A015441 -id:A015441 - cp's OEIS frontend

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

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

A080424 a(n) = 3a(n-1) + 18a(n-2), a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 27, 135, 891, 5103, 31347, 185895, 1121931, 6711903, 40330467, 241805655, 1451365371, 8706597903, 52244370387, 313451873415, 1880754287211, 11284396583103, 67706766919107, 406239439253175, 2437440122303451, 14624630273467503, 87747813021864627, 526486783988008935

Offset: 0

Paul Barry, Feb 24 2003

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
Index entries for linear recurrences with constant coefficients, signature (3,18).

Cf. A001045, A015441, A051958, A079773.

[(6^n-(-3)^n)/9: n in [0..30]]; // Vincenzo Librandi, Aug 13 2011

a[n_]:=(6^n - (-3)^n)/9; Array[a, 22, 0] (* Robert G. Wilson v, Aug 13 2011 *)
LinearRecurrence[{3,18}, {0,1}, 31] (* G. C. Greubel, Dec 22 2023 *)

a(n)=(6^n-(-3)^n)/9 \\ Charles R Greathouse IV, Jun 10 2011

[3^(n-1)*lucas_number1(n,1,-2) for n in range(31)] # G. C. Greubel, Dec 22 2023

G.f.: x/((1+3*x)*(1-6*x)).
a(n) = (6^n - (-3)^n)/9.
a(n+1) = 6*a(n) + (-3)^n. - Paul Curtz, Jun 07 2011
a(n) = 3^(n-1)*A001045(n). - R. J. Mathar, Mar 08 2021

A087451 G.f.: (2-x)/((1+2x)(1-3x)); e.g.f.: exp(3x)+exp(-2x); a(n)=3^n+(-2)^n.

Original entry on oeis.org

2, 1, 13, 19, 97, 211, 793, 2059, 6817, 19171, 60073, 175099, 535537, 1586131, 4799353, 14316139, 43112257, 129009091, 387682633, 1161737179, 3487832977, 10458256051, 31385253913, 94134790219, 282446313697, 847255055011

Offset: 0

Paul Barry, Sep 06 2003

Generalized Lucas-Jacobsthal numbers.

Pisano period lengths: 1, 1, 1, 2, 4, 1, 6, 2, 3, 4, 5, 2, 12, 6, 4, 4, 16, 3, 18, 4,... - R. J. Mathar, Aug 10 2012

Drexel University, Generation X and Y
G. Everest, Y. Puri and T. Ward, Integer sequences counting periodic points, arXiv:math/0204173 [math.NT], 2002.
OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (1,6)

Cf. A014551, A087452, A015441.

for i from 0 to 20 do print(3^i+(-2)^i) od; # Gary Detlefs, Dec 20 2009

a[0] = 2; a[1] = 1; a[n_] := a[n] = a[n - 1] + 6a[n - 2]; a /@ Range[0, 25] (* Robert G. Wilson v, Feb 02 2006 *)

[lucas_number2(n,1,-6) for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

a(0) = 2, a(1) = 1, a(n) = a(n-1)+6a(n-2).
a(n) = 3^n + (-2)^n. - Gary Detlefs, Dec 20 2009
The sequence 1, 13, 19... is a(n+1) = 3*3^n-2*(-2)^n.
exp( Sum_{n >= 1} a(n)*x^n/n ) = Sum_{n >= 0} A015441(n+1)*x^n. - Peter Bala, Mar 30 2015
a(n) = 2*A015441(n+1) - A015441(n), a formula given by Paul Curtz for autosequences of the 2nd kind. - Jean-François Alcover, Jun 02 2017

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).

A091005 Expansion of x^2/((1-2x)(1+3*x)).

Original entry on oeis.org

0, 0, 1, -1, 7, -13, 55, -133, 463, -1261, 4039, -11605, 35839, -105469, 320503, -953317, 2876335, -8596237, 25854247, -77431669, 232557151, -697147165, 2092490071, -6275373061, 18830313487, -56482551853, 169464432775, -508359743893, 1525146340543

Offset: 0

Paul Barry, Dec 13 2003

Inverse binomial transform of A091002.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,6).

Cf. A015441.

Concatenation([0], List([1..30], n -> (3*2^n + 2*(-3)^n)/30)); # G. C. Greubel, Feb 01 2019

[0] cat [(3*2^n + 2*(-3)^n)/30: n in [1..30]]; // G. C. Greubel, Feb 01 2019

a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[a[n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
Join[{0, 0}, LinearRecurrence[{-1, 6}, {1, -1}, 30]] (* G. C. Greubel, Feb 01 2019 *)
CoefficientList[Series[x^2/((1-2x)(1+3x)),{x,0,30}],x] (* Harvey P. Dale, Apr 30 2022 *)

vector(30, n, n--; (3*2^n + 2*(-3)^n - 5*0^n)/30) \\ G. C. Greubel, Feb 01 2019

[0] + [(3*2^n + 2*(-3)^n)/30 for n in (1..30)] # G. C. Greubel, Feb 01 2019

2^n = A091003(n) + 3*A091004(n) + 6*a(n).
a(n) = (3*2^n + 2*(-3)^n - 5*0^n)/30.
E.g.f.: (3*exp(2*x) + 2*exp(-3*x) - 5)/30. - G. C. Greubel, Feb 01 2019

A185409 T(n,k)=1/7 the number of nXk 0..6 arrays with every element equal to exactly one or two of its horizontal and vertical neighbors.

