A005409 -id:A005409 - cp's OEIS frontend

A114696 Expansion of (1+4x+x^2)/((1-x^2)(1-2*x-x^2)); a Pellian-related sequence.

1, 6, 15, 40, 97, 238, 575, 1392, 3361, 8118, 19599, 47320, 114241, 275806, 665855, 1607520, 3880897, 9369318, 22619535, 54608392, 131836321, 318281038, 768398399, 1855077840, 4478554081, 10812186006, 26102926095, 63018038200, 152139002497, 367296043198

Offset: 0

Creighton Dement, Feb 18 2006

Elements of odd index give match to A065113: Sum of the squares of the n-th and the (n+1)st triangular numbers (A000217) is a perfect square.

Generating floretion: - 1.5'i + 'j + 'k - .5i' + j' + k' + .5'ii' - .5'jj' - .5'kk' - 'ij' + 'ik' - 'ji' + .5'jk' + 2'ki' - .5'kj' + .5e

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

Cf. A000129, A002203, A005409, A100828, A111954, A113224.

Cf. A114647, A114688, A114689, A114695, A114697.

Q:= proc(n) option remember; # Q=A002203
    if n<2 then 2
  else 2*Q(n-1) + Q(n-2)
    fi; end:
seq((Q(n+2) -3 -(-1)^n)/2, n=0..40); # G. C. Greubel, May 24 2021

CoefficientList[Series[(1+4*x+x^2)/((1-x^2)*(1-2*x-x^2)), {x,0,30}], x] (* or *) LinearRecurrence[{2,2,-2,-1}, {1,6,15,40}, 30] (* Harvey P. Dale, Jan 23 2014 *)

Vec((1+4*x+x^2)/((1-x^2)*(1-2*x-x^2)) + O(x^30)) \\ Colin Barker, May 26 2016

[(lucas_number2(n+2,2,-1) -3 -(-1)^n)/2 for n in (0..30)] # G. C. Greubel, May 24 2021

G.f.: (1 +4*x +x^2)/((1-x)*(1+x)*(1-2*x-x^2)).
a(0)=1, a(1)=6, a(2)=15, a(3)=40, a(n) = 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4). - Harvey P. Dale, Jan 23 2014
a(n) = (-3 - (-1)^n + (3-2*sqrt(2))*(1-sqrt(2))^n + (1+sqrt(2))^n*(3+2*sqrt(2)))/2. - Colin Barker, May 26 2016
From G. C. Greubel, May 24 2021: (Start)
a(n) = 3*A000129(n+1) + A000129(n) - (3 + (-1)^n)/2.
a(n) = (1/2)*(A002203(n+2) - 3 - (-1)^n). (End)

A182188 A sequence of row differences for table A182119.

Original entry on oeis.org

1, -1, -11, -69, -407, -2377, -13859, -80781, -470831, -2744209, -15994427, -93222357, -543339719, -3166815961, -18457556051, -107578520349, -627013566047, -3654502875937, -21300003689579

Offset: 0

Kenneth J Ramsey, Apr 17 2012

This is a list of row differences corresponding to a difference of 1 in table A182119, column 0. If A181119(k+1,0) - A182119(k,0) = 1, then a(n) = A182119(k+1,n) - A182119(k,n).

If p is a prime of the form 8*n +- 3, then a(p) == 3 (mod p). If p is a prime of the form 8*n +- 1, then a(p) == -1 (mod p).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001109, A001333, A001542, A005409, A182189, A182190, A182119.

m = 13;n = 3; c = 0;
list3 = Reap[While[c < 22, t = 6 n - m - 4; Sow[t];m = n; n = t;c++]][[2,1]]
Table[1 -Fibonacci[2*n, 2], {n,0,40}] (* G. C. Greubel, May 24 2021 *)

[1 - lucas_number1(2*n,2,-1) for n in (0..40)] # G. C. Greubel, May 24 2021

a(n) = 6*a(n-1) - a(n-2) - 4. [corrected by Klaus Purath, Mar 19 2021]
a(n) = -(A182189(n-1) + 2*A182190(n-1)).
a(n) = 2 - A182189(n).
From Klaus Purath, Mar 19 2021: (Start)
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
a(n) = (-1)*Sum_{i=1..2*n-1} A001333(i) for n > 0.
a(n) = 1 - A001542(n) for n > 0.
a(n) = 1 - 2*A001109(n) for n > 0.
a(n) = (-1)*A005409(2*n) for n > 0. (End)
G.f.: (1 - 8*x + 3*x^2)/((1-x)*(1-6*x+x^2)). - Chai Wah Wu, Apr 08 2021
a(n) = 1 - Pell(2*n), where Pell(n) = A000129(n). - G. C. Greubel, May 24 2021

A210197 Triangle of coefficients of polynomials u(n,x) jointly generated with A210198; see the Formula section.

