A052924 -id:A052924 - cp's OEIS frontend

A001077 Numerators of continued fraction convergents to sqrt(5).

1, 2, 9, 38, 161, 682, 2889, 12238, 51841, 219602, 930249, 3940598, 16692641, 70711162, 299537289, 1268860318, 5374978561, 22768774562, 96450076809, 408569081798, 1730726404001, 7331474697802, 31056625195209

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001076(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = -1.
a(2*n) with b(2*n) := A001076(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 5*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,9) = A023039(n), n >= 0 and a(2*n+1) = 2*S(2*n, 2*sqrt(5)) = A075796(n+1), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
From Greg Dresden, May 21 2023: (Start)
For n >= 2, 8*a(n) is the number of ways to tile this T-shaped figure of length n-1 with four colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 8*a(6) = 23112 different tilings.
    _
   || _  
   |||_|||
   |_|
(End)

			1  2  9  38  161  (A001077)
-, -, -, --, ---, ...
0  1  4  17   72  (A001076)
1 + 2*x + 9*x^2 + 38*x^3 + 161*x^4 + 682*x^5 + 2889*x^6 + 12238*x^7 + ... - _Michael Somos_, Aug 11 2009

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
V. Thébault, Les Récréations Mathématiques, Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Ex. 1, pp. 237-238.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A001076, A023039, A049629, A052924, A078343, A164581, A179237, A180148, A329723, A374439.

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001077:=(-1+2*z)/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
with(combinat): a:=n->fibonacci(n+1, 4)-2*fibonacci(n, 4): seq(a(n), n=0..30); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{4, 1}, {1, 2}, 30]
Join[{1},Numerator[Convergents[Sqrt[5],30]]] (* Harvey P. Dale, Mar 23 2016 *)
CoefficientList[Series[(1-2*x)/(1-4*x-x^2), {x, 0, 30}], x] (* G. C. Greubel, Dec 19 2017 *)
LucasL[3*Range[0,30]]/2 (* Rigoberto Florez, Apr 03 2019 *)
a[ n_] := LucasL[n, 4]/2; (* Michael Somos, Nov 02 2021 *)

{a(n) = fibonacci(3*n) / 2 + fibonacci(3*n - 1)}; /* Michael Somos, Aug 11 2009 */

a(n)=if(n<2,n+1,my(t=4);for(i=1,n-2,t=4+1/t);numerator(2+1/t)) \\ Charles R Greathouse IV, Dec 05 2011

x='x+O('x^30); Vec((1-2*x)/(1-4*x-x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, May 14 2009

