] and 493 [111 101 101]. These vectors lead for the side squares to A126473 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1 - 2*x - (k+8)*x^2 - 2*k*x^3). Red king sequences that are members of this family are A083424 (k=0), A179604 (k=1), A179600 (k=2), A179596 (k=3; this sequence) and A086346 (k=4). Other members of this family are A015528 (k=5) and A179608 (k=-4).

Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Charles Krauthammer, Did Chess Make Him Crazy?, Time, April 26, 2005.
Johannes W. Meijer, The red king sequences.
Index entries for linear recurrences with constant coefficients, signature (2,11,6).

Cf. A180140 (berserker sequences).

Cf. Red king sequences corner squares [decimal value A[5]]: A086346 [495], A015525 [239], A179596 [367], A179600 [335], A015524 [95], A083858 [31], A179604 [327], A015523 [27], A179610 [85], A083424 [325], A015521 [11], A007482 [2], A014335 [16].

nmax:=22; m:=1; A[1]:= [0,1,0,1,1,0,0,0,0]: A[2]:= [1,0,1,1,1,1,0,0,0]: A[3]:= [0,1,0,0,1,1,0,0,0]: A[4]:=[1,1,0,0,1,0,1,1,0]: A[5]:= [1,0,1,1,0,1,1,1,1]: A[6]:= [0,1,1,0,1,0,0,1,1]: A[7]:= [0,0,0,1,1,0,0,1,0]: A[8]:= [0,0,0,1,1,1,1,0,1]: A[9]:= [0,0,0,0,1,1,0,1,0]: A:=Matrix([A[1],A[2],A[3],A[4],A[5], A[6],A[7],A[8],A[9]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

LinearRecurrence[{2,11,6},{1,3,17},30] (* Harvey P. Dale, May 18 2011 *)

Vec((1+x)/(1-2*x-11*x^2-6*x^3)+O(x^99)) \\ Charles R Greathouse IV, Jul 16 2011

G.f.: (1+x)/(1 - 2*x - 11*x^2 - 6*x^3).
a(n) = 2*a(n-1) + 11*a(n-2) + 6*a(n-3) with a(0)=1, a(1)=3 and a(2)=17.
a(n) = (-1)^(-n)*2^(n+1)/9 + ((49+17*sqrt(7))*A^(-n) + (49-17*sqrt(7))*B^(-n))/126 with A = (-2+sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->infinity} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7) - A108851(n)).

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

A014334 Exponential convolution of Fibonacci numbers with themselves.

Original entry on oeis.org

0, 0, 2, 6, 22, 70, 230, 742, 2406, 7782, 25190, 81510, 263782, 853606, 2762342, 8939110, 28927590, 93611622, 302933606, 980313702, 3172361830, 10265978470, 33221404262, 107506722406, 347899061862, 1125825013350, 3643246274150, 11789792601702

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
C. A. Church and Marjorie Bicknell, Exponential generating functions for Fibonacci identities, Fibonacci Quarterly 11, no. 3 (1973), 275-281.
Sergio Falcon, Half self-convolution of the k-Fibonacci sequence, Notes on Number Theory and Discrete Mathematics, Vol. 26, No. 3 (2020), pp. 96-106.
Helmut Prodinger, Convolution identities for Tribonacci numbers via the diagonal of a bivariate generating function, arXiv:1910.08323 [math.NT], 2019.
Charles R. Wall, Problem B-573, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 24, No. 2 (1986), p. 181; Solution to Problem B-573 by Bob Prielipp, ibid., Vol. 25, No. 2 (1987), p. 184.
Index entries for linear recurrences with constant coefficients, signature (3,2,-4).

Cf. A000032, A000045, A014335, A103435.

[(2^n*Lucas(n) -2)/5: n in [0..40]]; // Vincenzo Librandi, Jul 15 2018

LinearRecurrence[{3,2,-4}, {0,0,2}, 30] (* Harvey P. Dale, Oct 24 2015 *)
Table[(2^n LucasL[n] -2)/5, {n,0,100}] (* Vladimir Reshetnikov, May 18 2016 *)

a(n)=if(n<1,0,sum(k=0,n-1,fibonacci(k)*2^k))

[(2^n*lucas_number2(n,1,-1) -2)/5 for n in range(41)] # G. C. Greubel, Jan 06 2023

