A015441 -id:A015441 - cp's OEIS frontend

A051958 a(n) = 2a(n-1) + 24a(n-2), a(0)=0, a(1)=1.

0, 1, 2, 28, 104, 880, 4256, 29632, 161408, 1033984, 5941760, 36699136, 216000512, 1312780288, 7809572864, 47125872640, 281681494016, 1694383931392, 10149123719168, 60963461791744, 365505892843520, 2194134868688896

Offset: 0

Barry E. Williams, Jan 04 2000

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Felix P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, Preprint on ResearchGate, March 2014.
Index entries for linear recurrences with constant coefficients, signature (2,24).

Cf. A015441, A238801.

[(6^n-(-4)^n)/10: n in [0..30]]; // Vincenzo Librandi, Mar 08 2014

Table[(6^n-(-4)^n)/10, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
CoefficientList[Series[x/((1+4 x) (1-6 x)), {x,0,30}], x] (* Vincenzo Librandi, Mar 08 2014 *)
LinearRecurrence[{2,24},{0,1},30] (* Harvey P. Dale, May 08 2022 *)

PARI
```
a(n)=(6^n-(-4)^n)/10
```

A051958=BinaryRecurrenceSequence(2,24,0,1)
[A051958(n) for n in range(31)] # G. C. Greubel, Nov 11 2024

G.f.: x/((1+4*x)*(1-6*x)).
a(n) = (6^n - (-4)^n)/10.
a(n) = 2^(n-1)*A015441(n).
a(n+1) = Sum_{k = 0..n} A238801(n,k)*5^k. - Philippe Deléham, Mar 07 2014
Limit_{n -> oo} a(n+1)/a(n) = 6. - Felix P. Muga II, Mar 10 2014
E.g.f.: (1/10)*(exp(6*x) - exp(-4*x)). - G. C. Greubel, Nov 11 2024

A122117 a(n) = 3a(n-1) + 4a(n-2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 10, 38, 154, 614, 2458, 9830, 39322, 157286, 629146, 2516582, 10066330, 40265318, 161061274, 644245094, 2576980378, 10307921510, 41231686042, 164926744166, 659706976666, 2638827906662, 10555311626650, 42221246506598

Offset: 0

Philippe Deléham, Oct 19 2006

Inverse binomial transform of A005053. Binomial transform of [1, 1, 7, 13, 55, ...] = A015441(n+1).
Convolved with [1, 2, 2, 2, ...] = powers of 4: [1, 4, 16, 64, ...]. - Gary W. Adamson, Jun 02 2009
a(n) is the number of compositions of n when there are 2 types of 1 and 6 types of other natural numbers. - Milan Janjic, Aug 13 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A055380, A100088, A108981, A122016, A201455.

a:=[1,2];; for n in [3..30] do a[n]:=3*a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, May 18 2019

I:=[1, 2]; [n le 2 select I[n] else 3*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 06 2012

CoefficientList[Series[(1-x)/(1-3*x-4*x^2),{x,0,30}],x] (* Vincenzo Librandi, Jul 06 2012 *)

Vec((1-x)/(1-3*x-4*x^2)+O(x^30)) \\ Charles R Greathouse IV, Jan 11 2012

def A122117(n): return ((4<<(m:=n<<1))|2)//5-((1<Chai Wah Wu, Apr 22 2025

from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,2,3,4, lambda n: 0); [next(it) for i in range(24)] # Zerinvary Lajos, Jul 03 2008

((1-x)/(1-3*x-4*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 18 2019

a(n) = 2*A108981(n-1) for n > 0, with a(0) = 1.
a(2*n) = 4*a(2*n-1) + 2, a(2*n+1) = 4*a(2*n) - 2.
a(n) = Sum_{k=0..n} 2^(n-k)*A055380(n,k).
G.f.: (1-x)/(1-3*x-4*x^2).
Lim_{n->infinity} a(n+1)/a(n) = 4.
a(n) = Sum_{k=0..n} A122016(n,k)*2^k. - Philippe Deléham, Nov 05 2008
a(n) = A100088(2*n). - Chai Wah Wu, Apr 22 2025

Corrected by T. D. Noe, Nov 07 2006

A079773 a(n) = 2a(n-1)+15a(n-2) with n>0, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 19, 68, 421, 1862, 10039, 48008, 246601, 1213322, 6125659, 30451148, 152787181, 762341582, 3816490879, 19068105488, 95383574161, 476788730642, 2384331073699, 11920493107028, 59605952319541, 298019301244502

Offset: 0

Paul Barry, Feb 20 2003

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
Index entries for linear recurrences with constant coefficients, signature (2,15)

Cf. A051958, A015441.