G.f.: (1-2*x)/(1-4*x-x^2).
a(n) = 4*a(n-1) + a(n-2), a(0)=1, a(1)=2.
a(n) = ((2 + sqrt(5))^n + (2 - sqrt(5))^n)/2.
a(n) = A014448(n)/2.
Limit_{n->infinity} a(n)/a(n-1) = phi^3 = 2 + sqrt(5). - Gregory V. Richardson, Oct 13 2002
a(n) = ((-i)^n)*T(n, 2*i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
Binomial transform of A084057. - Paul Barry, May 10 2003
E.g.f.: exp(2x)cosh(sqrt(5)x). - Paul Barry, May 10 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)*5^k*2^(n-2k). - Paul Barry, Nov 15 2003
a(n) = 4*a(n-1) + a(n-2) when n > 2; a(1) = 1, a(2) = 2. - Alex Vinokur (alexvn(AT)barak-online.net), Oct 25 2004
a(n) = A001076(n+1) - 2*A001076(n) = A097924(n) - A015448(n+1); a(n+1) = A097924(n) + 2*A001076(n) = A097924(n) + 2(A048876(n) - A048875(n)). - Creighton Dement, Mar 19 2005
a(n) = F(3*n)/2 + F(3*n-1) where F() = Fibonacci numbers A000045. - Gerald McGarvey, Apr 28 2007
a(n) = A000032(3*n)/2.
For n >= 1: a(n) = (1/2)*Fibonacci(6*n)/Fibonacci(3*n) and a(n) = integer part of (2 + sqrt(5))^n. - Artur Jasinski, Nov 28 2011
a(n) = Sum_{k=0..n} A201730(n,k)*4^k. - Philippe Deléham, Dec 06 2011
a(n) = A001076(n) + A015448(n). - R. J. Mathar, Jul 06 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) is the (1,1)-entry of the matrix W^n with W=[2, sqrt(5); sqrt(5), 2]. - Carmine Suriano, Mar 21 2014
From Rigoberto Florez, Apr 03 2019: (Start)
a(n) = A099919(n) + A049651(n) if n > 0.
a(n) = 1 + Sum_{k=0..n-1} L(3*k + 1) if n >= 0, L(n) = n-th Lucas number (A000032). (End)
From Christopher Hohl, Aug 22 2021: (Start)
For n >= 2, a(2n-1) = A079962(6n-9) + A079962(6n-3).
For n >= 1, a(2n) = sqrt(20*A079962(6n-3)^2 + 1). (End)
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*4^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*4^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 4^n*Sum_{k=0..n} A374439(n, k)*(-1/4)^k. - Peter Luschny, Jul 26 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) + 5*(-1)^(n+1)/a(n)) = 3/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = b(n) - b(n+1), where b(n) = (1/4) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5*(-1)^(n+1)/a(n)) = 1/8, since 1/(a(n) + 5*(-1)^(n+1)/a(n)) = c(n) + c(n+1), where c(n) = (1/4) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A006189 Number of self-avoiding walks of any length from NW to SW corners of a grid or lattice with n rows and 3 columns.

Original entry on oeis.org

1, 1, 3, 11, 38, 126, 415, 1369, 4521, 14933, 49322, 162900, 538021, 1776961, 5868903, 19383671, 64019918, 211443426, 698350195, 2306494009, 7617832221, 25159990673, 83097804242, 274453403400, 906458014441, 2993827446721, 9887940354603, 32657648510531

Offset: 0

N. J. A. Sloane

a(n) = number of non-self-intersecting (or self-avoiding) paths from upper-left to lower-left of a grid of squares with 3 columns and n rows. E.g., for 3 columns and 2 rows, the paths are D, RDL, and RRDLL and the second a(n) = 3. The next a(n) = 11, which is the number of paths in a 3 X 3 grid: DD, DRDL, DRRDLL, DRURDDLL, RDDL, RDRDLL, RDLD, RRDDLL, RRDDLULD, RRDLDL, RRDLLD (where R=right, L=left, D=down, U=up). - Toby Gottfried, Mar 04 2013

H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Andrew Howroyd, Table of n, a(n) for n = 0..100
H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178. (Annotated scanned copy)
Index entries for linear recurrences with constant coefficients, signature (4,-3,2,1).

Column 3 of A271465.
Cf. A005409 (grids with 3 rows), A001333.
Cf. A214931 (grids with 4 rows).
Cf. A216211 (grids with 4 columns).

I:=[1,3,11,38]; [1] cat [n le 4 select I[n] else 4*Self(n-1) -3*Self(n-2) +2*Self(n-3) +Self(n-4): n in [1..41]]; // G. C. Greubel, May 24 2021

LinearRecurrence[{4,-3,2,1}, {1,1,3,11,38}, 100] (* Jean-François Alcover, Oct 08 2017 *)
With[{U = ChebyshevU}, Table[(1/2)*(U[n, 1/2] -U[n-1, 1/2] + I^n*(U[n, -3*I/2] + I*U[n-1, -3*I/2]) ), {n, 0, 40}]] (* G. C. Greubel, May 24 2021 *)

