A015446 -id:A015446 - cp's OEIS frontend

A109466 Riordan array (1, x(1-x)).

1, 0, 1, 0, -1, 1, 0, 0, -2, 1, 0, 0, 1, -3, 1, 0, 0, 0, 3, -4, 1, 0, 0, 0, -1, 6, -5, 1, 0, 0, 0, 0, -4, 10, -6, 1, 0, 0, 0, 0, 1, -10, 15, -7, 1, 0, 0, 0, 0, 0, 5, -20, 21, -8, 1, 0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1, 0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1, 0, 0, 0, 0, 0, 0, 1, -21, 70, -84, 45, -11, 1, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Aug 28 2005

Inverse is Riordan array (1, xc(x)) (A106566).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Modulo 2, this sequence gives A106344. - Philippe Deléham, Dec 18 2008
Coefficient array of the polynomials Chebyshev_U(n, sqrt(x)/2)*(sqrt(x))^n. - Paul Barry, Sep 28 2009

			Rows begin:
  1;
  0,  1;
  0, -1,  1;
  0,  0, -2,  1;
  0,  0,  1, -3,  1;
  0,  0,  0,  3, -4,   1;
  0,  0,  0, -1,  6,  -5,   1;
  0,  0,  0,  0, -4,  10,  -6,   1;
  0,  0,  0,  0,  1, -10,  15,  -7,  1;
  0,  0,  0,  0,  0,   5, -20,  21, -8,  1;
  0,  0,  0,  0,  0,  -1,  15, -35, 28, -9, 1;
From _Paul Barry_, Sep 28 2009: (Start)
Production array is
  0,    1,
  0,   -1,    1,
  0,   -1,   -1,   1,
  0,   -2,   -1,  -1,   1,
  0,   -5,   -2,  -1,  -1,  1,
  0,  -14,   -5,  -2,  -1, -1,  1,
  0,  -42,  -14,  -5,  -2, -1, -1,  1,
  0, -132,  -42, -14,  -5, -2, -1, -1,  1,
  0, -429, -132, -42, -14, -5, -2, -1, -1, 1 (End)

Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Tom Copeland, Addendum to Elliptic Lie Triad

Cf. A026729 (unsigned version), A000108, A030528, A124644.

/* As triangle */ [[(-1)^(n-k)*Binomial(k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jan 14 2016

(* The function RiordanArray is defined in A256893. *)
RiordanArray[1&, #(1-#)&, 13] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

Number triangle T(n, k) = (-1)^(n-k)*binomial(k, n-k).
T(n, k)*2^(n-k) = A110509(n, k); T(n, k)*3^(n-k) = A110517(n, k).
Sum_{k=0..n} T(n,k)*A000108(k)=1. - Philippe Deléham, Jun 11 2007
From Philippe Deléham, Oct 30 2008: (Start)
Sum_{k=0..n} T(n,k)*A144706(k) = A082505(n+1).
Sum_{k=0..n} T(n,k)*A002450(k) = A100335(n).
Sum_{k=0..n} T(n,k)*A001906(k) = A100334(n).
Sum_{k=0..n} T(n,k)*A015565(k) = A099322(n).
Sum_{k=0..n} T(n,k)*A003462(k) = A106233(n). (End)
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = -12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12 respectively. - Philippe Deléham, Oct 27 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A010892(n), A099087(n), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n) for x = 0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 28 2008
G.f.: 1/(1-y*x+y*x^2). - Philippe Deléham, Dec 15 2011
T(n,k) = T(n-1,k-1) - T(n-2,k-1), T(n,0) = 0^n. - Philippe Deléham, Feb 15 2012
Sum_{k=0..n} T(n,k)*x^(n-k) = F(n+1,-x) where F(n,x)is the n-th Fibonacci polynomial in x defined in A011973. - Philippe Deléham, Feb 22 2013
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Feb 26 2013
Sum_{k=0..n} T(n,k)*T(n+1,k) = -A110320(n). - Philippe Deléham, Feb 26 2013
For T(0,0) = 0, the signed triangle below has the o.g.f. G(x,t) = [t*x(1-x)]/[1-t*x(1-x)] = L[t*Cinv(x)] where L(x) = x/(1-x) and Cinv(x)=x(1-x) with the inverses Linv(x) = x/(1+x) and C(x)= [1-sqrt(1-4*x)]/2, an o.g.f. for the shifted Catalan numbers A000108, so the inverse o.g.f. is Ginv(x,t) = C[Linv(x)/t] = [1-sqrt[1-4*x/(t(1+x))]]/2 (cf. A124644 and A030528). - Tom Copeland, Jan 19 2016

A015447 Generalized Fibonacci numbers: a(n) = a(n-1) + 11*a(n-2).

