A060925 -id:A060925 - cp's OEIS frontend

A003947 Expansion of (1+x)/(1-4*x).

1, 5, 20, 80, 320, 1280, 5120, 20480, 81920, 327680, 1310720, 5242880, 20971520, 83886080, 335544320, 1342177280, 5368709120, 21474836480, 85899345920, 343597383680, 1374389534720, 5497558138880, 21990232555520, 87960930222080, 351843720888320

Offset: 0

N. J. A. Sloane

Coordination sequence for infinite tree with valency 5.
For n>=1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+1} and fixed y_1, y_2,...,y_n in {1,2,3,4,5} we have f(x_i)<>y_i, (i=1,2,...,n). - Milan Janjic, May 10 2007
Number of length-n strings of 5 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012
Create a rectangular prism with edges of lengths 2^(n-2), 2^(n-1), and 2^(n) starting at n=2; then the surface area = a(n). - J. M. Bergot, Aug 08 2013

T. D. Noe, Table of n, a(n) for n = 0..200
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 306
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (4).
Index entries for sequences related to trees

Cf. A003948, A003949. Column 5 in A265583.

Concatenation([1], List([1..30], n-> 5*4^(n-1) )); # G. C. Greubel, Aug 10 2019

[1] cat [5*4^(n-1): n in [1..30]]; // G. C. Greubel, Aug 10 2019

k := 5; if n = 0 then 1 else k*(k-1)^(n-1); fi;

q = 5; Join[{a = 1}, Table[If[n != 0, a = q*a - a, a = q*a], {n, 0, 25}]] (* and *) Join[{1}, 5*4^Range[0, 25]] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)
LinearRecurrence[{4},{1,5},30] (* Harvey P. Dale, Apr 19 2015 *)

a(n)=5*4^n\4 \\ Charles R Greathouse IV, Sep 08 2011

[1]+[5*4^(n-1) for n in (1..30)] # G. C. Greubel, Aug 10 2019

Binomial transform of A060925. Its binomial transform is A003463 (without leading zero). - Paul Barry, May 19 2003
From Paul Barry, May 19 2003: (Start)
a(n) = (5*4^n - 0^n)/4.
G.f.: (1+x)/(1-4*x).
E.g.f.: (5*exp(4*x) - exp(0))/4. (End)
a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 3. - Philippe Deléham, Jul 10 2005
a(n) = A146523(n)*A011782(n). - R. J. Mathar, Jul 08 2009
a(n) = 5*A000302(n-1), n>0.
a(n) = 4*a(n-1), n>1. - Vincenzo Librandi, Dec 31 2010
G.f.: 2+x- 2/G(0), where G(k)= 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013

Edited by N. J. A. Sloane, Dec 04 2009

A046717 a(n) = 2a(n-1) + 3a(n-2), a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 5, 13, 41, 121, 365, 1093, 3281, 9841, 29525, 88573, 265721, 797161, 2391485, 7174453, 21523361, 64570081, 193710245, 581130733, 1743392201, 5230176601, 15690529805, 47071589413, 141214768241, 423644304721, 1270932914165, 3812798742493, 11438396227481

Offset: 0

Gervais Deroo and M. Deroo

Form the digraph with matrix A = [0,1,1,1; 1,0,1,1; 1,1,0,1; 1,0,1,1]. Then the sequence 0,1,1,5,... or (3^(n-1)-(-1)^n)/2+0^n/3 with g.f. x(1-x)/(1-2x-3x^2) corresponds to the (1,2) term of A^n. - Paul Barry, Oct 02 2004
3*a(n+1) + a(n) = 4*A060925(n); a(n+1) = A015518(n) + A060925(n); a(n+1) - 6*A015518(n) = (-1)^n. - Creighton Dement, Nov 15 2004
The sequence corresponds to the (1,1) term of the matrix [1,2;2,1]^n. - Simone Severini, Dec 04 2004
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 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
a(n)^2 + (2*A015518(n))^2 = a(2n). E.g., a(3) = 13, 2*A015518(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
Equals INVERTi transform of A104934: (1, 2, 8, 28, 100, 356, 1268, ...). - Gary W. Adamson, Jul 21 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square just one A[5] vector, with decimal value 341, leads to this sequence (without the first leading 1). For the corner squares this vector leads to the companion sequence A015518 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 2, 1, 4, 2, 6, 4, 2, 4, 10, 2, 6, 6, 4, 8, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
a(n) is the number of words of length n over a ternary alphabet whose position in the lexicographic order is a multiple of two. - Alois P. Heinz, Apr 13 2022
a(n) is the sum, for k=0..3, of the number of walks of length n between two vertices at distance k of the cube graph. - Miquel A. Fiol, Mar 09 2024