Vec((1-x)*(1-2*x)/((1-x+x^2)*(1-3*x-x^2)) + O(x^40)) \\ Colin Barker, Nov 17 2017

u=chebyshev_U;
[(1/2)*( u(n, 1/2) - u(n-1, 1/2) + i^n*(u(n, -3*i/2) + i*u(n-1, -3*i/2)) ) for n in (0..30)] # G. C. Greubel, May 24 2021

a(n) = 4*a(n-1) - 3*a(n-2) + 2*a(n-3) + a(n-4) for n > 3. - Giovanni Resta, Mar 13 2013
G.f.: (1-x)*(1-2*x)/((1 - x + x^2)*(1 - 3*x - x^2)). - Colin Barker, Nov 17 2017
2*a(n) = A010892(n) + A052924(n). - R. J. Mathar, Sep 27 2020
a(n) = (1/2)*( ChebyshevU(n, 1/2) - ChebyshevU(n-1, 1/2) + i^n*( ChebyshevU(n, -3*i/2) + i*ChebyshevU(n-1, -3*i/2) ) ). - G. C. Greubel, May 24 2021

Based on upper-left to lower-left path-counting program, more terms from Toby Gottfried, Mar 04 2013
Name clarified, offset changed, a(16)-a(25) from Andrew Howroyd, Apr 07 2016
a(0)=1 prepended by Colin Barker, Nov 17 2017

A108300 a(n+2) = 3*a(n+1) + a(n), with a(0) = 1, a(1) = 5.

Original entry on oeis.org

1, 5, 16, 53, 175, 578, 1909, 6305, 20824, 68777, 227155, 750242, 2477881, 8183885, 27029536, 89272493, 294847015, 973813538, 3216287629, 10622676425, 35084316904, 115875627137, 382711198315, 1264009222082, 4174738864561, 13788225815765, 45539416311856

Offset: 0

Creighton Dement, Jul 24 2005

Binomial transform is A109114.
Invert transform is A109115.
Inverse invert transform is A016777.
Inverse binomial transform is A006130.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Sergio Falcon, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
Tanya Khovanova, Recursive Sequences
Vincent Vatter, Growth rates of permutation classes: from countable to uncountable, arXiv:1605.04297 [math.CO], 2016. (Mentions a signed version.)
Index entries for linear recurrences with constant coefficients, signature (3,1).

Row sums and main diagonal of A143972. - Gary W. Adamson, Sep 06 2008

Cf. A109114, A109115, A016777, A006130, A000004, A052924, A228916, A374439.

seriestolist(series((-2*x-1)/(x^2-1+3*x), x=0,25));

LinearRecurrence[{3,1},{1,5},40] (* Harvey P. Dale, Jul 04 2013 *)

Vec((1 + 2*x)/(1 - 3*x - x^2) + O(x^30)) \\ Andrew Howroyd, Jun 05 2021

G.f.: (1 + 2*x)/(1 - 3*x - x^2).
a(n) = A052924(n+1) - A052924(n).
a(n)*a(n-2) = a(n-1)^2 + 9*(-1)^n. - Roger L. Bagula, May 17 2010
a(n) = 3^n*Sum_{k=0..n} A374439(n, k)*(1/3)^k. - Peter Luschny, Jul 26 2024

A193723 Mirror of the fusion triangle A193722.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 18, 21, 8, 1, 54, 81, 45, 11, 1, 162, 297, 216, 78, 14, 1, 486, 1053, 945, 450, 120, 17, 1, 1458, 3645, 3888, 2295, 810, 171, 20, 1, 4374, 12393, 15309, 10773, 4725, 1323, 231, 23, 1, 13122, 41553, 58320, 47628, 24948, 8694, 2016, 300, 26, 1

Offset: 0

Clark Kimberling, Aug 04 2011

A193723 is obtained by reversing the rows of the triangle A193722.
Triangle T(n,k), read by rows, given by [2,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
From Philippe Deléham, Nov 14 2011: (Start)
Riordan array ((1-x)/(1-3x), x/(1-3x)).
Product A200139*A007318 as infinite lower triangular arrays. (End)

			First six rows:
    1;
    2,   1;
    6,   5,   1;
   18,  21,   8,   1;
   54,  81,  45,  11,   1;
  162, 297, 216,  78,  14,   1;

Cf. A084938, A193722, A052924 (antidiagonal sums), Diagonals: A000012, A016789, A081266, Columns: A025192, A081038.

z = 9; a = 1; b = 1; c = 1; d = 2;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193722 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)