Original entry on oeis.org

1, 1, 12, 23, 155, 408, 2113, 6601, 29844, 102455, 430739, 1557744, 6295873, 23431057, 92685660, 350427287, 1369969547, 5224669704, 20294334721, 77765701465, 301003383396, 1156426099511, 4467463316867, 17188150411488

Offset: 0

Olivier Gérard

The compositions of n in which each positive integer is colored by one of p different colors are called p-colored compositions of n. For n>=2, 12*a(n-2) equals the number of 12-colored compositions of n, with all parts >= 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,11).

Cf. A015446, A015443.

[n le 2 select 1 else Self(n-1) + 11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012

Join[{a=1,b=1},Table[c=b+11*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{1,11},{1,1},30] (* or *) CoefficientList[Series[ 1/(1-x-11 x^2),{x,0,50}],x] (* Harvey P. Dale, May 08 2011 *)

Vec(1/(1-x-11*x^2)+O(x^99)) \\ Charles R Greathouse IV, Feb 03 2014

[lucas_number1(n,1,-11) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

a(n) = ( ( (1+3*sqrt(5))/2 )^(n+1) - ( (1-3*sqrt(5))/2 )^(n+1) )/(3*sqrt(5)).
a(n-1) = (1/3)*(-1)^n*Sum_{i=0..n} (-3)^i*Fibonacci(i)*C(n, i). - Benoit Cloitre, Mar 06 2004
a(n) = Sum_{k=0..n} A109466(n,k)*(-11)^(n-k). - Philippe Deléham, Oct 26 2008
G.f.: 1/(1 - x - 11*x^2). - Harvey P. Dale, May 08 2011
a(n) = ( Sum_{1<=k<=n+1, k odd} C(n+1,k)*45^((k-1)/2) )/2^n. - Vladimir Shevelev, Feb 05 2014

A111006 Another version of Fibonacci-Pascal triangle A037027.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 0, 2, 3, 0, 0, 1, 5, 5, 0, 0, 0, 3, 10, 8, 0, 0, 0, 1, 9, 20, 13, 0, 0, 0, 0, 4, 22, 38, 21, 0, 0, 0, 0, 1, 14, 51, 71, 34, 0, 0, 0, 0, 0, 5, 40, 111, 130, 55, 0, 0, 0, 0, 0, 1, 20, 105, 233, 235, 89, 0, 0, 0, 0, 0, 0, 6, 65, 256, 474, 420, 144

Offset: 0

Philippe Deléham, Oct 02 2005

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.
Row sums are the Jacobsthal numbers A001045(n+1) and column sums form Pell numbers A000129.
Maximal column entries: A038149 = {1, 1, 2, 5, 10, 22, ...}.
T(n,k) gives a convolved Fibonacci sequence (A001629, A001872, ...).
Triangle read by rows: T(n,n-k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)). - Philippe Deléham, Feb 17 2014
Diagonal sums are A013979(n). - Philippe Deléham, Feb 17 2014
T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and 1 X 2 tiles. - Emeric Deutsch, Aug 14 2014

			Triangle begins:
  1;
  0, 1;
  0, 1, 2;
  0, 0, 2, 3;
  0, 0, 1, 5,  5;
  0, 0, 0, 3, 10,  8;
  0, 0, 0, 1,  9, 20, 13;
  0, 0, 0, 0,  4, 22, 38,  21;
  0, 0, 0, 0,  1, 14, 51,  71,  34;
  0, 0, 0, 0,  0,  5, 40, 111, 130,  55;
  0, 0, 0, 0,  0,  1, 20, 105, 233, 235,  89;
  0, 0, 0, 0,  0,  0,  6,  65, 256, 474, 420, 144;

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000045, A000129, A001045, A037027, A038112, A038149, A084938, A128100 (reversed version).

