A053404 -id:A053404 - cp's OEIS frontend

A180250 a(n) = 5a(n-1) + 10a(n-2), with a(1)=0 and a(2)=1.

0, 1, 5, 35, 225, 1475, 9625, 62875, 410625, 2681875, 17515625, 114396875, 747140625, 4879671875, 31869765625, 208145546875, 1359425390625, 8878582421875, 57987166015625, 378721654296875, 2473479931640625, 16154616201171875, 105507880322265625

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 16 2011

Nathaniel Johnston, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (5,10).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015440, A015441, A015443, A015444, A015445, A015447, A030195, A053404, A057087, A057088, A083858, A085939, A090017, A091914, A099012, A180222, A180226.

[n le 2 select n-1 else 5*Self(n-1) +10*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2018

Join[{a=0,b=1},Table[c=5*b+10*a;a=b;b=c,{n,100}]]
LinearRecurrence[{5,10}, {0,1}, 30] (* G. C. Greubel, Jan 16 2018 *)

a(n)=([0,1;10,5]^(n-1))[1,2] \\ Charles R Greathouse IV, Oct 03 2016

my(x='x+O('x^30)); concat([0], Vec(x^2/(1-5*x-10*x^2))) \\ G. C. Greubel, Jan 16 2018

A180250= BinaryRecurrenceSequence(5,10,0,1)
[A180250(n-1) for n in range(1,41)] # G. C. Greubel, Jul 21 2023

a(n) = ((5+sqrt(65))^(n-1) - (5-sqrt(65))^(n-1))/(2^(n-1)*sqrt(65)). - Rolf Pleisch, May 14 2011
G.f.: x^2/(1-5*x-10*x^2).
a(n) = (i*sqrt(10))^(n-1) * ChebyshevU(n-1, -i*sqrt(5/8)). - G. C. Greubel, Jul 21 2023

A015551 Expansion of x/(1 - 6x - 5x^2).

Original entry on oeis.org

0, 1, 6, 41, 276, 1861, 12546, 84581, 570216, 3844201, 25916286, 174718721, 1177893756, 7940956141, 53535205626, 360916014461, 2433172114896, 16403612761681, 110587537144566, 745543286675801, 5026197405777636

Offset: 0

Olivier Gérard

Let the generator matrix for the ternary Golay G_12 code be [I|B], where the elements of B are taken from the set {0,1,2}. Then a(n)=(B^n)1,2 for instance. - _Paul Barry, Feb 13 2004

Pisano period lengths: 1, 2, 4, 4, 1, 4, 42, 8, 12, 2, 10, 4, 12, 42, 4, 16, 96, 12, 360, 4, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Lucyna Trojnar-Spelina, Iwona Włoch, On Generalized Pell and Pell-Lucas Numbers, Iranian Journal of Science and Technology, Transactions A: Science (2019), 1-7.
Index entries for linear recurrences with constant coefficients, signature (6,5).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015440, A015441, A015443, A015444, A015445, A015447, A015548, A030195, A053404, A057087, A057088, A057089, A083858, A085939, A090017, A091914, A099012, A135030, A135032, A180222, A180226, A180250.

I:=[0,1]; [n le 2 select I[n] else 6*Self(n-1)+5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

Join[{a=0,b=1},Table[c=6*b+5*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
CoefficientList[Series[x/(1-6x-5x^2),{x,0,20}],x] (* or *) LinearRecurrence[ {6,5},{0,1},30] (* Harvey P. Dale, Oct 30 2017 *)

a(n)=([0,1; 5,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

[lucas_number1(n,6,-5) for n in range(0, 21)] # Zerinvary Lajos, Apr 24 2009

a(n) = 6*a(n-1) + 5*a(n-2).

a(n) = sqrt(14)*(3+sqrt(14))^n/28 - sqrt(14)*(3-sqrt(14))^n/28. - Paul Barry, Feb 13 2004

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.

A072024 Table by antidiagonals of T(n,k) = ((n+1)^k - (-n)^k)/(2*n+1).