[(5^n-(-3)^n)/8: n in [0..25]]; // Vincenzo Librandi, Aug 05 2013

Join[{a=0,b=1},Table[c=2*b+15*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
CoefficientList[Series[x / ((1 + 3 x) (1 - 5 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 05 2013 *)

[lucas_number1(n,2,-15) for n in range(0, 23)] # Zerinvary Lajos, Apr 22 2009

G.f.: x/((1+3*x)*(1-5*x)).
a(n) = (5^n-(-3)^n)/8.
a(n) = sum(k=1..n, binomial(n, 2*k-1)*4^(2*(k-1))).
E.g.f.: exp(x)*sinh(4*x)/4. - Paul Barry, Jul 09 2003
a(n+1) = Sum_{k = 0..n} A238801(n,k)*4^k. - Philippe Deléham, Mar 07 2014

A135030 Generalized Fibonacci numbers: a(n) = 6a(n-1) + 2a(n-2).

Original entry on oeis.org

0, 1, 6, 38, 240, 1516, 9576, 60488, 382080, 2413456, 15244896, 96296288, 608267520, 3842197696, 24269721216, 153302722688, 968355778560, 6116740116736, 38637152257536, 244056393778688, 1541612667187200

Offset: 0

Rolf Pleisch, Feb 10 2008, Feb 14 2008

For n>0, a(n) equals the number of words of length n-1 over {0,1,...,7} in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Jan 08 2017

Joshua Zucker and Robert Israel, Table of n, a(n) for n = 0..1000 (n=0..51 from Joshua Zucker).
Index entries for linear recurrences with constant coefficients, signature (6, 2).

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, A083858, A085939, A090017, A091914, A099012, A180222, A180226, A180250.

[n le 2 select n-1 else 6*Self(n-1) + 2*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Sep 18 2016

A:= gfun:-rectoproc({a(0) = 0, a(1) = 1, a(n) = 2*(3*a(n-1) + a(n-2))},a(n),remember):
seq(A(n),n=1..30); # Robert Israel, Sep 16 2014

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

a(n)=([0,1; 2,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

[lucas_number1(n,6,-2) for n in range(0, 21)] # Zerinvary Lajos, Apr 24 2009

a(0) = 0; a(1) = 1; a(n) = 2*(3*a(n-1) + a(n-2)).
a(n) = 1/(2*sqrt(11))*( (3 + sqrt(11))^n - (3 - sqrt(11))^n ).
G.f.: x/(1 - 6*x - 2*x^2). - Harvey P. Dale, Jun 20 2011
a(n+1) = Sum_{k=0..n} A099097(n,k)*2^k. - Philippe Deléham, Sep 16 2014
E.g.f.: (1/sqrt(11))*exp(3*x)*sinh(sqrt(11)*x). - G. C. Greubel, Sep 17 2016

More terms from Joshua Zucker, Feb 23 2008

A053428 a(n) = a(n-1) + 20*a(n-2), n >= 2; a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 21, 41, 461, 1281, 10501, 36121, 246141, 968561, 5891381, 25262601, 143090221, 648342241, 3510146661, 16476991481, 86679924701, 416219754321, 2149818248341, 10474213334761, 53470578301581, 262954844996801

Offset: 0

Barry E. Williams, Jan 10 2000

Hankel transform is 1,20,0,0,0,0,0,0,0,0,0,0,... - Philippe Deléham, Nov 02 2008

Zero followed by this sequence gives the inverse binomial transform of A080424. - Paul Curtz, Jun 07 2011

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..400
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
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 (1,20).