Write w(n,k) for the triangle at A193722.  The triangle at A193723 is then given by w(n,n-k).
T(n,k) = T(n-1,k-1) + 3*T(n-1,k) with T(0,0)=T(1,1)=1 and T(1,0)=2. - Philippe Deléham, Oct 05 2011
From Philippe Deléham, Nov 14 2011: (Start)
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for x=-2,-1,0,1,2,3,4,5,6,7,8,9 respectively.
T(n,k) = Sum_{j>=0} T(n-1-j,k-1)*3^j.
G.f.: (1-x)/(1-(3+y)*x). (End)

A213892 Fixed points of a sequence h(n) defined by the minimum number of 3's in the relation n*[n,3,3,...,3,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 7, 19, 31, 47, 67, 71, 83, 151, 163, 167, 223, 227, 271, 307, 331, 359, 379, 431, 463, 479, 487, 499, 631, 643, 683, 691, 743, 787, 811, 839, 863, 947, 967, 1019, 1051, 1087, 1103, 1123, 1163, 1259, 1279, 1307, 1319, 1399, 1423, 1451, 1471

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,3,3,...,3,n] and increase the number of 3's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 3, 3, 2] = [4, 1, 1, 1, 1, 4],
3 * [3, 3, 3] = [9, 1, 9],
4 * [4, 3, 3, 3, 3, 3, 4] = [17, 4, 1, 2, 1, 4, 17],
5 * [5, 3, 3, 5] = [26, 1, 1, 26],
6 * [6, 3, 3, 3, 3, 3, 6] = [37, 1, 4, 2, 4, 1, 37],
7 * [7, 3, 3, 3, 3, 3, 3, 3, 7] = [51, 8, 2, 1, 2, 8, 51].
The number of 3's needed defines the sequence h(n) = 2, 1, 5, 2, 5, 7, 5, 9, 2, ... (n>=2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n) = 3*f(n-1) + f(n-2), A006190, A003688, A052924, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 3*f(n-1) + f(n-2).
The above sequence h() is recorded as A262213. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891, A213893 - A213899, A261311; A213358.

Cf. A213648, A262212 - A262220, A213900, A262211.

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[3, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,3), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 57, 296, 1537, 7981, 41442, 215191, 1117397, 5802176, 30128277, 156443561, 812346082, 4218173971, 21903215937, 113734253656, 590574484217, 3066606674741, 15923607857922, 82684645964351, 429346837679677, 2229418834362736, 11576441009493357

Offset: 0

Vincenzo Librandi, Aug 17 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A000032, A001077, A052924, A078343, A179237, A180148, A329723.

[ n le 2 select (n) else 5*Self(n-1)+Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Sep 12 2013

LinearRecurrence[{5, 1}, {1, 2}, 40] (* or *) Rest[CoefficientList[Series [x (1 - 3 x) / (1 - 5 x - x^2), {x, 0, 40}], x]] (* Harvey P. Dale, May 02 2011 *)

Vec((1-3*x)/(1-5*x-x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

a(n) = 5*a(n-1)+a(n-2) = A052918(n)-3*A052918(n-1).
G.f.: (1-3*x)/(1-5*x-x^2).
a(n) = A052918(n) + A015449(n). - R. J. Mathar, Jul 06 2012
a(n) = (2^(-1-n)*((5-sqrt(29))^n*(1+sqrt(29))+(-1+sqrt(29))*(5+sqrt(29))^n))/sqrt(29). - Colin Barker, Oct 13 2015
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*5^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*5^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = A052918(n) -3*A052918(n-1). - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) - 7*(-1)^n/a(n)) = 3/10, since 1/(a(n) - 7*(-1)^n/a(n)) = b(n) - b(n+1), where b(n) = (1/5) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 7*(-1)^n/a(n)) = 1/10, since 1/(a(n) - 7*(-1)^n/a(n)) = c(n) + c(n+1), where c(n) = (1/5) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

A100229 Triangle, read by rows, of the coefficients of [x^k] in G100228(x)^n such that the row sums are 4^n-1 for n>0, where G100228(x) is the g.f. of A100228.