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
P. D. Jarvis and J. G. Sumner, Matrix group structure and Markov invariants in the strand symmetric phylogenetic substitution model, arXiv preprint arXiv:1307.5574 [q-bio.PE], 2013.
Index entries for linear recurrences with constant coefficients, signature (2,3).

The first difference sequence of A015518.
Row sums of triangle A080928.
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. A015518.
Cf. A104934. - Gary W. Adamson, Jul 21 2010

[n le 2 select 1 else 2*Self(n-1)+3*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Jul 21 2013

[(3^n + (-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jan 07 2018

a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+3*a[n-2] od: seq(a[n], n=0..33); # Zerinvary Lajos, Dec 14 2008
seq(denom(((-2)^(2*n)+6^(2*n))/((-2)^n+6^n)),n=0..26)

Table[(3^n + (-1)^n)/2, {n, 0, 30}] (* Artur Jasinski, Dec 10 2006 *)
CoefficientList[ Series[(1 - x)/(1 - 2x - 3x^2), {x, 0, 30}], x]  (* Robert G. Wilson v, Apr 04 2011 *)
Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, Apr 04 2011 *)

{a(n) = (3^n+(-1)^n)/2};
for(n=0,30, print1(a(n), ", ")) /* modified by G. C. Greubel, Jan 07 2018 */

x='x+O('x^30); Vec((1-x)/((1+x)*(1-3*x))) \\ G. C. Greubel, Jan 07 2018

[lucas_number2(n,2,-3)/2 for n in range(0, 27)] # Zerinvary Lajos, Apr 30 2009

G.f.: (1-x)/((1+x)*(1-3*x)).
a(n) = (3^n + (-1)^n)/2.
a(n) = Sum_{k=0..n} binomial(n, 2k)2^(2k). - Paul Barry, Feb 26 2003
Binomial transform of A000302 (powers of 4) with interpolated zeros. Inverse binomial transform of A081294. - Paul Barry, Mar 17 2003
E.g.f.: exp(x)cosh(2x). - Paul Barry, Mar 17 2003
a(n) = ceiling(3^n/4) + floor(3^n/4) = ceiling(3^n/4)^2 - floor(3^n/4)^2. - Paul Barry, Jan 17 2005
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,j)C(n-j,k)*(1+(-1)^(j-k))/2. - Paul Barry, May 21 2006
a(n) = Sum_{k=0..n} A098158(n,k)*4^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = (3^n + (-1)^n)/2. - M. F. Hasler, Mar 20 2008
a(n) = 2 A015518(n) + (-1)^n; for n > 0, a(n) = A080925(n). - M. F. Hasler, Mar 20 2008
((1 + sqrt4)^n + (1 - sqrt4)^n)/2. The offset is 0. a(3)=13. - Al Hakanson (hawkuu(AT)gmail.com), Nov 22 2008
If p[1]=1 and p[i]=4 (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (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*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
G.f.: G(0)/2, where G(k) = 1 + (-1)^k/(3^k - 3*9^k*x/(3*3^k*x + (-1)^k/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013

Description corrected by and more terms from Michael Somos

A084221 a(n+2) = 4*a(n), with a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 4, 12, 16, 48, 64, 192, 256, 768, 1024, 3072, 4096, 12288, 16384, 49152, 65536, 196608, 262144, 786432, 1048576, 3145728, 4194304, 12582912, 16777216, 50331648, 67108864, 201326592, 268435456, 805306368, 1073741824, 3221225472, 4294967296, 12884901888

Offset: 0

Paul Barry, May 21 2003

Binomial transform is A060925. Binomial transform of A084222.
Sequences with similar recurrence rules: A016116 (multiplier 2), A038754 (multiplier 3), A133632 (multiplier 5). See A133632 for general formulas. - Hieronymus Fischer, Sep 19 2007
Equals A133080 * A000079. A122756 is a companion sequence. - Gary W. Adamson, Sep 19 2007