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 3, 1, 1, 0, 1, 5, 7, 1, 1, 0, 1, 11, 13, 13, 1, 1, 0, 1, 21, 55, 25, 21, 1, 1, 0, 1, 43, 133, 181, 41, 31, 1, 1, 0, 1, 85, 463, 481, 461, 61, 43, 1, 1, 0, 1, 171, 1261, 2653, 1281, 991, 85, 57, 1, 1, 0, 1, 341, 4039, 8425, 10501, 2821, 1891, 113, 73, 1, 1, 0

Offset: 0

Henry Bottomley, Jun 06 2002

Rows of the array have g.f. x/((1+k*x)*(1-(k+1)*x)). - Philippe Deléham, Nov 24 2013

			Rows start:
0 1 1  1   1    1     1      1       1        1 ...
0 1 1  3   5   11    21     43      85      171 ...
0 1 1  7  13   55   133    463    1261     4039 ...
0 1 1 13  25  181   481   2653    8425    40261 ...
0 1 1 21  41  461  1281  10501   36121   246141 ...
0 1 1 31  61  991  2821  32551  117181  1093711 ...
0 1 1 43  85 1891  5461  84883  314245  3879331 ...
0 1 1 57 113 3305  9633 194713  734161 11638089 ...
...

Andrew Howroyd, Table of n, a(n) for n = 0..1275

Rows include A057427, A001045, A015441, A053404, A053428, A053430, A065874, etc. Columns include A000004, A000012, A000012, A002061, A001844, A072025, etc.

Cf. A081297.

[((k+1)^(n-k) - (-k)^(n-k))/(2*k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 27 2020

seq(seq( ((k+1)^(n-k) - (-k)^(n-k))/(2*k+1), k=0..n), n=0..12); # G. C. Greubel, Jan 27 2020

T[n_, k_]:= ((n + 1)^k - (-n)^k)/(2n + 1); Flatten[Join[{0}, Table[T[k, n- k], {n, 1, 15}, {k, 0, n}]]] (* Indranil Ghosh, Mar 27 2017 *)

for(n=0, 10, for(k=0, 9, print1(((n+1)^k-(-n)^k)/(2*n+1), ", "); ); print(); ) \\ Andrew Howroyd, Mar 26 2017

def T(n, k): return ((n+1)^k - (-n)^k)/(2*n+1)
[[T(k,n-k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 27 2020

T(n, k) = T(n, k-1) + n*(n+1)*T(n, k-2) = A060959(A002378(n), k).

T(k, 2n) = (2n+1)*A047969(n, k+1).

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).

A053430 a(n) = (6^(n+1) - (-5)^(n+1))/11.

Original entry on oeis.org

1, 1, 31, 61, 991, 2821, 32551, 117181, 1093711, 4609141, 37420471, 175694701, 1298308831, 6569149861, 45518414791, 242592910621, 1608145354351, 8885932672981, 57130293303511, 323708273492941, 2037617072598271

Offset: 0

Barry E. Williams, Jan 10 2000

Hankel transform is := 1,30,0,0,0,0,0,0,0,0,0,0,... - Philippe Deléham, Nov 02 2008

The ratio a(n+1)/a(n) converges to 6 as n approaches infinity. - Felix P. Muga II, Mar 10 2014

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
Index entries for linear recurrences with constant coefficients, signature (1,30).

Cf. A001045, A015441, A053404.

[(6^(n+1)-(-5)^(n+1))/11: n in [0..30]]; // Vincenzo Librandi, Mar 11 2014

A053430:=n->( 6^(n+1)-(-5)^(n+1) )/11; seq(A053430(n), n=0..20); # Wesley Ivan Hurt, Mar 11 2014

Join[{a=1,b=1},Table[c=b+30*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
Table[(6^(n+1)-(-5)^(n+1))/11,{n,0,20}] (* Harvey P. Dale, May 09 2012 *)
LinearRecurrence[{1,30},{1,1},21] (* Harvey P. Dale, May 09 2012 *)
CoefficientList[Series[-1/(5 x + 1)/(6 x - 1), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 11 2014 *)

a(n) = ( 6^(n+1)-(-5)^(n+1) )/11; \\ Joerg Arndt, Mar 10 2014

G.f.: -1/(5*x+1)/(6*x-1). - R. J. Mathar, Nov 16 2007

a(0)=1, a(1)=1, a(n) = a(n-1) + 30*a(n-2). - Harvey P. Dale, May 09 2012

More terms from James Sellers, Feb 02 2000

A015541 Expansion of x/(1 - 5x - 7x^2).

Original entry on oeis.org

0, 1, 5, 32, 195, 1199, 7360, 45193, 277485, 1703776, 10461275, 64232807, 394392960, 2421594449, 14868722965, 91294775968, 560554940595, 3441838134751, 21133075257920, 129758243232857, 796722742969725, 4891921417478624, 30036666288181195

Offset: 0

Olivier Gérard

Pisano period lengths: 1, 3, 8, 6, 8, 24, 6, 6, 24, 24, 5, 24, 12, 6, 8, 12, 16, 24, 120, 24, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,7).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015443, A015447, A030195, A053404, A057087, A057088, A083858, A085939, A090017, A091914, A099012, A180222, A180226.

