A152187 -id:A152187 - cp's OEIS frontend

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

1, 5, 23, 107, 497, 2309, 10727, 49835, 231521, 1075589, 4996919, 23214443, 107848529, 501037445, 2327695367, 10813893803, 50238661313, 233396326661, 1084301290583, 5037394142315, 23402480441009, 108722104190981, 505095858086951, 2346549744920747

Offset: 0

R. H. Hardin, Dec 27 2006

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.
From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.
The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)
Inverse binomial transform of A154244. (End)
Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010
Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Shanzhen Gao and Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.
Index entries for linear recurrences with constant coefficients, signature (4,3).

Cf. 5 symbol differing by two or less A126392, one or less A057960.
Cf. Red king sequences side squares [numerical value A[5]]: A086347 [495], A179598 [239], A126473 [367], A123347 [335], A179602 [95], A154964 [31], A015448 [327], A152187 [27], A003947 [325], A108981 [11], A007483 [2]. - Johannes W. Meijer, Aug 01 2010
Cf. A055099.

with(LinearAlgebra): nmax:=19; m:=2; A[5]:= [1,0,1,1,0,1,1,1,1]: A:=Matrix([[0,1,0,1,1,0,0,0,0],[1,0,1,1,1,1,0,0,0],[0,1,0,0,1,1,0,0,0],[1,1,0,0,1,0,1,1,0],A[5],[0,1,1,0,1,0,0,1,1],[0,0,0,1,1,0,0,1,0],[0,0,0,1,1,1,1,0,1],[0,0,0,0,1,1,0,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Aug 01 2010
# second Maple program:
a:= n-> (M-> M[1,2]+M[2,2])(<<0|1>, <3|4>>^n):
seq(a(n), n=0..24);  # Alois P. Heinz, Jun 28 2021

LinearRecurrence[{4, 3}, {1, 5}, 24] (* Jean-François Alcover, Dec 10 2024 *)

a(n)=([0,1; 3,4]^n*[1;5])[1,1] \\ Charles R Greathouse IV, May 10 2016

From Johannes W. Meijer, Aug 01 2010: (Start)
G.f.: (1+x)/(1-4*x-3*x^2).
a(n) = 4*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = ((1+3/sqrt(7))/2)*(A)^(-n) + ((1-3/sqrt(7))/2)*(B)^(-n) with A = (-2 + sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7)-A108851(n))
(End)
a(n) = A015330(n)+A015330(n+1). - R. J. Mathar, May 09 2023

Edited by Johannes W. Meijer, Aug 10 2010

A015523 a(n) = 3a(n-1) + 5a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 3, 14, 57, 241, 1008, 4229, 17727, 74326, 311613, 1306469, 5477472, 22964761, 96281643, 403668734, 1692414417, 7095586921, 29748832848, 124724433149, 522917463687, 2192374556806, 9191710988853, 38537005750589

Offset: 0

Olivier Gérard

From Johannes W. Meijer, Aug 01 2010: (Start)
a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 and 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For n >= 1, the sequence above corresponds to 24 red king vectors, i.e., A[5] vectors, with decimal values 27, 30, 51, 54, 57, 60, 90, 114, 120, 147, 150, 153, 156, 177, 180, 210, 216, 240, 282, 306, 312, 402, 408 and 432. These vectors lead for the side squares to A152187 and for the central square to A179606.
This sequence belongs to a family of sequences with g.f. 1/(1-3*x-k*x^2). Red king sequences that are members of this family are A007482 (k=2), A015521 (k=4), A015523 (k=5; this sequence), A083858 (k=6), A015524 (k=7) and A015525 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A049072 (k=-4), A057083 (k=-3), A000225 (k=-2), A001906 (k=-1), A000244 (k=0), A006190 (k=1), A030195 (k=3), A099012 (k=9), A015528 (k=10) and A015529 (k=11).
Inverse binomial transform of A052918 (with extra leading 0).
(End)
First differences in A197189. - Bruno Berselli, Oct 11 2011
Pisano period lengths:  1, 3, 4, 6, 4, 12, 3, 12, 12, 12, 120, 12, 12, 3, 4, 24, 288, 12, 72, 12, ... - R. J. Mathar, Aug 10 2012
This is the Lucas U(P=3, Q=-5) sequence, and hence for n >= 0, a(n+2)/a(n+1) equals the continued fraction 3 + 5/(3 + 5/(3 + 5/(3 + ... + 5/3))) with n 5's. - Greg Dresden, Oct 06 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A072263, A072264, A152187, A179606, A197189.

[ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+5*Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 23 2011

Join[{a = 0, b = 1}, Table[c = 3 * b + 5 * a; a = b; b = c, {n, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
a[0] := 0; a[1] := 1; a[n_] := a[n] = 3a[n - 1] + 5a[n - 2]; Table[a[n], {n, 0, 49}] (* Alonso del Arte, Jan 16 2011 *)

x='x+O('x^30); concat([0], Vec(x/(1-3*x-5*x^2))) \\ G. C. Greubel, Jan 01 2018

[lucas_number1(n,3,-5) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

a(n) = 3*a(n-1) + 5*a(n-2).
From Paul Barry, Jul 20 2004: (Start)
a(n) = ((3/2 + sqrt(29)/2)^n - (3/2 - sqrt(29)/2)^n)/sqrt(29).
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1,k)*5^k*3^(n-2*k-1). (End)
G.f.: x/(1 - 3*x - 5*x^2). - R. J. Mathar, Nov 16 2007
From Johannes W. Meijer, Aug 01 2010: (Start)
Limit_{k->oo} a(n+k)/a(k) = (A072263(n) + a(n)*sqrt(29))/2.
Limit_{n->oo} A072263(n)/a(n) = sqrt(29). (End)
G.f.: G(0)*x/(2-3*x), where G(k) = 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
E.g.f.: 2*exp(3*x/2)*sinh(sqrt(29)*x/2)/sqrt(29). - Stefano Spezia, Oct 06 2019

A015449 Expansion of (1-4x)/(1-5x-x^2).

Original entry on oeis.org

1, 1, 6, 31, 161, 836, 4341, 22541, 117046, 607771, 3155901, 16387276, 85092281, 441848681, 2294335686, 11913527111, 61861971241, 321223383316, 1667978887821, 8661117822421, 44973567999926, 233528957822051

Offset: 0

Olivier Gérard

Row m=5 of A135597.
Binomial transform of A152187. - Johannes W. Meijer, Aug 01 2010
For n>=1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....5
.2..|..1.....5....25
.3..|..1....10....25.....125
.4..|..1....10....75.....125....625
.5..|..1....15....75.....500....625....3125
.6..|..1....15...150.....500...3125....3125...15625
.7..|..1....20...150....1250...3125...18750...15625...78125
which is triangle for numbers 5^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n+1) is (for n>=0) the number of length-n strings of 6 letters {0,1,2,3,4,5} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
With offset 1, the sequence is the INVERT transform (1, 5, 5*4, 5*4^2, 5*4^3, ...); i.e., of A003947. The sequence can also be obtained by taking powers of the matrix [(1,5); (1,4)] and extracting the upper left terms. - Gary W. Adamson, Jul 31 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, Matters Computational (The Fxtbook)
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 16.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A084057, A108306, A164549.

a:=[1,1];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 23 2019

[n le 2 select 1 else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 06 2012

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..21); # Zerinvary Lajos, Jul 26 2006

Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{1,5},#]}]&, {1,1},40]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
LinearRecurrence[{5,1}, {1,1}, 30] (* Vincenzo Librandi, Nov 06 2012 *)
CoefficientList[Series[(1-4*x)/(1-5*x-x^2), {x,0,30}], x] (* G. C. Greubel, Dec 19 2017 *)
Sum[Fibonacci[Range[30] +k-2, 5], {k,0,1}] (* G. C. Greubel, Oct 23 2019 *)

Vec((1-4*x)/(1-5*x-x^2) +O('x^30)) \\ _G. C. Greubel, Dec 19 2017

def A015449_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-4*x)/(1-5*x-x^2)).list()
A015449_list(30) # G. C. Greubel, Oct 23 2019

a(n) = 5*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 4^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-4*x)/(1-5*x-x^2). - Philippe Deléham, Nov 20 2008
For n >= 2, a(n) = F_n(5) + F_(n+1)(5), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = Sum_{k=0..n} A046854(n-1,k)*5^k. - R. J. Mathar, Feb 10 2024

A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3x - 5x^2).