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A114197, A162741, A228074.

a111006 n k = a111006_tabl !! n !! k
a111006_row n = a111006_tabl !! n
a111006_tabl =  map fst $ iterate (\(us, vs) ->
   (vs, zipWith (+) (zipWith (+) ([0] ++ us ++ [0]) ([0,0] ++ us))
                    ([0] ++ vs))) ([1], [0,1])
-- Reinhard Zumkeller, Aug 15 2013

T(0, 0) = 1, T(n, k) = 0 for k < 0 or for n < k, T(n, k) = T(n-1, k-1) + T(n-2, k-1) + T(n-2, k-2).
T(n, k) = A037027(k, n-k). T(n, n) = A000045(n+1). T(3n, 2n) = (n+1)*A001002(n+1) = A038112(n).
G.f.: 1/(1-yx(1-x)-x^2*y^2). - Paul Barry, Oct 04 2005
Sum_{k=0..n} x^k*T(n,k) = (-1)^n*A053524(n+1), (-1)^n*A083858(n+1), (-1)^n*A002605(n), A033999(n), A000007(n), A001045(n+1), A083099(n) for x = -4, -3, -2, -1, 0, 1, 2 respectively. - Philippe Deléham, Dec 02 2006
Sum_{k=0..n} T(n,k)*x^(n-k) = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 respectively. - Philippe Deléham, Feb 17 2014

A186446 Expansion of 1/(1 - 7x + 2x^2).

Original entry on oeis.org

1, 7, 47, 315, 2111, 14147, 94807, 635355, 4257871, 28534387, 191224967, 1281505995, 8588092031, 57553632227, 385699241527, 2584787426235, 17322113500591, 116085219651667, 777952310560487, 5213495734620075, 34938565521219551

Offset: 0

Bruno Berselli, Feb 21 2011

The first differences are in A122074.

a(n+1) equals the number of words of length n over {0,1,2,3,4,5,6} avoiding 01 and 02. - Milan Janjic, Dec 17 2015

Bruno Berselli, Table of n, a(n) for n = 0..800
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607.
Index entries for linear recurrences with constant coefficients, signature (7,-2).

For similar closed formulas: A015446 [((1+sqrt(41))^(1+n)-(1-sqrt(41))^(1+n))/(2^(1+n)*sqrt(41))], A015525 [((3+sqrt(41))^n-(3-sqrt(41))^n)/(2^n*sqrt(41))], A015537 [((5+sqrt(41))^n-(5-sqrt(41))^n)/(2^n*sqrt(41))], A178869 [((9+sqrt(41))^n-(9-sqrt(41))^n)/(2^n*sqrt(41))].

Same recurrence as in A122074, A003771.

m:=21; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-7*x+2*x^2)));

I:=[1,7]; [n le 2 select I[n] else 7*Self(n-1)-2*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 19 2013

CoefficientList[Series[1 / (1 - 7 x + 2 x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{7,-2},{1,7},30] (* Harvey P. Dale, Aug 06 2017 *)

Vec(1/(1-7*x+2*x^2) + O(x^100)) \\ Altug Alkan, Dec 17 2015

G.f.: 1/(1-7*x+2*x^2).
a(n) = ((7+sqrt(41))^(1+n)-(7-sqrt(41))^(1+n))/(2^(1+n)*sqrt(41)).
a(n) = 7*a(n-1)-2*a(n-2), with a(0)=1, a(1)=7.

A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 5, 5, 1, 5, 7, 11, 8, 1, 6, 9, 19, 21, 13, 1, 7, 11, 29, 40, 43, 21, 1, 8, 13, 41, 65, 97, 85, 34, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1, 10, 17, 71, 133, 301, 441, 508, 341, 89, 1, 11, 19, 89, 176, 463, 781, 1165, 1159, 683, 144, 1, 12, 21, 109, 225, 673

Offset: 1

R. H. Hardin, Jul 24 2011

Transposed variant of A083856. - R. J. Mathar, Aug 23 2011

As to the sequences by columns beginning (1, N, ...), let m = (N-1). The g.f. for the sequence (1, N, ...) is 1/(1 - x - m*x^2). Alternatively, the corresponding matrix generator is [[1,1], [m,0]]. Another equivalency is simply: The sequence beginning (1, N, ...) is the INVERT transform of (1, m, 0, 0, 0, ...). Convergents to the sequences a(n)/a(n-1) are (1 + sqrt(4*m+1))/2. - Gary W. Adamson, Feb 25 2014