Cf. A001045, A015441, A053404, A000302, A053573, A080424.

[((5^(n+1))-(-4)^(n+1)) div 9: n in [0..40]]; // Vincenzo Librandi, Jun 07 2011

Join[{a=1,b=1},Table[c=b+20*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)

a(n)=((5^(n+1))-(-4)^(n+1))/9 \\ Charles R Greathouse IV, Jun 10 2011

a(n) = ((5^(n+1)) - (-4)^(n+1))/9.

G.f.: 1/((1+4*x)*(1-5*x)). - R. J. Mathar, Nov 16 2007

More terms from James Sellers, Feb 02 2000

A081297 Array T(k,n), read by antidiagonals: T(k,n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 7, 5, 1, 1, 1, 13, 13, 11, 1, 1, 1, 21, 25, 55, 21, 1, 1, 1, 31, 41, 181, 133, 43, 1, 1, 1, 43, 61, 461, 481, 463, 85, 1, 1, 1, 57, 85, 991, 1281, 2653, 1261, 171, 1, 1, 1, 73, 113, 1891, 2821, 10501, 8425, 4039, 341, 1, 1, 1, 91, 145, 3305

Offset: 0

Paul Barry, Mar 17 2003

Square array of solutions of a family of recurrences.
Rows of the array give solutions to the recurrences a(n)=a(n-1)+k(k-1)a(n-2), a(0)=a(1)=1.
Subarray of array in A072024. - Philippe Deléham, Nov 24 2013

			Rows begin
  1, 1,  1,  1,   1,    1, ...
  1, 1,  3,  5,  11,   21, ...
  1, 1,  7, 13,  55,  133, ...
  1, 1, 13, 25, 181,  481, ...
  1, 1, 21, 41, 461, 1281, ...

Andrew Howroyd, Table of n, a(n) for n = 0..1274

Cf. A059259, A072024.
Rows include A001045, A015441, A053404, A053428, A053430.
Columns include A002061, A001844, A072025.
Diagonals include A081298, A081299, A081300, A081301, A081302.

T[n_, k_]:=((n + 1)^(k + 1) - (-n)^(k + 1)) / (2n + 1); Flatten[Table[T[n - k, k], {n, 0, 10}, {k, 0, n}]] (* Indranil Ghosh, Mar 27 2017 *)

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

def T(n, k): return ((n + 1)**(k + 1) - (-n)**(k + 1)) // (2*n + 1)
for n in range(11):
    print([T(n - k, k) for k in range(n + 1)]) # Indranil Ghosh, Mar 27 2017

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

Rows of the array have g.f. 1/((1+kx)(1-(k+1)x)).

Name clarified by Andrew Howroyd, Mar 27 2017

A083856 Square array T(n,k) of generalized Fibonacci numbers, read by antidiagonals upwards (n, k >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 5, 5, 1, 0, 1, 1, 5, 7, 11, 8, 1, 0, 1, 1, 6, 9, 19, 21, 13, 1, 0, 1, 1, 7, 11, 29, 40, 43, 21, 1, 0, 1, 1, 8, 13, 41, 65, 97, 85, 34, 1, 0, 1, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1

Offset: 0

Paul Barry, May 06 2003

Row n >= 0 of the array gives the solution to the recurrence b(k) = b(k-1) + n*b(k-2) for k >= 2 with b(0) = 0 and b(1) = 1.

			Array T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
  0, 1, 1,  1,  1,   1,   1,    1,    1,     1, ... [A057427]
  0, 1, 1,  2,  3,   5,   8,   13,   21,    34, ... [A000045]
  0, 1, 1,  3,  5,  11,  21,   43,   85,   171, ... [A001045]
  0, 1, 1,  4,  7,  19,  40,   97,  217,   508, ... [A006130]
  0, 1, 1,  5,  9,  29,  65,  181,  441,  1165, ... [A006131]
  0, 1, 1,  6, 11,  41,  96,  301,  781,  2286, ... [A015440]
  0, 1, 1,  7, 13,  55, 133,  463, 1261,  4039, ... [A015441]
  0, 1, 1,  8, 15,  71, 176,  673, 1905,  6616, ... [A015442]
  0, 1, 1,  9, 17,  89, 225,  937, 2737, 10233, ... [A015443]
  0, 1, 1, 10, 19, 109, 280, 1261, 3781, 15130, ... [A015445]
  ...