From Benoit Cloitre, May 29 2003: (Start)
a(n) = 3*a(n-1) + 2*a(n-2) - 4*a(n-3), a(0)=0, a(1)=0, a(2)=2.
a(n) = Sum_{k=0..n-1} 2^k*Fibonacci(k) for n > 0.
a(n) = (-2 + ((1+sqrt(5))^n + (1-sqrt(5))^n))/5. (End)
a(n) = Sum_{k=0..n} Fibonacci(k)*Fibonacci(n-k)*binomial(n, k). - Benoit Cloitre, May 11 2005
From R. J. Mathar, Sep 29 2010: (Start)
a(n) = 2*A014335(n).
G.f.: 2*x^2/((1-x)*(1-2*x-4*x^2)).
a(n) = Sum_{k=1..n-1} A103435(k). (End)
a(n) = (2^n*A000032(n) - 2)/5. - Vladimir Reshetnikov, May 18 2016
E.g.f.: 2*(cosh(sqrt(5)*x)-1)*exp(x)/5. - Ilya Gutkovskiy, May 18 2016
a(n) = ((Sum_{k=0..n} Lucas(k)*Lucas(n-k)*binomial(n, k)) - 4)/5 (Wall, 1986). - Amiram Eldar, Jan 27 2022

A346415 Triangle T(n,k), n>=0, 0<=k<=n, read by rows, where column k is (1/k!) times the k-fold exponential convolution of Fibonacci numbers with themselves.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 3, 11, 6, 1, 0, 5, 35, 35, 10, 1, 0, 8, 115, 180, 85, 15, 1, 0, 13, 371, 910, 630, 175, 21, 1, 0, 21, 1203, 4494, 4445, 1750, 322, 28, 1, 0, 34, 3891, 22049, 30282, 16275, 4158, 546, 36, 1, 0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1

Offset: 0

Alois P. Heinz, Jul 15 2021

The sequence of column k>0 satisfies a linear recurrence with constant coefficients of order k+1.

			Triangle T(n,k) begins:
  1;
  0,  1;
  0,  1,     1;
  0,  2,     3,      1;
  0,  3,    11,      6,      1;
  0,  5,    35,     35,     10,      1;
  0,  8,   115,    180,     85,     15,     1;
  0, 13,   371,    910,    630,    175,    21,    1;
  0, 21,  1203,   4494,   4445,   1750,   322,   28,   1;
  0, 34,  3891,  22049,  30282,  16275,  4158,  546,  36,  1;
  0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-4 give: A000007, A000045, A014335, A014337, A014341.
T(n+j,n) for j=0-2 give: A000012, A000217, A000914.
Row sums give A256180.

b:= proc(n) option remember; `if`(n=0, 1, add(expand(x*b(n-j)
      *binomial(n-1, j-1)*(<<0|1>, <1|1>>^j)[1, 2]), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..12);
# second Maple program:
b:= proc(n, k) option remember; `if`(k=0, 0^n, `if`(k=1,
       combinat[fibonacci](n), (q-> add(binomial(n, j)*
       b(j, q)*b(n-j, k-q), j=0..n))(iquo(k, 2))))
    end:
T:= (n, k)-> b(n, k)/k!:
seq(seq(T(n, k), k=0..n), n=0..12);

b[n_, k_] := b[n, k] = If[k == 0, 0^n, If[k == 1, Fibonacci[n], With[{q = Quotient[k, 2]}, Sum[Binomial[n, j] b[j, q] b[n-j, k-q], {j, 0, n}]]]];
T[n_, k_] := b[n, k]/k!;
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Nov 06 2021, after 2nd Maple program *)

A203579 Exponential (or binomial) convolution of A000032 (Lucas) with itself, divided by 2.

Original entry on oeis.org

2, 2, 7, 17, 57, 177, 577, 1857, 6017, 19457, 62977, 203777, 659457, 2134017, 6905857, 22347777, 72318977, 234029057, 757334017, 2450784257, 7930904577, 25664946177, 83053510657, 268766806017, 869747654657, 2814562533377, 9108115685377, 29474481504257

Offset: 0

Wolfdieter Lang, Jan 14 2012

			With A000032 = {2,1,3,4,7,...},
  2*a(4) = 1*2*7 + 4*1*4 + 6*3*3 + 4*4*1 + 1*7*2 = 114.

Michael De Vlieger, Table of n, a(n) for n = 0..1961
Sergio Falcon, Half self-convolution of the k-Fibonacci sequence, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 96-106.

Cf. A000032, A014335.