Original entry on oeis.org

1, 4, 17, 71, 298, 1249, 5237, 21956, 92053, 385939, 1618082, 6783941, 28442233, 119246404, 499950377, 2096083151, 8788001338, 36844419769, 154473265997, 647641896836, 2715292020493, 11384085545659, 47728716739442

Offset: 0

Johannes W. Meijer, Jul 28 2010

a(n) represents the number of n-move routes of a fairy chess piece starting in the central square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
The sequence above corresponds to 24 red king vectors, i.e., A[5] vectors, with decimal values 27, 30, 51, 54, 57, 60, 90, 114, 120, 147, 150, 153, 156, 177, 180, 210, 216, 240, 282, 306, 312, 402, 408 and 432. These vectors lead for the corner squares to A015523 and for the side squares to A152187.
This sequence belongs to a family of sequences with g.f. (1 + (k-4)*x)/(1 - 3*x - k*x^2). Red king sequences that are members of this family are A007483 (k= 2), A015521 (k=4), A179606 (k=5; this sequence), A154964 (k=6), A179603 (k=7) and A179599 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A006190 (k=1), A133494 (k=0) and A168616 (k=-2).
Inverse binomial transform of A052918.
The sequence b(n+1) = 6*a(n), n >= 0 with b(0)=1, is a berserker sequence, see A180147. The b(n) sequence corresponds to 16 A[5] vectors with decimal values between 111 and 492. These vectors lead for the corner squares to sequence c(n+1)=4*A179606(n), n >= 0 with c(0)=1, and for the side squares to A180140. - Johannes W. Meijer, Aug 14 2010
Equals the INVERT transform of A063782: (1, 3, 10, 32, 104, ...). Example: a(3) = 71 = (1, 1, 4, 7) dot (32, 10, 3, 1) = (32 + 10 + 12 + 17). - Gary W. Adamson, Aug 14 2010

Indranil Ghosh, Table of n, a(n) for n = 0..1603
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A179597 (central square).

Cf. A072263, A072264, A152187, A197189.

with(LinearAlgebra): nmax:=22; m:=5; A[1]:= [0,1,0,1,1,0,0,0,0]: A[2]:= [1,0,1,1,1,1,0,0,0]: A[3]:= [0,1,0,0,1,1,0,0,0]: A[4]:= [1,1,0,0,1,0,1,1,0]: A[5]:= [0,0,0,1,1,1,0,0,1]: A[6]:= [0,1,1,0,1,0,0,1,1]: A[7]:= [0,0,0,1,1,0,0,1,0]: A[8]:= [0,0,0,1,1,1,1,0,1]: A[9]:= [0,0,0,0,1,1,0,1,0]: A:=Matrix([A[1],A[2],A[3],A[4],A[5],A[6],A[7],A[8],A[9]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

CoefficientList[Series[(1+x)/(1-3*x-5*x^2), {x, 0, 22}],x] (* or *) LinearRecurrence[{3,5,0},{1,4},23] (* Indranil Ghosh, Mar 05 2017 *)

print(Vec((1 + x)/(1- 3*x - 5*x^2) + O(x^23))); \\ Indranil Ghosh, Mar 05 2017

G.f.: (1+x)/(1 - 3*x - 5*x^2).
a(n) = A015523(n) + A015523(n+1).
a(n) = 3*a(n-1) + 5*a(n-2) with a(0) = 1 and a(1) = 4.
a(n) = ((29 + 7*sqrt(29))*A^(-n-1) + (29-7*sqrt(29))*B^(-n-1))/290 with A = (-3+sqrt(29))/10 and B = (-3-sqrt(29))/10
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000351(n)*A130196(n)/(A015523(n)*sqrt(29) - A072263(n)) for n >= 1.

A197189 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 43, 184, 767, 3221, 13498, 56599, 237287, 994856, 4171003, 17487289, 73316882, 307387091, 1288745683, 5403172504, 22653245927, 94975600301, 398193030538, 1669457093119, 6999336432047, 29345294761736, 123032566445443, 515824173145009, 2162635351662242

Offset: 0

Bruno Berselli, Oct 11 2011

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. for type of recurrence: A015523, A072263, A072264, A152187, A179606 and also A180140.