Original entry on oeis.org

1, 3, 7, 1, 15, 5, 31, 17, 1, 63, 49, 7, 127, 129, 31, 1, 255, 321, 111, 9, 511, 769, 351, 49, 1, 1023, 1793, 1023, 209, 11, 2047, 4097, 2815, 769, 71, 1, 4095, 9217, 7423, 2561, 351, 13, 8191, 20481, 18943, 7937, 1471, 97, 1, 16383, 45057, 47103

Offset: 1

Clark Kimberling, Mar 18 2012

Column 1:  -1+2^n
Row sums:  A048739
Alternating row sums: triangular numbers, A000217
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
3
7....1
15...5
31...17...1
First three polynomials u(n,x): 1, 3, 7 + x.

K. Dilcher, K. B. Stolarsky, Nonlinear recurrences related to Chebyshev polynomials, The Ramanujan Journal, 2014, Online Oct. 2014, pp. 1-23. See Table 1.

Cf. A210198, A208510.

Essentially the same as the triangle in A257597.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*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[%]   (* A210197 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210198 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A048739 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A005409 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A000217 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A000027 *)

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

A210560 Triangle of coefficients of polynomials v(n,x) jointly generated with A210559; see the Formula section.

Original entry on oeis.org

1, 3, 1, 5, 4, 2, 7, 9, 9, 3, 9, 16, 23, 16, 5, 11, 25, 46, 48, 30, 8, 13, 36, 80, 110, 101, 54, 13, 15, 49, 127, 215, 257, 203, 97, 21, 17, 64, 189, 378, 552, 570, 401, 172, 34, 19, 81, 268, 616, 1057, 1337, 1228, 776, 303, 55, 21, 100, 366, 948, 1862, 2772

Offset: 1

Clark Kimberling, Mar 22 2012

Column 1: odd positive integers (A005408)
Column 2: squares (A000290)
Row n ends with F(n), where F=A000045 (Fibonacci numbers)
Row sums: A005409
Alternating row sums: 1,2,3,4,5,6,7,8,...(A000027)
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
3...1
5...4...2
7...9...9...3
9...16...23...16...5
First three polynomials v(n,x): 1, 3 + x , 5 + 4x + 2x^2.

Cf. A210559, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*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[%]  (* A210559 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210560 *)

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

A335711 The number of free polyominoes of width 2 and height n.

Original entry on oeis.org

2, 6, 12, 30, 65, 158, 362, 875, 2064, 4984, 11914, 28764, 69155, 166956, 402372, 971413, 2343518, 5657754, 13654968, 32966010, 79577189, 192116330, 463786190, 1119678911, 2703086892, 6525829036, 15754607062, 38034986040, 91824246215, 221683340568, 535190123592

Offset: 2

R. J. Mathar, Jun 18 2020

nonn

The second column of A268371.

			a(2)=2, bounding box 2 X 2, counts the L-shaped 3-omino and the full block 4-omino.
a(3)=6, bounding box 2 X 3, counts three 4-ominoes, two 5-omioes, and the full 2 X 3 block 6-omino.
a(4)=12, bounding box 2 X 4, counts three 5-ominoes, six 6-ominoes, two 7-ominoes, and the full 2 X 4 block 8-omino.

Cf. A268371, A107769 (asymmetric), A005409 (C_2 symmetry and higher), A352720 (width 2 and size n).