			Binary...............Decimal
1..........................1
11.........................3
100........................4
1100......................12
10000.....................16
110000....................48
1000000...................64
11000000.................192
100000000................256
1100000000...............768
10000000000.............1024
110000000000............3072, etc. - _Philippe Deléham_, Mar 21 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Hester Graves, The Minimal Euclidean Function on the Gaussian Integers, arXiv:1802.08281 [math.NT], 2018. See Definition 2.3 p.3.
Index entries for linear recurrences with constant coefficients, signature (0,4)

For partial sums see A133628. Partial sums for other multipliers p: A027383(p=2), A087503(p=3), A133629(p=5).
Other related sequences: A132666, A132667, A132668, A132669.
Cf. A000302, A133080, A133087, A164346.

[(5*2^n-(-2)^n)/4: n in [0..40]]; // Vincenzo Librandi, Aug 13 2011

CoefficientList[Series[(-3*x - 1)/(4*x^2 - 1), {x, 0, 200}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)

a(n)=([0,1; 4,0]^n*[1;3])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

a(n) = (5*2^n-(-2)^n)/4.
G.f.: (1+3*x)/((1-2*x)(1+2*x)).
E.g.f.: (5*exp(2*x) - exp(-2*x))/4.
a(n) = A133628(n) - A133628(n-1) for n>1. - Hieronymus Fischer, Sep 19 2007
Equals A133080 * [1, 2, 4, 8, ...]. Row sums of triangle A133087. - Gary W. Adamson, Sep 08 2007
a(n+1)-2a(n) = A000079 signed. a(n)+a(n+2)=5*a(n). First differences give A135520. - Paul Curtz, Apr 22 2008
a(n) = A074323(n+1)*A016116(n). - R. J. Mathar, Jul 08 2009
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = Sum_{k=0..n+1} A181650(n+1,k)*2^k. - Philippe Deléham, Nov 19 2011
a(2*n) = A000302(n); a(2*n+1) = A164346(n). - Philippe Deléham, Mar 21 2014

Edited by N. J. A. Sloane, Dec 14 2007

A135522 a(n) = 2a(n-1) + 3a(n-2), with a(0) = 2 and a(1) = 3.

Original entry on oeis.org

2, 3, 12, 33, 102, 303, 912, 2733, 8202, 24603, 73812, 221433, 664302, 1992903, 5978712, 17936133, 53808402, 161425203, 484275612, 1452826833, 4358480502, 13075441503, 39226324512, 117678973533, 353036920602, 1059110761803

Offset: 0

Paul Curtz, Feb 19 2008

Also: inverse binomial transform of A135520. - R. J. Mathar, Apr 17 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3).

Cf. A060925.

[(5*3^n+3*(-1)^n)/4: n in [0..40]]; // Vincenzo Librandi, Jun 02 2011

f[n_]:=3/(n+2);x=2;Table[x=f[x];Numerator[x],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 11 2010 *)
Transpose[NestList[Join[Rest[#],ListCorrelate[{3,2},#]]&, {2,3},30]][[1]]  (* Harvey P. Dale, Mar 14 2011 *)
CoefficientList[Series[(x-2)/(3x^2+2x-1),{x,0,30}],x]  (* Harvey P. Dale, Mar 14 2011 *)

a(n)=(5*3^n+3*(-1)^n)/4 \\ Charles R Greathouse IV, Jun 01 2011

From R. J. Mathar, Feb 23 2008: (Start)
O.g.f.: (5/(1-3*x) + 3/(1+x))/4.
a(n) = (5*3^n + 3*(-1)^n)/4. (End)
G.f.: (x-2)/(3*x^2 + 2*x - 1). - Harvey P. Dale, Mar 14 2011
E.g.f.: (1/4)*(5*exp(3*x) + 3*exp(-x)). - G. C. Greubel, Oct 17 2016

More terms from R. J. Mathar, Feb 23 2008

A060922 Convolution triangle for Lucas numbers A000032(n+1), n >= 0.