			Array T(n,k) (with rows n >= 1 and column k >= 1) begins as follows:
  ..1...1....1....1.....1.....1.....1......1......1......1......1......1...
  ..2...3....4....5.....6.....7.....8......9.....10.....11.....12.....13...
  ..3...5....7....9....11....13....15.....17.....19.....21.....23.....25...
  ..5..11...19...29....41....55....71.....89....109....131....155....181...
  ..8..21...40...65....96...133...176....225....280....341....408....481...
  .13..43...97..181...301...463...673....937...1261...1651...2113...2653...
  .21..85..217..441...781..1261..1905...2737...3781...5061...6601...8425...
  .34.171..508.1165..2286..4039..6616..10233..15130..21571..29844..40261...
  .55.341.1159.2929..6191.11605.19951..32129..49159..72181.102455.141361...
  .89.683.2683.7589.17621.35839.66263.113993.185329.287891.430739.624493...
  ...
Some solutions for n = 5 and k = 3 with colors = 1, 2, 3 and empty = 0:
..0....2....3....2....0....1....0....0....2....0....0....2....3....0....0....0
..0....2....3....2....2....1....2....3....2....1....0....2....3....1....1....1
..1....0....0....0....2....0....2....3....2....1....0....1....0....1....1....1
..1....2....2....0....3....2....2....3....2....0....3....1....3....3....2....1
..0....2....2....0....3....2....2....3....0....0....3....0....3....3....2....1

R. H. Hardin, Table of n, a(n) for n = 1..9999
Ron H. Hardin, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.
Robert Israel, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.

Column 1 is A000045(n+1), column 2 is A001045(n+1), column 3 is A006130, column 4 is A006131, column 5 is A015440, column 6 is A015441(n+1), column 7 is A015442(n+1), column 8 is A015443, column 9 is A015445, column 10 is A015446, column 11 is A015447, and column 12 is A053404,
Row 2 is A000027(n+1), row 3 is A004273(n+1), row 4 is A028387, row 5 is A000567(n+1), and row 6 is A106734(n+2).
Diagonal is A171180, superdiagonal 1 is A083859(n+1), and superdiagonal 2 is A083860(n+1).

T:= proc(n,k) option remember; `if`(n<0, 0,
      `if`(n<2 or k=0, 1, k*T(n-2, k) +T(n-1, k)))
    end;
seq(seq(T(n, d+1-n), n=1..d), d=1..12); # Alois P. Heinz, Jul 29 2011

T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 2 || k == 0, 1, k*T[n-2, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 12}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. Thus, T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n = 0, 1, ..., z-1.
The solution is T(n,k) = Sum_r r^(-n-1)/(1 + z*k*r^(z-1)), where the sum is over the roots r of the polynomial k*x^z + x - 1.
For z = 2, T(n,k) = ((2*k / (sqrt(1 + 4*k) - 1))^(n+1) - (-2*k/(sqrt(1 + 4*k) + 1))^(n+1)) / sqrt(1 + 4*k).
T(n,k) = Sum_{s=0..[n/2]} binomial(n-s,s) * k^s.
For z X 1 tiles, T(n,k,z) = Sum_{s = 0..[n/z]} binomial(n-(z-1)*s, s) * k^s. - R. H. Hardin, Jul 31 2011

Formula and proof from Robert Israel in the Sequence Fans mailing list.

A350470 Array read by ascending antidiagonals. T(n, k) = J(k, n) where J are the Jacobsthal polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 11, 1, 1, 1, 9, 13, 29, 21, 1, 1, 1, 11, 17, 55, 65, 43, 1, 1, 1, 13, 21, 89, 133, 181, 85, 1, 1, 1, 15, 25, 131, 225, 463, 441, 171, 1, 1, 1, 17, 29, 181, 341, 937, 1261, 1165, 341, 1