Original entry on oeis.org

1, 1, 2, 1, 4, 10, 1, 6, 21, 35, 1, 8, 36, 92, 118, 1, 10, 55, 185, 380, 392, 1, 12, 78, 322, 879, 1506, 1297, 1, 14, 105, 511, 1715, 3948, 5803, 4286, 1, 16, 136, 760, 3004, 8536, 17020, 21904, 14158, 1, 18, 171, 1077, 4878, 16344, 40395, 71109, 81387, 46763

Offset: 0

Paul D. Hanna, Nov 29 2004

The main diagonal forms A100230. Secondary diagonal is T(n+1,n) = (n+1)*A052924(n). More generally, if g.f. F(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]F(x)^n, then F(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](F(x)+z*x)^n for all z and F(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(1-2*x*y+m*x^2*y^2)/((1-x*y)*(1-(m-1)*x*y-x^2*y^2-x*(1-x*y))).

			Rows begin:
[1],
[1,2],
[1,4,10],
[1,6,21,35],
[1,8,36,92,118],
[1,10,55,185,380,392],
[1,12,78,322,879,1506,1297],
[1,14,105,511,1715,3948,5803,4286],
[1,16,136,760,3004,8536,17020,21904,14158],...
where row sums form 4^n-1 for n>0:
4^1-1 = 1+2 = 3
4^2-1 = 1+4+10 = 15
4^3-1 = 1+6+21+35 = 63
4^4-1 = 1+8+36+92+118 = 255
4^5-1 = 1+10+55+185+380+392 = 1023.
The main diagonal forms A100230 = [1,2,10,35,118,392,1297,...],
where Sum_{n>=1} A100230(n)/n*x^n = log((1-x)/(1-3*x-x^2)).

Cf. A100228, A100230, A052924.

PARI
```
T(n,k,m=4)=if(n
				
```

G.f.: A(x, y)=(1-2*x*y+4*x^2*y^2)/((1-x*y)*(1-3*x*y-x^2*y^2-x*(1-x*y))).

A110522 Riordan array (1/(1+x), x(1-2x)/(1+x)^2).

Original entry on oeis.org

1, -1, 1, 1, -5, 1, -1, 12, -9, 1, 1, -22, 39, -13, 1, -1, 35, -115, 82, -17, 1, 1, -51, 270, -344, 141, -21, 1, -1, 70, -546, 1106, -773, 216, -25, 1, 1, -92, 994, -2954, 3199, -1466, 307, -29, 1, -1, 117, -1674, 6888, -10791, 7461, -2487, 414, -33, 1, 1, -145, 2655, -14484, 31179, -30645, 15060, -3900, 537, -37, 1

Offset: 0

Paul Barry, Jul 24 2005

Inverse of A110519.

Product of inverse binomial transform matrix (1/(1+x), x/(1+x)) and (1, x*(1-3*x)) (A110517).

			Rows begin
   1;
  -1,    1;
   1,   -5,    1;
  -1,   12,   -9,    1;
   1,  -22,   39,  -13,    1;
  -1,   35, -115,   82,  -17,    1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, J. Int. Seq. 16 (2013) #13.5.4.

Cf. A110519 (inverse), A110523 (row sums), A110524 (diagonal sums).

Cf. A000326, A016813, A033999, A052924, A078005, A110517, A236267.

A110522:= func< n,k | (-1)^(n+k)*(&+[ 3^(j-k)*Binomial(k,j-k)*Binomial(n,j) : j in [0..n]] ) >;
[A110522(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 28 2023

T[n_,k_]:= Sum[(-1)^(n-j)*(-3)^(j-k)*Binomial[k, j- k]*Binomial[n, j], {j,0,n}];
Table[T[n,k], {n,0,20}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 30 2017 *)