Original entry on oeis.org

1, 3, 1, 4, 6, 1, 7, 17, 9, 1, 11, 38, 39, 12, 1, 18, 80, 120, 70, 15, 1, 29, 158, 315, 280, 110, 18, 1, 47, 303, 753, 905, 545, 159, 21, 1, 76, 566, 1687, 2568, 2120, 942, 217, 24, 1, 123, 1039, 3612, 6666, 7043, 4311

Offset: 0

Wolfdieter Lang, Apr 20 2001

In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. G.f. for row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) is (1+2*z)/(1-(1+x)*z-(1+2*x)*z^2).
Row sums give A060925. Column sequences (without leading zeros) are, for m=0..6: A000032(n+1)= A000204(n+1) (Lucas), A004799(n+1), A060929-33.
Bisection of this triangle gives triangles A060923 (even part) and A060924 (odd part).
For the m-th column sequence (without leading zeros) one has: a(n+m,m)= (pL1(m,n)*L(n+2)+pL2(m,n)*L(n+1))/(m!*5^m), m >= 0, with the Lucas numbers L(n)=A000032(n), n >= 0 and the row polynomials pL1(n,x) := sum(A061188(n,m)*x^n,m=0..n) and pL2(n,x) := sum(A061189(n,m)*x^m,m=0..n).
Riordan array ((1+2*x)/(1-x-x^2), x*(1+2*x)/(1-x-x^2)). - Philippe Deléham, Jan 21 2014
T is the convolution triangle of A000204 (see A357368). - Peter Luschny, Oct 19 2022

			p(2,x) = 4+6*x+x^2.
Triangle begins:
1 ;
3, 1;
4, 6, 1;
7, 17, 9, 1;
11, 38, 39, 12, 1;
18, 80, 120, 70, 15, 1;
29, 158, 315, 280, 110, 18, 1;
47, 303, 753, 905, 545, 159, 21, 1;

Cf. A000032.

# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, A000204); # Peter Luschny, Oct 19 2022

a(n, m)=((n-m+1)*a(n, m-1)+2*(2*n-m)*a(n-1, m-1)+4*(n-1)*a(n-2, m-1))/(5*m), n >= m >= 1, a(n, 0)= A000204(n+1)= A000032(n+1).
G.f. for m-th column: ((1+2*x)/(1-x-x^2))* ((x*(1+2*x))/(1-x-x^2))^m.
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 21 2014

Example improved by Philippe Deléham, Jan 21 2014

A087206 a(n) = 2a(n-1) + 4a(n-2); with a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888

Offset: 0

Paul Barry, Aug 25 2003

Binomial transform of A056487. Unsigned version of A152174.
Number of words of length n over the alphabet {1,2,3,4} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 18 2017
From Sean A. Irvine, Jun 06 2025: (Start)
Also, the number of walks of length n starting at 0 in the following graph:
  1---2
  |\ /|
  | 0 |
  |/ \|
  4---3. (End)

Jens Christian Claussen, Time-evolution of the Rule 150 cellular automaton activity from a Fibonacci iteration, arXiv:math/0410429 [math.CO], 2004. See Table II, p. 4.
Sean A. Irvine, Walks on Graphs.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A060925, A063782, A253064.

Equals (1/2) * A063727(n-1). Cf. A006483.

LinearRecurrence[{2, 4}, {1, 4}, 25] (* Jean-François Alcover, Sep 21 2017 *)

G.f.: (1+2x)/(1-2x-4x^2).
a(n) = (1-sqrt(5))^n*(1/2-3*sqrt(5)/10)+(1+sqrt(5))^n*(1/2+3*sqrt(5)/10).
a(n) = 2^n*Fibonacci(n+2). - Paul Barry, Mar 22 2004
a(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/sqrt(80). Offset 2. a(4)=12. - Al Hakanson (hawkuu(AT)gmail.com), Apr 11 2009
G.f.: 1/(-2x-1/(-2x-1)). - Paul Barry, Mar 24 2010

Comment corrected by Philippe Deléham, Nov 27 2008

A209763 Triangle of coefficients of polynomials u(n,x) jointly generated with A209764; see the Formula section.

Original entry on oeis.org

1, 1, 2, 2, 5, 4, 3, 9, 13, 8, 4, 15, 31, 35, 16, 5, 23, 61, 97, 85, 32, 6, 33, 107, 219, 279, 203, 64, 7, 45, 173, 433, 717, 761, 469, 128, 8, 59, 263, 779, 1583, 2195, 1991, 1067, 256, 9, 75, 381, 1305, 3141, 5361, 6381, 5049, 2389, 512, 10, 93, 531

Offset: 1

Clark Kimberling, Mar 14 2012

Row n begins with n and ends with 2^(n-1).
Row sums:  1,3,11,33,101,303,911,... A081250
Alternating row sums: 1,-1,1,-1,1,.. A033999
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...2
2...5....4
3...9....13...8
4...15...31...35...16
First three polynomials u(n,x): 1, 1 + 2x, 2 + 5x + 4x^2.