[n le 2 select n else 3*Self(n-1)+5*Self(n-2): n in [1..26]];

a = {1, 2}; Do[AppendTo[a, 3 a[[-1]] + 5 a[[-2]]], {24}]; a (* Bruno Berselli, Dec 26 2012 *)

v=vector(26); v[1]=1; v[2]=2; for(i=3, #v, v[i]=3*v[i-1]+5*v[i-2]); v

G.f.: (1-x)/(1-3*x-5*x^2).
a(n) = ((29+sqrt(29))*(3+sqrt(29))^n+(29-sqrt(29))*(3-sqrt(29))^n)/(58*2^n).
a(n) = A015523(n+1)-A015523(n).
G.f.: G(0)*(1-x)/(2-3*x), where G(k)= 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013

A072263 a(n) = 3a(n-1) + 5a(n-2), with a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 19, 72, 311, 1293, 5434, 22767, 95471, 400248, 1678099, 7035537, 29497106, 123669003, 518492539, 2173822632, 9113930591, 38210904933, 160202367754, 671661627927, 2815996722551, 11806298307288, 49498878534619, 207528127140297

Offset: 0

Miklos Kristof, Jul 08 2002

Inverse binomial transform of A087130. - Johannes W. Meijer, Aug 01 2010
Pisano period lengths: 1, 3, 4, 6, 4, 12, 3, 12, 12, 12, 120, 12, 12, 3, 4, 24, 288, 12, 72, 12... - R. J. Mathar, Aug 10 2012
This is the Lucas sequence V(3,-5). - Bruno Berselli, Jan 09 2013

			a(5)=5*b(4)+b(6): 1293=5*57+1008.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Wikipedia, Lucas sequence: Specific names.
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A072264, A152187, A197189.

Appears in A179606 and A015523. - Johannes W. Meijer, Aug 01 2010

a:=[2,3];; for n in [3..40] do a[n]:=3*a[n-1]+5*a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[2,3]; [n le 2 select I[n] else 3*Self(n-1) +5*Self(n-2): n in [1..40]]; // G. C. Greubel, Jan 14 2020

seq(coeff(series((2-3*x)/(1-3*x-5*x^2), x, n+1), x, n), n = 0..40); # G. C. Greubel, Jan 14 2020

LinearRecurrence[{3,5},{2,3},40] (* Harvey P. Dale, Nov 23 2018 *)

my(x='x+O('x^40)); Vec((2-3*x)/(1-3*x-5*x^2)) \\ G. C. Greubel, Jan 14 2020

[lucas_number2(n,3,-5) for n in range(0, 16)] # Zerinvary Lajos, Apr 30 2009

a(n) = 2*A015523(n+1) - 3*A015523(n).
a(n) = ((3 + sqrt(29))/2)^n + ((3 - sqrt(29))/2)^n.
G.f.: (2-3*x)/(1-3*x-5*x^2). - R. J. Mathar, Feb 06 2010
From Johannes W. Meijer, Aug 01 2010: (Start)
Limit_{k -> Infinity} a(n+k)/a(k) = (A072263(n) + A015523(n)*sqrt(29))/2.
Limit_{n -> Infinity} (A072263(n)/A015523(n)) = sqrt(29). (End)
G.f.: G(0), where G(k)= 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 29*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = 5^((n-1)/2)*( 2*sqrt(5)*Fibonacci(n+1, 3/sqrt(5)) - 3*Fibonacci(n, 3/sqrt(5)) ). - G. C. Greubel, Jan 14 2020

Offset changed and terms added by Johannes W. Meijer, Jul 19 2010

A072264 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 8, 29, 127, 526, 2213, 9269, 38872, 162961, 683243, 2864534, 12009817, 50352121, 211105448, 885076949, 3710758087, 15557659006, 65226767453, 273468597389, 1146539629432, 4806961875241, 20153583772883, 84495560694854, 354254600948977, 1485241606321201

Offset: 0

Miklos Kristof, Jul 08 2002

			a(5)=3*a(4)+5*a(3): 127=3*29+5*8=87+40.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A015523, A072263, A152187, A179606, A197189.

a:=[1,1];; for n in [3..30] do a[n]:=3*a[n-1]+5*a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