Conjecture: a(n) = A107769(n-1) + A005409(floor((n+3)/2)).
Conjectures from Colin Barker, Jun 24 2020: (Start)
G.f.: x^2*(2 - 8*x^2 + 2*x^3 - x^4 + x^5 + x^6) / ((1 - x)*(1 - 2*x - x^2)*(1 - 2*x^2 - x^4)).
a(n) = 3*a(n-1) + a(n-2) - 7*a(n-3) + 3*a(n-4) - a(n-5) + a(n-6) + a(n-7) for n>8.
(End)
a(n) = (2*r(n) + 2*m(n) + A078057(n) + 1) / 4, where r(n) = A078057(floor((n-1)/2) - 1)/2, and m(n) = A078057(floor((n+1)/2) - 3)/2. - John Mason, Feb 28 2022

a(12)-a(20) from Jean-Luc Manguin, Jun 23 2020
a(21)-a(28) from John Mason, Feb 27 2022
a(29)-a(32) from John Mason, Feb 28 2022

A364312 Irregular triangle T read by rows, giving in row n the nonnegative coefficients of polynomials of height n and degree k (of decreasing powers), for k = 1, 2, ..., n-1, used for Cantor's counting of algebraic numbers, written for m = 1, 2, ..., A364313(n), for n >= 2, and for n = 1 the degree is k = 1.

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 1, 2, 1, 0, 1, 3, 1, 1, 3, 2, 0, 1, 1, 0, 2, 1, 1, 1, 4, 1, 1, 4, 3, 2, 2, 3, 3, 0, 1, 1, 0, 3, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 2, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 5, 1, 1, 5, 4, 0, 1, 1, 0, 4, 3, 0, 2, 2, 0, 3, 3, 1, 1, 1, 3, 1, 1, 1, 3, 2, 2, 1, 2, 1, 2, 1, 2, 2

Offset: 1

Wolfdieter Lang, Jul 19 2023

The length of row n is A364313(n). Different orders k are separated by a semicolon in the examples below.
The number of polynomials given from row n is A364314(n).
The entries for k = n-1 are only present for n >= 2 and A001227(n-1) = 1, that is, n = 2^q + 1 = A000051(q), for q >= 0. This is because otherwise x^(n-1) + 1 and x^(n-1) - 1 are both reducible (factorize over the integers).
For Cantor's counting (and determination) of algebraic numbers these polynomials have later to be signed, keeping the positive leading coefficient. See the example for n = 4 below. Complex solutions are omitted if real algebraic numbers are counted.
The polynomials with nonnegative coefficients recorded here are sometimes reducible over the integers. But in this case irreducible signed versions exist. E.g., for n = 6 and k = 2 the polynomial x^2 + 3*x + 2 = (x + 1)*(x + 2) is recorded as [1,3,2] (falling powers of x), because x^2 + 3*x - 2 and x^2 - 3*x - 2 are irreducible, each having two real solutions.
The number of distinct real solutions of the signed polynomials of degree k and height n is given in A364315(n, k). The total number is A364316(n). Note that no repetition of real solutions already obtained for lower heights can appear due to irreducibility. For the list of all real algebraic numbers for heights 1 to 7 see the W. Lang link.
The coefficients of the polynomials are determined from the relative prime compositions of K = n - (k-1). The order is taken from the corresponding partitions, with rising number of parts m, and for given m the order is anti-lexicographic (e.g., [4,1,1], [3,2,1] for K = 6 and m = 3). For each partition the compositions are ordered also anti-lexicographically, not considering the possible 0 parts which are distributed according to decreasing powers of x (e.g., [3,1,0,1], [3,0,1,1], [1,3,0,1], [1,0,3,1], [1,1,0,3], [1,0,1,3]).