Offset: 0

Peter Luschny, Mar 19 2022

			Array starts:
n\k 0, 1,  2,  3,   4,    5,    6,     7,      8,      9, ...
---------------------------------------------------------------------
[0] 1, 1,  1,  1,   1,    1,    1,     1,      1,      1, ... A000012
[1] 1, 1,  3,  5,  11,   21,   43,    85,    171,    341, ... A001045
[2] 1, 1,  5,  9,  29,   65,  181,   441,   1165,   2929, ... A006131
[3] 1, 1,  7, 13,  55,  133,  463,  1261,   4039,  11605, ... A015441
[4] 1, 1,  9, 17,  89,  225,  937,  2737,  10233,  32129, ... A015443
[5] 1, 1, 11, 21, 131,  341, 1651,  5061,  21571,  72181, ... A015446
[6] 1, 1, 13, 25, 181,  481, 2653,  8425,  40261, 141361, ... A053404
[7] 1, 1, 15, 29, 239,  645, 3991, 13021,  68895, 251189, ... A350468
[8] 1, 1, 17, 33, 305,  833, 5713, 19041, 110449, 415105, ... A168579
[9] 1, 1, 19, 37, 379, 1045, 7867, 26677, 168283, 648469, ... A350469
      A005408 | A082108 |
           A016813   A014641

Rows: A000012, A001045, A006131, A015441, A015443, A015446, A053404, A350468, A168579, A350469.
Columns: A000012, A005408, A016813, A082108, A014641.
Cf. A350467 (main diagonal), A352361 (Fibonacci polynomials), A352362 (Lucas polynomials).

J := (n, x) -> add(2^k*binomial(n - k, k)*x^k, k = 0..n):
seq(seq(J(k, n-k), k = 0..n), n = 0..10);

T[n_, k_] := Hypergeometric2F1[(1 - k)/2, -k/2, -k, -8 n];
Table[T[n, k], {n, 0, 9}, {k, 0, 9}] // TableForm
(* or *)
T[n_, k_] := With[{s = Sqrt[8*n+1]}, ((1+s)^(k+1) - (1-s)^(k+1)) / (2^(k+1)*s)];
Table[Simplify[T[n, k]], {n, 0, 9}, {k, 0, 9}] // TableForm

T(n, k) = ([1, 2; k, 0]^n)[1, 1] ;
export(T)
for(k = 0, 9, print(parvector(10, n, T(n - 1, k))))

T(n, k) = Sum_{j=0..k} binomial(k - j, j)*(2*n)^j.
T(n, k) = ((1+s)^(k+1) - (1-s)^(k+1)) / (2^(k+1)*s) where s = sqrt(8*n + 1).
T(n, k) = [x^k] (1 / (1 - x - 2*n*x^2)).
T(n, k) = hypergeom([1/2 - k/2, -k/2], [-k], -8*n).

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 44, 312, 2224, 15840, 112832, 803712, 5724928, 40779264, 290475008, 2069084160, 14738305024, 104982503424, 747801460736, 5326668791808, 37942424436736, 270267896954880, 1925146777223168, 13713023838978048, 97679317251653632, 695780094221746176

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,8).

Sequences of the form a(n) = c*a(n-1) + d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A000045,A001045,A006130,A006131,A015440,A015441,A015442,A015443,A015445,A015446
.A000129,A002605,A015518,A063727,A002532,A083099,A015519,A003683,A002534,A083102
.A006190,A007482,A030195,A015521,A015523,A083858,A015524,A015525,A099012,A015528
.A001076,A090017,A015530,A057087,A015531,A085939,A015532,A180222,A015533,A180226
.A052918,A015535,A015536,A015537,A057088,A015540,A015541,A015544,A015545,A180250
.A005668,A135030,A090018,A135032,A015551,A057089,A015552,A189800,A189801,A190005
.A054413,A015555,A015559,A015561,A015562,A015564,A057090,A015565,A015566,A015568
.A041025,A190331,A015574,A190510,A015575,A190560,A015576,A057091,A015577,A190953
.A099371,A015579,A181353,A015580,A015581,A153191,A015583,A015584,A057092,A015585
A041041,A191014,A015588,A190954,A190955,A190956,A015589,A190957,A015591,A057093

I:=[0,1]; [n le 2 select I[n] else 6*Self(n-1)+8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

LinearRecurrence[{6, 8}, {0, 1}, 50]
CoefficientList[Series[-(x/(-1+6 x+8 x^2)),{x,0,50}],x] (* Harvey P. Dale, Jul 26 2011 *)

a(n)=([0,1; 8,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

G.f.: x/(1 - 2*x*(3+4*x)). - Harvey P. Dale, Jul 26 2011

A060959 Table by antidiagonals of generalized Fibonacci numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=0 and T(n,1)=1.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 5, 5, 4, 1, 1, 0, 1, 8, 11, 7, 5, 1, 1, 0, 1, 13, 21, 19, 9, 6, 1, 1, 0, 1, 21, 43, 40, 29, 11, 7, 1, 1, 0, 1, 34, 85, 97, 65, 41, 13, 8, 1, 1, 0, 1, 55, 171, 217, 181, 96, 55, 15, 9, 1, 1, 0, 1, 89, 341, 508, 441, 301, 133, 71, 17, 10, 1, 1, 0