[n le 2 select 1 else 3*Self(n-1)+5*Self(n-2): n in [1..26]];  // Bruno Berselli, Oct 11 2011

seq(coeff(series((1-2*x)/(1-3*x-5*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Jan 14 2020

LinearRecurrence[{3,5},{1,1},30] (* Harvey P. Dale, Feb 17 2018 *)

my(x='x+O('x^30)); Vec((1-2*x)/(1-3*x-5*x^2)) \\ G. C. Greubel, Jan 14 2020

def A072264_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-2*x)/(1-3*x-5*x^2) ).list()
A072264_list(30) # G. C. Greubel, Jan 14 2020

G.f.: (1-2*x)/(1-3*x-5*x^2). - Jaume Oliver Lafont, Mar 06 2009
G.f.: G(0)*(1-2*x)/(2-3*x), where G(k)= 1 + 1/(1 - x*(29*k-9)/(x*(29*k+20) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 17 2013
a(n) = 5^((n-1)/2)*( sqrt(5)*Fibonacci(n+1, 3/sqrt(5)) - 2*Fibonacci(n, 3/sqrt(5)) ). - G. C. Greubel, Jan 14 2020

Offset changed and more terms added by Bruno Berselli, Oct 11 2011

A152185 a(n) = -3a(n-1) + 5a(n-2), n > 1; a(0)=1, a(1)=-5.

Original entry on oeis.org

1, -5, 20, -85, 355, -1490, 6245, -26185, 109780, -460265, 1929695, -8090410, 33919705, -142211165, 596232020, -2499751885, 10480415755, -43940006690, 184222098845, -772366329985, 3238209484180, -13576460102465

Offset: 0

Philippe Deléham, Nov 28 2008

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-3,5).

Cf. A152163, A152166, A152167, A152174.

LinearRecurrence[{-3, 5}, {1, -5}, 25] (* Paolo Xausa, Jan 19 2024 *)

Vec((1-2*x)/(1+3*x-5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jan 11 2012

G.f.: (1-2x)/(1+3x-5x^2).
a(n) = Sum_{k=0..n} A147703(n,k)*(-6)^k.
a(n) = (-1)^n*A152187(n). - Philippe Deléham, Nov 29 2008

A152240 a(n)=5a(n-1)+7a(n-2), n>1 ; a(0)=1, a(1)=7 .

Original entry on oeis.org

1, 7, 42, 259, 1589, 9758, 59913, 367871, 2258746, 13868827, 85155357, 522858574, 3210380369, 19711911863, 121032221898, 743144492531, 4562948015941, 28016751527422, 172024393748697, 1056239229435439, 6485366903418074

Offset: 0

Philippe Deléham, Nov 30 2008

Unsigned version of A152239.

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

Cf. A152187, A152224, A152239.

LinearRecurrence[{5, 7}, {1, 7}, 25] (* Paolo Xausa, Jan 19 2024 *)

Vec((1+2*x)/(1-5*x-7*x^2)+O(x^99)) \\ Charles R Greathouse IV, Jan 17 2012

G.f.: (1+2*x)/(1-5*x-7*x^2).

Several terms corrected by Johannes W. Meijer, Aug 17 2010

cp's OEIS Frontend

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

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A015523 a(n) = 3*a(n-1) + 5*a(n-2), with a(0)=0, a(1)=1.

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A015449 Expansion of (1-4*x)/(1-5*x-x^2).

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A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3*x - 5*x^2).

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A197189 a(n) = 3*a(n-1) + 5*a(n-2), with a(0)=1, a(1)=2.

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A072263 a(n) = 3*a(n-1) + 5*a(n-2), with a(0)=2, a(1)=3.

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A015523 a(n) = 3a(n-1) + 5a(n-2), with a(0)=0, a(1)=1.

A015449 Expansion of (1-4x)/(1-5x-x^2).

A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3x - 5x^2).

A197189 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=2.

A072263 a(n) = 3a(n-1) + 5a(n-2), with a(0)=2, a(1)=3.

A072264 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=1.

A152185 a(n) = -3a(n-1) + 5a(n-2), n > 1; a(0)=1, a(1)=-5.

A152240 a(n)=5a(n-1)+7a(n-2), n>1 ; a(0)=1, a(1)=7 .