			The irregular triangle T, with entries T(n, m), begins: (increasing k >= 1 values are separated by ;)
n\m   1  2    3  4    5  6  7    8  9 10   11 12 13 ...
1:   [1, 0]
2:   [1, 1]
3:   [2, 1], [1, 2]; [1, 0, 1]
4:   [3, 1], [1, 3]; [2, 0, 1], [1, 0, 2], [1, 1, 1]
...
n = 5: [4, 1], [1, 4], [3, 2], [2, 3]; [3, 0, 1], [1, 0, 3], [2, 1, 1], [1, 2, 1], [1, 1, 2]; [2, 0, 0, 1], [1, 0, 0, 2], [1, 1, 0, 1], [1, 0, 1, 1]; [1, 0, 0, 0, 1]
---------
n = 6: [5, 1], [1, 5]; [4, 0, 1], [1, 0, 4], [3, 0, 2], [2, 0, 3], [3, 1, 1], [1, 3, 1], [1, 1, 3], [2, 2, 1], [2, 1, 2], [1, 2, 2]; [3, 0, 0, 1], [1, 0, 0, 3], [2, 1, 0, 1], [2, 0, 1, 1], [1, 2, 0, 1], [1, 0, 2, 1], [1, 1, 0, 2], [1, 0, 1, 2], [1, 1, 1, 1]; [2, 0, 0, 0, 1], [1, 0, 0, 0, 2], [1, 1, 0, 0, 1], [1, 0, 1, 0, 1], [1, 0, 0, 1, 1]
---------
n = 7: [6, 1], [1, 6], [5, 2], [2, 5], [4, 3], [3, 4]; [5, 0, 1], [1, 0, 5], [4, 1, 1], [1, 4, 1], [1, 1, 4], [3, 2, 1], [3, 1, 2], [2, 3, 1], [2, 1, 3], [1, 3, 2], [1, 2, 3]; [4, 0, 0, 1], [1, 0, 0, 4], [3, 0, 0, 2], [2, 0, 0, 3], [3, 1, 0, 1], [3, 0, 1, 1], [1, 3, 0, 1], [1, 0, 3, 1], [1, 1, 0, 3], [1, 0, 1, 3], [2, 2, 0, 1], [2, 0, 2, 1], [2, 1, 0, 2], [2, 0, 1, 2], [1, 2, 0, 2], [1, 0, 2, 2], [2, 1, 1, 1], [1, 2, 1, 1], [1, 1, 2, 1], [1, 1, 1, 2]; [3, 0, 0, 0, 1], [1, 0, 0, 0, 3], [2, 1, 0, 0, 1], [2, 0, 1, 0, 1], [2, 0, 0, 1, 1], [1, 2, 0, 0, 1], [1, 0, 2, 0, 1], [1, 0, 0, 2, 1], [1, 1, 0, 0, 2], [1, 0, 1, 0, 2], [1, 0, 0, 1, 2], [1, 1, 1, 0, 1], [1, 1, 0, 1, 1], [1, 0, 1, 1, 1]; [2, 0, 0, 0, 0, 1],  [1, 0, 0, 0, 0, 2], [1, 1, 0, 0, 0, 1], [1, 0, 1, 0, 0, 1], [1, 0, 0, 1, 0, 1], [1, 0, 0, 0, 1, 1]
-------
x^6 + 1 = (x^2 + 1)*(x^4 - x^2 + 1), hence no [1, 0, 0, 0, 0, 0, 1] recorded, because x^6 - 1 also factorizes.
...
---------------------------------------------------------------------------
Polynomials: n = 4, degree k = 1:  3*x + 1, x + 3; k = 2: 2*x^2 + 1, x^2 + 2, x^2 + x + 1; k = 3: no entry [1, 0, 0, 1], because x^3 + 1 factorizes, as well as x^3 - 1.
---------------------------------------------------------------------------
Height n = 4, degree k = 2, with signed polynomials:
[2, 0, 1] for 2*x^2 + 1, 2*x^2 - 1, [1, 0, 2] for x^2 + 2, x^2 - 2, and [1, 1, 1] for x^2 + x + 1, x^2 + x - 1, x^2 - x + 1, x^2 - x - 1. The corresponding real algebraic numbers come in signed pairs only from 2*x^2 - 1, x^2 - 2, x^2 + x - 1, and x^2 - x - 1, namely, -sqrt(1/2), +sqrt(1/2), -sqrt(2), +sqrt(2), -phi = -A001622, phi - 1, and -(phi - 1), phi. So Cantor's phi (our Phi) is Phi(4, 2) = 8. Together with the four real k = 1 roots from the signed polynomials for [3, 1] and [1, 3] one finds Phi(4) = 12. See A362364.

Georg Cantor, Ueber eine Eigenschaft des Inbegriffes aller reellen algebraischen Zahlen, Journal f. d. reine u. angew. Math. 77 (1874), 258-262.
Georg Cantor, On a Property of the Class of all Real Algebraic Numbers, (English version).
Wolfdieter Lang, Cantor's List of Real Algebraic Numbers of Heights 1 to 7, arXiv:2307.10645 [math.NT], 2023.