Original entry on oeis.org

0, 1, 1, 1, 13, 1, 7, 150, 150, 7, 13, 1848, 5040, 1848, 13, 55, 22656, 279918, 279918, 22656, 55, 133, 278832, 12205368, 53808534, 12205368, 278832, 133, 463, 3430776, 595974240, 9560464068, 9560464068, 595974240, 3430776, 463, 1261, 42221928

Offset: 1

R. H. Hardin Jan 26 2011

Table starts
....0........1...........1.............7............13............55
....1.......13.........150..........1848.........22656........278832
....1......150........5040........279918......12205368.....595974240
....7.....1848......279918......53808534....9560464068.1755419411964
...13....22656....12205368....9560464068.6409594156632
...55...278832...595974240.1755419411964
..133..3430776.27536905674
..463.42221928
.1261

			Some solutions for 4X3 with a(1,1)=0
..0..3..6....0..0..0....0..0..4....0..0..3....0..3..6....0..3..3....0..0..0
..0..3..6....4..4..4....0..4..4....3..3..3....0..3..6....0..2..2....4..4..4
..2..4..4....1..5..4....0..5..5....3..2..2....5..0..6....1..1..1....5..5..6
..2..2..2....1..5..5....4..4..4....1..1..1....5..0..6....6..6..6....0..0..6

R. H. Hardin, Table of n, a(n) for n = 1..49

Column 1 is A015441(n-1)

A254667 The nonnegative numbers with 2 instead of 1.

Original entry on oeis.org

0, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62

Offset: 0

Paul Curtz, Feb 04 2015

An autosequence of the first kind is a sequence whose main diagonal in the difference table is A000004 = 0's.
This is the case for a(n).
Difference table of a(n):
   0,  2,  2,  3,  4,  5, ...
   2,  0,  1,  1,  1,  1, ...
  -2,  1,  0,  0,  0,  0, ...
   3, -1,  0,  0,  0,  0, ...
  -4,  1,  0,  0,  0,  0, ...
   5, -1,  0,  0,  0,  0, ...
  etc.
The inverse binomial transform of a(n) is (-1)^(n+1)*a(n).
0 followed by A000012(n) is not in the OEIS. See A054977.
What is the meaning of a(n)?
Among many others, A015441 is an autosequence of the first kind.
General form for such autosequence.
Starting from the first upper diagonal s0, s1, s2, s3, s4, ...,
the autosequence is
0, s0, s0, s0 + s1, s0 + 2*s1, s0 + 3*s1 + s2, s0 + 4*s1 + 3*s2, ... .
After 0, the corresponding coefficients are A011973(n).

			G.f. = 2*x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + 8*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..2500
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000004, A001477, A011973, A015441, A027641, A027642, A054977, A063524, A164555, A164558, A233583.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x-2*x^2+x^3)/(1-x)^2)); // G. C. Greubel, Aug 03 2018

CoefficientList[Series[(2*x-2*x^2+x^3)/(1-x)^2, {x, 0, 60}], x] (* G. C. Greubel, Aug 03 2018 *)
a[ n_] := n + Boole[n == 1]; (* Michael Somos, Aug 19 2018 *)
Join[{0,2},Range[2,70]] (* Harvey P. Dale, Oct 10 2024 *)

{a(n) = n + (n==1)}; /* Michael Somos, Feb 09 2015 */

a(n) = (A164558(n) + (-1)^(n+1)*A164555(n))/A027642(n).
a(n) = A063524(n) + A001477(n). - David A. Corneth, Aug 03 2018
G.f.: (2*x - 2*x^2 + x^3) / (1 - x)^2. - Michael Somos, Feb 09 2015

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

cp's OEIS Frontend

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

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A080424 a(n) = 3*a(n-1) + 18*a(n-2), a(0)=0, a(1)=1.

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A087451 G.f.: (2-x)/((1+2x)(1-3x)); e.g.f.: exp(3x)+exp(-2x); a(n)=3^n+(-2)^n.

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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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A091005 Expansion of x^2/((1-2*x)*(1+3*x)).

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

A080424 a(n) = 3a(n-1) + 18a(n-2), a(0)=0, a(1)=1.

A091005 Expansion of x^2/((1-2x)(1+3*x)).