A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, 21(2) (2015), 35-42.

Rows include A057427 (n=0), A000045 (n=1), A001045 (n=2), A006130 (n=3), A006131 (n=4), A015440 (n=5), A015441 (n=6), A015442 (n=7), A015443 (n=8), A015445 (n=9).
Columns include A000012 (k=1,2), A000027 (k=3), A005408 (k=4), A028387 (k=5), A000567 (k=6), A106734 (k=7).
Cf. A083857 (binomial transform), A083859 (main diagonal), A083860 (first subdiagonal), A083861 (second binomial transform), A110112, A110113 (diagonal sums), A193376 (transposed variant), A172237 (transposed variant).

function generalized_fibonacci(r, n)
   F = BigInt[1 r; 1 0]
   Fn = F^n
   Fn[2, 1]
end
for r in 0:6 println([generalized_fibonacci(r, n) for n in 0:9]) end # Peter Luschny, Mar 06 2017

A083856_row := proc(r, n) local R; R := proc(n) option remember;
if n<=1 then n else R(n-1)+r*R(n-2) fi end: R(n) end:
for r from 0 to 9 do seq(A083856_row(r, n), n=0..9) od; # Peter Luschny, Mar 06 2017

T[, 0] = 0; T[, 1|2] = 1; T[n_, k_] := T[n, k] = T[n, k-1] + n T[n, k-2];
Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *)

T(n, k) = (((1 + sqrt(4*n + 1))/2)^k - ((1 - sqrt(4*n + 1))/2)^k)/sqrt(4*n + 1). [corrected by Michel Marcus, Jun 25 2018]
From Thomas Baruchel, Jun 25 2018: (Start)
The g.f. for row n >= 0 is x/(1 - x - n*x^2).
The g.f. for column k >= 1 is g(k,x) = 1/(1-x) + Sum_{m = 1..floor((k-1)/2)} (1 - x)^(-1 - m) * binomial(k - 1 - m, m) * Sum_{i = 0..m} x^i * Sum_{j = 0..i} (-1)^j * (i - j)^m * binomial(1 + m, j).
The g.f. for column k >= 1 is also g(k,x) = 1 + Sum_{m = 1..floor((k+1)/2)} ((1 - x)^(-m) * binomial(k-m, m-1) * Sum_{j = 0..m} (-1)^j * binomial(m, j) *  x^m * Phi(x, -m+1, -j+m)) + Sum_{s = 0..floor((k-1)/2)} binomial(k-s-1, s) * PolyLog(-s, x), where Phi is the Lerch transcendent function. (End)
T(n,k) = Sum_{i = 0..k} (-1)^(k+i) * binomial(k,i) * A083857(n,i). - Petros Hadjicostas, Dec 24 2019

Various sections edited by Petros Hadjicostas, Dec 24 2019

A096951 Sum of odd powers of 2 and of 3 divided by 5.

Original entry on oeis.org

1, 7, 55, 463, 4039, 35839, 320503, 2876335, 25854247, 232557151, 2092490071, 18830313487, 169464432775, 1525146340543, 13726182847159, 123535108753519, 1111813831298023, 10006315891747615, 90056808665990167, 810511140554958031, 7294599715238808391

Offset: 0

Wolfdieter Lang, Jul 16 2004

Sequence appears in A096952 (upper bounds for Lagrange remainder in Taylor expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2)).
Divisibility of 2^(2*n+1) + 3^(2*n+1) by 5 is proved by induction.
The sequence a(n+1), with g.f. (7-36*x)/(1-13*x+36*x^2) and formula (27*9^n + 8*4^n)/5, is the Hankel transform of C(n) + 6*C(n+1), where C(n) is A000108(n). - Paul Barry, Dec 06 2006

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,-36).

