A015447 -id:A015447 - 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.

A063092 a(0)=1, a(1)=2 and, for n>1, a(n) = a(n-1) + 11*a(n-2).

Original entry on oeis.org

1, 2, 13, 35, 178, 563, 2521, 8714, 36445, 132299, 533194, 1988483, 7853617, 29726930, 116116717, 443112947, 1720396834, 6594639251, 25519004425, 98060036186, 378769084861, 1457429482907, 5623889416378, 21655613728355, 83518397308513, 321730148320418

Offset: 0

John W. Layman, Apr 29 2003

nonn

The binomial transform of this sequence is {(3^n)F(n+1)} where {F(n)} is the Fibonacci sequence A000045.

Harry J. Smith, Table of n, a(n) for n=0..200
Index entries for linear recurrences with constant coefficients, signature (1,11).

Cf. A000045.

LinearRecurrence[{1,11},{1,2},30] (* Harvey P. Dale, Oct 25 2020 *)

{ for (n=0, 200, if (n>1, a=a1 + 11*a2; a2=a1; a1=a, if (n, a=2; a2=1; a1=2, a=1)); write("b063092.txt", n, " ", a) ) } \\ Harry J. Smith, Aug 18 2009

G.f.: (1+x)/(1-x-11*x^2). - Jaume Oliver Lafont, Sep 07 2009
a(n) = A015447(n) + A015447(n-1), n>0. - Ralf Stephan, Jul 21 2013
a(n)^2 - (3*A015447(n-1))^2 - (3*A015447(n-1))*a(n) = (-11)^n. - Philippe Deléham, Mar 17 2023

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

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).

A099134 Expansion of x/(1-2x-19x^2).

Original entry on oeis.org

0, 1, 2, 23, 84, 605, 2806, 17107, 87528, 500089, 2663210, 14828111, 80257212, 442248533, 2409384094, 13221490315, 72221278416, 395650872817, 2163506035538, 11844378654599, 64795371984420, 354633938406221

Offset: 0

Paul Barry, Sep 29 2004

Binomial transform is A099133. Binomial transform of x/(1-20x^2), or (0,1,0,20,0,400,0,8000,....). The inverse binomial transform of k^(n-1)Fib(n) has g.f. x/(1-(k-2)x-(k^2+k-1)x^2).

4*a(n) = (-1)^(n+1)*b(n;4) = 3^n*b(n;4/3), where b(n;d), n=0,1,..., d \in C, denote one of the delta-Fibonacci numbers defined in comments to A014445 (see also Witula-Slota's paper). Our first identity is equivalent to the second formula given below. We note that the sequence (4/3)^n*F(n) is the binomial transform of the sequence 3^(-n)*b(n;4). - Roman Witula, Jul 24 2012

R. Witula, D. Slota, \delta-Fibonacci Numbers, Appl. Anal. Discrete Math., 3 (2009), 310-329.

Index entries for linear recurrences with constant coefficients, signature (2,19).

Cf. A015447.

Join[{a=0,b=1},Table[c=2*b+19*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
CoefficientList[Series[x/(1-2x-19x^2),{x,0,30}],x] (* or *) LinearRecurrence[ {2,19},{0,1},30] (* Harvey P. Dale, Dec 25 2019 *)

a(n) = 2a(n-1) + 19a(n-2).
a(n) = sum{k=0..n, (-1)^(n-k)binomial(n, k)4^(k-1)*Fib(k)}.
a(n) = sum{k=0..n, binomial(n, 2k+1)20^k}.

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

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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A063092 a(0)=1, a(1)=2 and, for n>1, a(n) = a(n-1) + 11*a(n-2).

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

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

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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.

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

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

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