Cf. A000051, A001227, A001622, A005409, A364313, A364314, A364315, A364316.

A048745 Partial sums of A048654.

Original entry on oeis.org

1, 5, 14, 36, 89, 217, 526, 1272, 3073, 7421, 17918, 43260, 104441, 252145, 608734, 1469616, 3547969, 8565557, 20679086, 49923732, 120526553, 290976841, 702480238, 1695937320, 4094354881, 9884647085, 23863649054, 57611945196, 139087539449, 335787024097

Offset: 0

Barry E. Williams

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-1).

Cf. A000129, A005409, A098790.

I:=[1,5,14]; [n le 3 select I[n] else 3*Self(n-1) -Self(n-2) -Self(n-3): n in [1..31]]; // G. C. Greubel, May 23 2021

t={1,5}; Do[AppendTo[t, t[[-2]] + 2*t[[-1]] + 3], {n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Jan 27 2012 *)
Accumulate[LinearRecurrence[{2,1},{1,4},30]] (* or *) LinearRecurrence[{3,-1,-1},{1,5,14},30] (* Harvey P. Dale, Aug 03 2020 *)

a(n)=polcoeff((1+2*x)/(1-3*x+x^2+x^3)+x*O(x^n),n) \\ Paul D. Hanna

[(5*lucas_number1(n+1,2,-1) + 3*lucas_number1(n,2,-1) -3)/2 for n in (0..30)] # G. C. Greubel, May 23 2021

a(n) = 2*a(n-1) + a(n-2) + 3, a(0)=1, a(1)=5.
a(n) = ( ((4+(5/2)*sqrt(2))*(1+sqrt(2))^n - (4-(5/2)*sqrt(2))*(1-sqrt(2))^n)/ 2*sqrt(2) ) - 3/2.
G.f.: (1+2*x)/((1-x)*(1-2*x-x^2)). - Paul D. Hanna, Feb 22 2005
a(n) = 3*a(n-1) - a(n-2) - a(n-3), n>2, a(0)=1, a(1)=5, a(2)=14. - Philippe Deléham, Dec 16 2008
2*a(n) = A135532(n+2) - 3. - R. J. Mathar, Mar 06 2013
a(n) = (1/2)*( 5*P(n+1) + 3*P(n) - 3), where P(n) = A000129(n). - G. C. Greubel, May 23 2021

A210198 Triangle of coefficients of polynomials v(n,x) jointly generated with A210197; see the Formula section.

Original entry on oeis.org

1, 3, 1, 7, 4, 15, 12, 1, 31, 32, 6, 63, 80, 24, 1, 127, 192, 80, 8, 255, 448, 240, 40, 1, 511, 1024, 672, 160, 10, 1023, 2304, 1792, 560, 60, 1, 2047, 5120, 4608, 1792, 280, 12, 4095, 11264, 11520, 5376, 1120, 84, 1, 8191, 24576, 28160, 15360, 4032

Offset: 1

Clark Kimberling, Mar 18 2012

Row sums:  A005409
Column 1:  -1+2^n
Alternating row sums: 1, 2,3,4,5,6,..., A000027
For a discussion and guide to related arrays, see A208510.

			First five rows:
1
3....1
15...12...1
31...32...6
63...80...24...1
First three polynomials v(n,x): 1, 3 + x , 15 + 12x + x^2.

Cf. A210197, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*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[%]   (* A210197 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210198 *)
Table[u[n, x] /. x -> 1, {n, 1, z}]  (* A048739 *)
Table[v[n, x] /. x -> 1, {n, 1, z}]  (* A005409 *)
Table[u[n, x] /. x -> -1, {n, 1, z}] (* A000217 *)
Table[v[n, x] /. x -> -1, {n, 1, z}] (* A000027 *)

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

A210562 Triangle of coefficients of polynomials v(n,x) jointly generated with A210561; see the Formula section.