Array[Sum[Binomial[#, k] LucasL[k] LucasL[# - k], {k, 0, #}]/2 &, 28, 0] (* Michael De Vlieger, Dec 28 2020 *)

a(n) = sum(binomial(n,k)*L(k)*L(n-k),k=0..n)/2, n>=0, with L(n)=A000032(n).
E.g.f.: (1/2)*(exp(phi*x)+exp(-(phi-1)*x))^2 =
  exp(x)*(cosh(sqrt(5)*x)+1), with the golden section phi:=(1+sqrt(5))/2. (See the e.g.f. of A000032).
a(n) = 2^(n-1)*L(n) + 1.
a(n) = 5*A014335(n) + 2. - Vladimir Reshetnikov, Oct 06 2016

A099013 a(n) = Sum_{k=0..n} 3^(k-1)*Fibonacci(k).

Original entry on oeis.org

0, 1, 4, 22, 103, 508, 2452, 11929, 57856, 280930, 1363495, 6618856, 32128024, 155953777, 757013548, 3674624638, 17836995847, 86582609284, 420280790476, 2040085854985, 9902784679240, 48069126732586, 233332442310919

Offset: 0

Paul Barry, Sep 22 2004

Partial sums of A099012. Binomial transform of A063092 (with leading 0).

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

Cf. A000045, A014335.

I:=[0, 1, 4]; [n le 3 select I[n] else 4*Self(n-1)+6*Self(n-2)-9*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 25 2012

Join[{a=0,b=1},Table[c=3*b+9*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Table[Sum[3^(k-1) Fibonacci[k],{k,0,n}],{n,0,30}] (* or *) LinearRecurrence[{4,6,-9},{0,1,4},30] (* Harvey P. Dale, Dec 09 2011 *)
CoefficientList[Series[x/((1-x)(1-3x-9x^2)),{x,0,40}],x] (* Vincenzo Librandi, Jun 25 2012 *)

x='x+O('x^30); concat([0], Vec(x/((1-x)*(1 - 3*x - 9*x^2)))) \\ G. C. Greubel, Dec 31 2017

G.f.: x/((1-x)*(1 - 3*x - 9*x^2)).
a(n) = 4*a(n-1) + 6*a(n-2) - 9*a(n-3).
a(n) = 3^(n-1)*Sum_{k=0..n} Fibonacci(n-k)*3^(-k).
a(n) = (3/2 + 3*sqrt(5)/2)^n*(1/22 + 7*sqrt(5)/110) + (1/22 - 7*sqrt(5)/110)*(3/2 - 3*sqrt(5)/2)^n - 1/11.
a(n) = (3^n*A000285(n) - 1)/11, the case m = 3 of Sum_{k=0..n} m^(k-1)*F(k) = (m^n*(m*F(n) + F(n+1)) - 1)/(m^2 + m - 1), F=A000045. - Ehren Metcalfe, Apr 29 2018

Sign in second formula corrected by Harvey P. Dale, Dec 09 2011

A110291 Riordan array (1/(1-x), x(1+2x)).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 3, 9, 7, 1, 1, 3, 9, 19, 9, 1, 1, 3, 9, 27, 33, 11, 1, 1, 3, 9, 27, 65, 51, 13, 1, 1, 3, 9, 27, 81, 131, 73, 15, 1, 1, 3, 9, 27, 81, 211, 233, 99, 17, 1, 1, 3, 9, 27, 81, 243, 473, 379, 129, 19, 1, 1, 3, 9, 27, 81, 243, 665, 939, 577, 163, 21, 1

Offset: 0

Paul Barry, Jul 18 2005

Inverse is A110292.

			Rows begin
  1;
  1, 1;
  1, 3, 1;
  1, 3, 5,  1;
  1, 3, 9,  7,  1;
  1, 3, 9, 19,  9,   1;
  1, 3, 9, 27, 33,  11,  1;
  1, 3, 9, 27, 65,  51, 13,  1;
  1, 3, 9, 27, 81, 131, 73, 15, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000244, A001045, A001047, A005408, A014335.
Cf. A066810, A077890, A077912, A110292, A180146.
Cf. A000975 (row sums), A052947 (diagonal sums).