A110522(n,k) = if(n==0, 1, sum(j=0,n, (-1)^(n-j)*(-3)^(j-k)*binomial(n,j)*binomial(k, j-k)));
for(n=0,12, for(k=0,n, print1(A110522(n,k), ", "))) \\ G. C. Greubel, Aug 30 2017; Dec 28 2023

def A110522(n,k): return (-1)^(n+k)*sum(3^(j-k)*binomial(k,j-k)*binomial(n,j) for j in range(n+1))
flatten([[A110522(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 28 2023

Number triangle T(n, k) = Sum_{j=0..n} (-1)^(n-j)*C(n, j)*(-3)^(j-k)*C(k, j-k).
T(n, k) = Sum_{j=0..n} Sum_{i=0..k} C(k, i)*C(n+k-i-j-1, n-k-i-j)*(-1)^(n-k)*2^i.
Sum_{k=0..n} T(n, k) = A110523(n) (row sums).
Sum_{k=0..floor(n/2)} T(n-k, k) = A110524(n) (diagonal sums).
T(n,k) = T(n-1,k-1) - 2*T(n-1,k) - T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(1,0) = -1, T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 12 2014
From G. C. Greubel, Dec 28 2023: (Start)
T(n, 0) = A033999(n).
T(n, 1) = (-1)^(n-1)*A000326(n), n >= 1.
T(n, n) = 1.
T(n, n-1) = -A016813(n-1), n >= 1.
T(n, n-2) = A236267(n-2), n >= 2.
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)^n*A052924(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (-1)^n*A078005(n). (End)

A249579 List of quadruples (r,s,t,u): the matrix M = [[4,12,9][2,5,3][1,2,1]] is raised to successive powers, then (r,s,t,u) are the square roots of M[3,1], M[3,3], M[1,1], M[1,3] respectively.

Original entry on oeis.org

0, 1, 1, 0, 1, 1, 2, 3, 3, 4, 7, 9, 10, 13, 23, 30, 33, 43, 76, 99, 109, 142, 251, 327, 360, 469, 829, 1080, 1189, 1549, 2738, 3567, 3927, 5116, 9043, 11781, 12970, 16897, 29867, 38910, 42837, 55807, 98644, 128511, 141481, 184318, 325799, 424443, 467280

Offset: 0

Russell Walsmith, Nov 02 2014

			M^0 = [[1,0,0][0,1,0][0,0,1]]: r = sqrt(M[3,1]) = a(0) = 0, s = sqrt(M[3,3]) = a(1) = 1, t = sqrt(M[1,1]) = a(2) = 1, u = sqrt(M[1,3])u = a(3) = 0.
M^2 = [[49, 126, 81][21, 55, 36][9, 24, 16]]: r = sqrt(M[3, 1]) = a(8) = 3, s = sqrt(M[3, 3]) = a(9) = 4, t = sqrt(M[1, 1]) = a(10) = 7, u = sqrt(M[1, 3]) = a(11) = 9.

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

Cf. A249580
a(4n) = A006190
a(4n+2) = A052924.

I:=[0,1,1,0,1,1,2,3]; [n le 8 select I[n] else 3*Self(n-4)+Self(n-8): n in [1..50]]; // Vincenzo Librandi, Nov 14 2014

CoefficientList[Series[- x (3 x^6 - x^5 - 2 x^4 + x^3 + x + 1) / (x^8 + 3 x^4 - 1), {x, 0, 50}], x] (* Vincenzo Librandi, Nov 14 2014 *)

concat(0, Vec(-x*(3*x^6-x^5-2*x^4+x^3+x+1)/(x^8+3*x^4-1) + O(x^100))) \\ Colin Barker, Nov 13 2014

Some identities:
a4(n-1) + a(4n) = a(4n+1),
a(4n) + a(4n+1) = a(4n+2),
3a(4n) = a(4n+3).
a(n) = 3*a(n-4)+a(n-8). - Colin Barker, Nov 13 2014
G.f.: -x*(3*x^6-x^5-2*x^4+x^3+x+1) / (x^8+3*x^4-1). - Colin Barker, Nov 13 2014

More terms from Colin Barker, Nov 13 2014

cp's OEIS Frontend

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A110522 Riordan array (1/(1+x), x(1-2x)/(1+x)^2).