Offset: 0

Henry Bottomley, May 10 2001

			Square array begins as:
  0, 1, 1, 1,  1,  1,  1, ...
  0, 1, 1, 2,  3,  5,  8, ...
  0, 1, 1, 3,  5, 11, 21, ...
  0, 1, 1, 4,  7, 19, 40, ...
  0, 1, 1, 5,  9, 29, 65, ...
  0, 1, 1, 6, 11, 41, 96, ...

G. C. Greubel, Antidiagonals n = 0..100, flattened

Rows include A057427, A000045, A001045, A006130, A006131, A015440, A015441, A015442, A015443, A015445, A015446, A015447, A053404.

Columns include A000004, A000012 (twice), A000027, A000567 A005408, A028387.

Flat(List([0..12], n-> List([0..n], k-> (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k))/Sqrt(1+4*k) ))); # G. C. Greubel, Jan 15 2020

[Round( (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k) )/Sqrt(1+4*k) ): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 15 2020

seq(seq( round((((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k)), k=0..n), n=0..12); # G. C. Greubel, Jan 15 2020

T[n_, k_]:= If[n==k==0, 0, Round[(((1+Sqrt[1+4n])/2)^k - ((1-Sqrt[1+4n])/2)^k)/Sqrt[1+4n]]]; Table[T[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 15 2020 *)

T(n,k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n);
for(n=0,12, for(k=0,n, print1( round(T(k,n-k)), ", "))) \\ G. C. Greubel, Jan 15 2020

[[ round( (((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 15 2020

T(n, k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n).

A109447 Binomial coefficients C(n,k) with n-k odd, read by rows.

Original entry on oeis.org

1, 2, 1, 3, 4, 4, 1, 10, 5, 6, 20, 6, 1, 21, 35, 7, 8, 56, 56, 8, 1, 36, 126, 84, 9, 10, 120, 252, 120, 10, 1, 55, 330, 462, 165, 11, 12, 220, 792, 792, 220, 12, 1, 78, 715, 1716, 1287, 286, 13, 14, 364, 2002, 3432, 2002, 364, 14, 1, 105, 1365, 5005, 6435, 3003, 455, 15

Offset: 1

Philippe Deléham, Aug 27 2005

The same as A119900 without 0's. A reflected version of A034867 or A202064. - Alois P. Heinz, Feb 07 2014
From Vladimir Shevelev, Feb 07 2014: (Start)
Also table of coefficients of polynomials  P_1(x)=1, P_2(x)=2, for n>=2, P_(n+1)(x) = 2*P_n(x)+(x-1)* P_(n-1)(x).  The polynomials P_n(x)/2^(n-1) are connected with sequences A000045 (x=5), A001045 (x=9), A006130 (x=13), A006131 (x=17), A015440 (x=21), A015441 (x=25), A015442 (x=29), A015443 (x=33), A015445 (x=37), A015446 (x=41), A015447 (x=45), A053404 (x=49); also the polynomials P_n(x) are connected with sequences A000129, A002605, A015518, A063727, A085449, A002532, A083099, A015519, A003683, A002534, A083102, A015520. (End)

			Starred terms in Pascal's triangle (A007318), read by rows:
1;
1*, 1;
1, 2*, 1;
1*, 3, 3*, 1;
1, 4*, 6, 4*, 1;
1*, 5, 10*, 10, 5*, 1;
1, 6*, 15, 20*, 15, 6*, 1;
1*, 7, 21*, 35, 35*, 21, 7*, 1;
1, 8*, 28, 56*, 70, 56*, 28, 8*, 1;
1*, 9, 36*, 84, 126*, 126, 84*, 36, 9*, 1;
Triangle T(n,k) begins:
1;
2;
1,    3;
4,    4;
1,   10,  5;
6,   20,  6;
1,   21,  35,   7;
8,   56,  56,   8;
1,   36, 126,  84,  9;
10, 120, 252, 120, 10;

Alois P. Heinz, Rows n = 1..200, flattened

Cf. A109446.