R:=PowerSeriesRing(Rationals(), 30);
F:= func< k | Coefficients(R!( x^k*(1+2*x)^k/(1-x) )) >;
A110291:= func< n,k | F(k)[n-k+1] >;
[A110291(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Jan 05 2023

F[k_]:= CoefficientList[Series[x^k*(1+2*x)^k/(1-x), {x,0,40}], x];
A110291[n_, k_]:= F[k][[n+1]];
Table[A110291[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 05 2023 *)

def p(k,x): return x^k*(1+2*x)^k/(1-x)
def A110291(n,k): return ( p(k,x) ).series(x, 30).list()[n]
flatten([[A110291(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 05 2023

T(n, k) = [x^n]( x^k*(1+2*x)^k/(1-x) ).
Sum_{k=0..n} T(n, k) = A000975(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A052947(n+1).
From G. C. Greubel, Jan 05 2023: (Start)
T(n, 0) = T(n, n) = 1.
T(n, n-1) = A005408(n-1).
T(2*n, n) = T(2*n+1, n) = A000244(n).
T(2*n, n+1) = A066810(n+1).
T(2*n, n-1) = A000244(n-1).
T(2*n+1, n+1) = A001047(n+1).
Sum_{k=0..n} (-1)^k * T(n, k) = A077912(n).
Sum_{k=0..n} 2^k * T(n, k) = A014335(n+2).
Sum_{k=0..n} 3^k * T(n, k) = A180146(n).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = A077890(n). (End)

a(30) and following corrected by Georg Fischer, Oct 11 2022

A203578 Exponential (or binomial) half-convolution of A000045 (Fibonacci) with itself.

Original entry on oeis.org

0, 0, 2, 3, 14, 35, 155, 371, 1518, 3891, 15745, 40755, 161459, 426803, 1671175, 4469555, 17301630, 46805811, 179569163, 490156851, 1865624365, 5132989235, 19404565567, 53753361203, 201986220339, 562912506675, 2103942223775, 5894896300851, 21927151270703, 61732155503411

Offset: 0

Wolfdieter Lang, Jan 13 2012

For the definition of the exponential (also known as binomial) half-convolution of a sequence with itself see a comment on A203576 where also the rule for the e.g.f. is given.

Michael De Vlieger, Table of n, a(n) for n = 0..1962
Sergio Falcon, Half self-convolution of the k-Fibonacci sequence, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 96-106.

Cf. A000045, 2*A014335 (exponential convolution), A032443.

Table[Sum[Binomial[n,k]Fibonacci[k]Fibonacci[n-k],{k,0,Floor[n/2]}],{n,0,30}] (* Harvey P. Dale, Mar 04 2013 *)

a(n) = sum(binomial(n,k)*F(k)*F(n-k),k=0..floor(n/2)), n>=0, with F(n)=A000045(n).
E.g.f.: (f(x)^2 + Fs2(x^2))/2, with the e.g.f. f(x) of A000045 and the o.g.f. Fs2(x):=sum((F(n)/n!)^2*x^n,n=0..infty) of the scaled squares. f(x)^2 = 2*exp(x)*(cosh((2*phi-1)*x)-1)/5 (see A000045 for f(x)) and  Fs2(x^2) = (BesselI(0,2*phi*x) + BesselI(0,2*(phi-1)*x) - 2*BesselI(0,2*i*x))/5, with the golden section phi:=(1+sqrt(5))/2, the complex unit i, and for BesselI see Abramowitz-Stegun (reference and link given in A008277, p. 375, eq. 9.6.10). BesselI(0,2*sqrt(y)) = hypergeom([],[1],y) is the e.g.f. of the sequence {1/n!}.
Bisection:
a(2*k) = (A032443(k)*L(2*k) - (1 + (-1)^k*binomial(2*k,k)))/5 and a(2*k) = (2^(2*k)*L(2*k+1) - 1)/5, k>=0, with the Lucas numbers L(n)=A000032(n), and A032443(k)=(2^(2*k) + binomial(2*k,k))/2. - Wolfdieter Lang, Jan 16 2012.

cp's OEIS Frontend

A204451 2*A014335 - A203578. Difference of the exponential convolution of A000045 (Fibonacci) with itself and the corresponding exponential half-convolution.

Views

Author

Keywords

Comments

Crossrefs

Formula

A084057 a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Sage

Formula

A179596 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2*x - 11*x^2 - 6*x^3).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Sage

Formula

Extensions

A014334 Exponential convolution of Fibonacci numbers with themselves.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A346415 Triangle T(n,k), n>=0, 0<=k<=n, read by rows, where column k is (1/k!) times the k-fold exponential convolution of Fibonacci numbers with themselves.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A203579 Exponential (or binomial) convolution of A000032 (Lucas) with itself, divided by 2.

Views

Author

Keywords

Examples

A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

A179596 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2x - 11x^2 - 6*x^3).

A110291 Riordan array (1/(1-x), x(1+2x)).