[n le 2 select n-1 else 5*Self(n-1) + 7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 13 2012

Join[{a=0,b=1},Table[c=5*b+7*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{5, 7}, {0, 1}, 30] (* Vincenzo Librandi, Nov 13 2012 *)

x='x+O('x^30); concat([0], Vec(x/(1-5*x-7*x^2))) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,5,-7) for n in range(0, 21)] # Zerinvary Lajos, Apr 24 2009

a(n) = 5*a(n-1) + 7*a(n-2).

A015544 Lucas sequence U(5,-8): a(n+1) = 5a(n) + 8a(n-1), a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 5, 33, 205, 1289, 8085, 50737, 318365, 1997721, 12535525, 78659393, 493581165, 3097180969, 19434554165, 121950218577, 765227526205, 4801739379641, 30130517107845, 189066500576353, 1186376639744525, 7444415203333449, 46713089134623445

Offset: 0

Olivier Gérard

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,8).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015537, A015441, A015443, A015447, A030195, A053404, A057087, A057088, A083858, A085939, A090017, A091914, A099012, A180222, A180226, A015555 (binomial transform).

[n le 2 select n-1 else 5*Self(n-1) + 8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 13 2012

a[n_]:=(MatrixPower[{{1,2},{1,-6}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{5, 8}, {0, 1}, 30] (* Vincenzo Librandi, Nov 13 2012 *)

A015544(n)=imag((2+quadgen(57))^n) \\ M. F. Hasler, Mar 06 2009

x='x+O('x^30); concat([0], Vec(x/(1 - 5*x - 8*x^2))) \\ G. C. Greubel, Jan 01 2018

[lucas_number1(n,5,-8) for n in range(0, 21)] # Zerinvary Lajos, Apr 24 2009

a(n) = 5*a(n-1) + 8*a(n-2).

G.f.: x/(1 - 5*x - 8*x^2). - M. F. Hasler, Mar 06 2009

More precise definition by M. F. Hasler, Mar 06 2009

A053455 a(n) = ((8^n) - (-6)^n)/14.

Original entry on oeis.org

0, 1, 2, 52, 200, 2896, 15392, 169792, 1078400, 10306816, 72376832, 639480832, 4753049600, 40201179136, 308548739072, 2546754076672, 19903847628800, 162051890937856, 1279488468058112, 10337467701133312, 82090381869056000, 660379213392510976, 5261096756499709952, 42220395755839946752

Offset: 0

Barry E. Williams, Jan 13 2000

Previous name was: A linear recursive sequence.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (2,48).

Cf. A053404, A051958, A015441, A080921.

A053455:=n->((8^n)-(-6)^n)/14; seq(A053455(n), n=0..30); # Wesley Ivan Hurt, Mar 07 2014

LinearRecurrence[{2,48},{1,2},30] (* Harvey P. Dale, Nov 28 2011 *)
CoefficientList[Series[x / (1 - 2 x - 48 x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Mar 08 2014 *)

a(n) = ((8^n)-(-6)^n)/14; \\ Joerg Arndt, Mar 08 2014

a(n) = 2*a(n-1) + 48*a(n-2), n>=2; a(0)=0, a(1)=1.
a(n) = ((8^n)-(-6)^n)/14 = (2^(n-1))*((4^n) - (-3)^n)/7 = 2^(n-1)*A053404(n).
G.f.: x/((1+6*x)*(1-8*x)). - Harvey P. Dale, Nov 28 2011
a(n) = A080921(n). - Philippe Deléham, Mar 05 2014
a(n+1) = Sum_{k=0..n} A238801(n,k)*7^k. - Philippe Deléham, Mar 07 2014

More terms from James Sellers, Feb 02 2000

New name (from formula), Joerg Arndt, Mar 05 2014

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).

cp's OEIS Frontend

A180250 a(n) = 5*a(n-1) + 10*a(n-2), with a(1)=0 and a(2)=1.

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A015551 Expansion of x/(1 - 6*x - 5*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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A072024 Table by antidiagonals of T(n,k) = ((n+1)^k - (-n)^k)/(2*n+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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A053430 a(n) = (6^(n+1) - (-5)^(n+1))/11.

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A180250 a(n) = 5a(n-1) + 10a(n-2), with a(1)=0 and a(2)=1.

A015551 Expansion of x/(1 - 6x - 5x^2).

A015541 Expansion of x/(1 - 5x - 7x^2).

A015544 Lucas sequence U(5,-8): a(n+1) = 5a(n) + 8a(n-1), a(0)=0, a(1)=1.