T:= (n, k)-> binomial(n, 2*k+1-irem(n, 2)):
seq(seq(T(n, k), k=0..ceil((n-2)/2)), n=1..20);  # Alois P. Heinz, Feb 07 2014

Flatten[ Table[ If[ OddQ[n - k], Binomial[n, k], {}], {n, 0, 15}, {k, 0, n}]] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Aug 30 2005

Corrected offset by Alois P. Heinz, Feb 07 2014

A128100 Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 5, 1, 8, 10, 3, 13, 20, 9, 1, 21, 38, 22, 4, 34, 71, 51, 14, 1, 55, 130, 111, 40, 5, 89, 235, 233, 105, 20, 1, 144, 420, 474, 256, 65, 6, 233, 744, 942, 594, 190, 27, 1, 377, 1308, 1836, 1324, 511, 98, 7, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 987, 3970

Offset: 0

Emeric Deutsch, Feb 18 2007

Row sums are the Jacobsthal numbers (A001045). Column 0 yields the Fibonacci numbers (A000045); the other columns yield convolved Fibonacci numbers (A001629, A001628, A001872, A001873, etc.). Sum_{k=0..floor(n/2)} k*T(n,k) = A073371(n-2).
Triangle T(n,k), with zeros omitted, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 24 2012
Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - Philippe Deléham, Feb 06 2012
Diagonal sums are A000073(n+2) (tribonacci numbers). - Philippe Deléham, Feb 16 2014
Number of induced subgraphs of the Fibonacci cube Gamma(n-1) that are isomorphic to the hypercube Q_k. Example: row n=4 is 5, 5, 1; indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; it has 5 vertices (i.e., Q_0's), 5 edges (i.e., Q_1's) and 1 square (i.e., Q_2). - Emeric Deutsch, Aug 12 2014
Row n gives the coefficients of the polynomial p(n,x) defined as the numerator of the rational function given by f(n,x) = 1 + (x + 1)/f(n-1,x), where f(x,0) = 1. Conjecture: for n > 2, p(n,x) is irreducible if and only if n is a (prime - 2). - Clark Kimberling, Oct 22 2014

			Triangle starts:
   1;
   1;
   2,  1;
   3,  2;
   5,  5,  1;
   8, 10,  3;
  13, 20,  9,  1;
  21, 38, 22,  4;
From _Philippe Deléham_, Jan 24 2012: (Start)
Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
   1;
   1,  0;
   2,  1,  0;
   3,  2,  0,  0;
   5,  5,  1,  0,  0;
   8, 10,  3,  0,  0,  0;
  13, 20,  9,  1,  0,  0,  0;
  21, 38, 22,  4,  0,  0,  0,  0; (End)
From _Clark Kimberling_, Oct 22 2014: (Start)
Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
  1
  2 +  x
  3 + 2x
  5 + 5x + x^2. (End)

C.-P. Chou and H. A. Witek, An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014

Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.

Cf. A109466, A119473.

G:=1/(1-z-(1+t)*z^2): Gser:=simplify(series(G,z=0,19)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form

p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
Numerator[Table[Factor[p[x, n]], {n, 1, 20}]]  (* Clark Kimberling, Oct 22 2014 *)

G.f.: 1/(1-z-(1+t)z^2).
Sum_{k=0..n} T(n,k)*x^k = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, and -13, respectively. - Philippe Deléham, Jan 24 2012
T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Jan 24 2012
G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 06 2013
T(n,k) = Sum_{i=k..floor(n/2)} binomial(n-i,i)*binomial(i,k). See Corollary 3.3 in the Klavzar et al. link. - Emeric Deutsch, Aug 12 2014

cp's OEIS Frontend

A109466 Riordan array (1, x(1-x)).

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A015447 Generalized Fibonacci numbers: a(n) = a(n-1) + 11*a(n-2).

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A111006 Another version of Fibonacci-Pascal triangle A037027.

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A186446 Expansion of 1/(1 - 7*x + 2*x^2).

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A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

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A350470 Array read by ascending antidiagonals. T(n, k) = J(k, n) where J are the Jacobsthal polynomials.

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A189800 a(n) = 6*a(n-1) + 8*a(n-2), with a(0)=0, a(1)=1.

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A186446 Expansion of 1/(1 - 7x + 2x^2).

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.