Cf. A209764, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209763 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209764 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A081250 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A060925 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A033999 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A004442 *)

u(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x),
v(n,x)=2x*u(n-1,x)+v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A209764 Triangle of coefficients of polynomials v(n,x) jointly generated with A209763; see the Formula section.

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 4, 8, 14, 8, 5, 14, 32, 34, 16, 6, 22, 62, 96, 86, 32, 7, 32, 108, 218, 280, 202, 64, 8, 44, 174, 432, 718, 760, 470, 128, 9, 58, 264, 778, 1584, 2194, 1992, 1066, 256, 10, 74, 382, 1304, 3142, 5360, 6382, 5048, 2390, 512, 11, 92, 532

Offset: 1

Clark Kimberling, Mar 14 2012

Row n begins with n and ends with 2^(n-1).
Row sums:  1,4,11,34,101,304,911,2734,... A060925.
Alternating row sums: 1,0,3,2,5,4,7,6,... A060925.
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
2...2
3...4....4
4...8....14...8
5...14...32...34...16
First three polynomials v(n,x): 1, 2 + 2x , 3 + 4x + 4x^2.

Cf. A209663, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209763 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209764 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A081250 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A060925 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A033999 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A004442*)

u(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x),
v(n,x)=2x*u(n-1,x)+v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A209765 Triangle of coefficients of polynomials u(n,x) jointly generated with A209766; see the Formula section.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 5, 15, 12, 1, 5, 21, 45, 29, 1, 5, 21, 77, 129, 70, 1, 5, 21, 89, 265, 361, 169, 1, 5, 21, 89, 353, 865, 991, 408, 1, 5, 21, 89, 377, 1325, 2717, 2681, 985, 1, 5, 21, 89, 377, 1549, 4733, 8281, 7169, 2378, 1, 5, 21, 89, 377, 1597, 6125

Offset: 1

Clark Kimberling, Mar 14 2012

Limiting row: F(2+3k), where F=A000045 (Fibonacci numbers)
Coefficient of x^n in u(n,x): 1,2,5,12,.... A000129(n)
Row sums:  1,3,11,33,101,303,911,2733,..... A081250
Alternating row sums: 1,-1,1,-1,1,-1,,..... A033999
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...2
1...5...5
1...5...15...12
1...5...21...45...29
First three polynomials u(n,x): 1, 1 + 2x, 1 + 5x + 5x^2.

Cf. A209766, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209765 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209766 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A081250 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A060925 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A033999 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A042963 signed *)

u(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x),
v(n,x)=2x*u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

A209766 Triangle of coefficients of polynomials v(n,x) jointly generated with A209765; see the Formula section.

Original entry on oeis.org

1, 1, 3, 1, 3, 7, 1, 3, 13, 17, 1, 3, 13, 43, 41, 1, 3, 13, 55, 133, 99, 1, 3, 13, 55, 209, 391, 239, 1, 3, 13, 55, 233, 739, 1113, 577, 1, 3, 13, 55, 233, 939, 2469, 3095, 1393, 1, 3, 13, 55, 233, 987, 3589, 7903, 8457, 3363, 1, 3, 13, 55, 233, 987, 4085

Offset: 1

Clark Kimberling, Mar 14 2012

Limiting row: F(1+3k), where F=A000045 (Fibonacci numbers)
Coefficient of x^n in u(n,x): A001333(n)
Row sums:  1,4,11,34,101,304,... A060925.
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
1...3
1...3...7
1...3...13...17
1...3...13...43...41
First three polynomials v(n,x): 1, 1 + 3x , 1 + 3x + 7x^2.

Cf. A209665, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := 2 x*u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209765 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209766 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A081250 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A060925 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A033999 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A042963 signed *)

u(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x),
v(n,x)=2x*u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

cp's OEIS Frontend

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A046717 a(n) = 2a(n-1) + 3a(n-2), a(0) = a(1) = 1.

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