Cf. A074614 (sum of even powers of 2 and of 3), A007689 (sum of powers of 2 and powers of 3).

Cf. A000108, A015441, A096952, A138233.

[(2^(2*n+1) + 3^(2*n+1))/5: n in [0..30]]; // Vincenzo Librandi, May 31 2011

LinearRecurrence[{13, -36},{1, 7},19] (* Ray Chandler, Jul 14 2017 *)

a(n) = (2^(2*n+1) + 3^(2*n+1))/5.
G.f.: (1-6*x)/((1-4*x)*(1-9*x)).
From Reinhard Zumkeller, Mar 07 2008: (Start)
a(n+1) = 4*a(n) + 3^(2*n+1), a(0) = 1.
a(n) = A138233(n)/5. (End)
From Elmo R. Oliveira, Aug 02 2025: (Start)
E.g.f.: exp(4*x)*(2 + 3*exp(5*x))/5.
a(n) = 13*a(n-1) - 36*a(n-2).
a(n) = A015441(2*n+1). (End)

A102901 a(n) = a(n-1) + 6*a(n-2), a(0)=1, a(1)=0.

Original entry on oeis.org

1, 0, 6, 6, 42, 78, 330, 798, 2778, 7566, 24234, 69630, 215034, 632814, 1923018, 5719902, 17258010, 51577422, 155125482, 464590014, 1395342906, 4182882990, 12554940426, 37652238366, 112981880922, 338895311118, 1016786596650

Offset: 0

Paul Barry, Jan 17 2005

Binomial transform is A102900.

Hankel transform is = 1,6,0,0,0,0,0,0,0,0,0,0,... - Philippe Deléham, Nov 02 2008

			a(6) = 330; (2*3^6 + 3*(-2)^6)/5 = (1458 + 192)/5 = 330.

Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,6).

Cf. A015441, A102900.

[(2*3^n+3*(-2)^n)/5: n in [0..50]]; // Vincenzo Librandi, Jul 20 2013

A102901:=n->(2*3^n+3*(-2)^n)/5; seq(A102901(k), k=0..60); # Wesley Ivan Hurt, Nov 05 2013

CoefficientList[Series[(1-x)/((1+2x)(1-3x)), {x,0,50}], x] (* Vincenzo Librandi, Jul 20 2013 *)

a(n)=([0,1; 6,1]^n*[1;0])[1,1] \\ Charles R Greathouse IV, Mar 28 2016

A102901=BinaryRecurrenceSequence(1,6,1,0)
[A102901(n) for n in range(51)] # G. C. Greubel, Dec 09 2022

G.f.: (1-x)/((1+2*x)*(1-3*x)).
a(n) = (2*3^n + 3*(-2)^n)/5.
a(n) = 6*A015441(n-1), for n>0.

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

Original entry on oeis.org

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

cp's OEIS Frontend

A051958 a(n) = 2*a(n-1) + 24*a(n-2), a(0)=0, a(1)=1.

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A122117 a(n) = 3*a(n-1) + 4*a(n-2), with a(0)=1, a(1)=2.

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A079773 a(n) = 2*a(n-1)+15*a(n-2) with n>0, a(0)=0, a(1)=1.

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A135030 Generalized Fibonacci numbers: a(n) = 6*a(n-1) + 2*a(n-2).

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A053428 a(n) = a(n-1) + 20*a(n-2), n >= 2; a(0)=1, a(1)=1.

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A081297 Array T(k,n), read by antidiagonals: T(k,n) = ((k+1)^(n+1)-(-k)^(n+1))/(2k+1).

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A051958 a(n) = 2a(n-1) + 24a(n-2), a(0)=0, a(1)=1.

A122117 a(n) = 3a(n-1) + 4a(n-2), with a(0)=1, a(1)=2.

A079773 a(n) = 2a(n-1)+15a(n-2) with n>0, a(0)=0, a(1)=1.

A135030 Generalized Fibonacci numbers: a(n) = 6a(n-1) + 2a(n-2).

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