Original entry on oeis.org

1, 2, 2, 2, 5, 4, 2, 6, 12, 8, 2, 6, 17, 28, 16, 2, 6, 18, 46, 64, 32, 2, 6, 18, 53, 120, 144, 64, 2, 6, 18, 54, 152, 304, 320, 128, 2, 6, 18, 54, 161, 424, 752, 704, 256, 2, 6, 18, 54, 162, 474, 1152, 1824, 1536, 512, 2, 6, 18, 54, 162, 485, 1372, 3056, 4352

Offset: 1

Clark Kimberling, Mar 22 2012

Last term in row n:  2^(n-1)
Limiting row:  2*3^(n-1)
Alternating row sums: 1,0,1,0,1,0,1,0,...
For a discussion and guide to related arrays, see A208510.

			First five rows:
  1
  2   2
  2   5   4
  2   6   12   8
  2   6   17   28   16
First three polynomials v(n,x): 1, 2 + 2*x, 2 + 5*x + 4*x^2.

P. Bala, A note on the diagonals of a proper Riordan Array

Row sums A005409. Cf. A208510, A210561.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := (x + 1)*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[%]  (* A210561 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A210562 *)

u(n,x) = x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x) = (x+1)*u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
From Peter Bala, Mar 06 2017: (Start)
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1).
E.g.f. for the n-th subdiagonal: exp(2*x)*(2 + 2*x + 2*x^2/2! + 2*x^3/3! + ... + 2*x^(n-1)/(n-1)! + x^n/n!).
Riordan array ((1 + x)/(1 - x), x*(2 + x)).
Row sums A005409 (except for the initial term). (End)

A210565 Triangle of coefficients of polynomials u(n,x) jointly generated with A210595; see the Formula section.

Original entry on oeis.org

1, 2, 2, 3, 5, 3, 4, 9, 10, 5, 5, 14, 22, 20, 8, 6, 20, 40, 51, 38, 13, 7, 27, 65, 105, 111, 71, 21, 8, 35, 98, 190, 256, 233, 130, 34, 9, 44, 140, 315, 511, 594, 474, 235, 55, 10, 54, 192, 490, 924, 1295, 1324, 942, 420, 89, 11, 65, 255, 726, 1554, 2534, 3130, 2860, 1836, 744, 144

Offset: 1

Clark Kimberling, Mar 23 2012

Row n starts with n and ends with F(n+1), where F=A000045 (Fibonacci numbers).
Row sums: A005409.
Alternating row sums: 1,0,1,0,1,0,1,0,1,0,1,0, ...
For a discussion and guide to related arrays, see A208510.

			First five rows:
  1;
  2,  2;
  3,  5,  3;
  4,  9, 10,  5;
  5, 14, 22, 20, 8;
First three polynomials u(n,x):
u(1, x) = 1;
u(2, x) = 2 + 2*x;
u(3, x) = 3 + 5*x + 3*x^2.

G. C. Greubel, Rows n = 1..30 of the triangle, flattened

Cf. A208510, A210595.

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 16;
u[n_, x_]:= x*u[n-1, x] + (x+1)*v[n-1, x] + 1;
v[n_, x_]:= 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[%]    (* A210565 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210595 *)
(* Second program *)
u[n_, x_]:= u[n, x]= If[n<2, (n+1)*(1+x)^n, (1+x)*u[n-1, x] +x^2*u[n-2, x] +1+x];
T[n_]:= CoefficientList[Series[u[n, x], {x, 0, n}], x];
Table[T[n-1], {n,12}] (* G. C. Greubel, May 23 2021 *)

@CachedFunction
def u(n,x): return (n+1)*(1+x)^n if (n<2) else (1+x)*u(n-1,x) + x^2*u(n-2,x) +1+x
def T(n): return taylor( u(n,x) , x,0,n).coefficients(x, sparse=False)
flatten([T(n-1) for n in (1..12)]) # G. C. Greubel, May 23 2021

u(n,x) = x*u(n-1,x) + (x+1)*v(n-1,x) + 1,
v(n,x) = x*u(n-1,x) + v(n-1,x) + 1,
where u(1,x) = 1, v(1,x) = 1.
T(n, k) = [x^k]( u(n, x) ), where u(n, x) = (1+x)*u(n-1,x) + x^2*u(n-2,x) + 1 + x, u(1, x) = 1, and u(2, x) = 2 + 2*x. - G. C. Greubel, May 24 2021

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