A001076 -id:A001076 - cp's OEIS frontend

A104763 Triangle read by rows: Fibonacci(1), Fibonacci(2), ..., Fibonacci(n) in row n.

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 8, 1, 1, 2, 3, 5, 8, 13, 1, 1, 2, 3, 5, 8, 13, 21, 1, 1, 2, 3, 5, 8, 13, 21, 34, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233

Offset: 1

Gary W. Adamson, Mar 23 2005

Triangle of A104762, Fibonacci sequence in each row starts from the right.
The triangle or chess sums, see A180662 for their definitions, link the Fibonacci(n) triangle to sixteen different sequences, see the crossrefs. The knight sums Kn14 - Kn18 have been added. As could be expected all sums are related to the Fibonacci numbers. - Johannes W. Meijer, Sep 22 2010
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A104763 is reluctant sequence of Fibonacci numbers (A000045), except 0. - Boris Putievskiy, Dec 13 2012

			First few rows of the triangle are:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 2, 3;
  1, 1, 2, 3, 5;
  1, 1, 2, 3, 5, 8;
  1, 1, 2, 3, 5, 8, 13; ...

Reinhard Zumkeller, Rows n = 1..100 of table, flattened
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A104762, A000045.
Cf. A000071 (row sums). - R. J. Mathar, Jul 22 2009
Triangle sums (see the comments): A000071 (Row1; Kn4 & Ca1 & Ca4 & Gi1 & Gi4); A008346 (Row2); A131524 (Kn11); A001911 (Kn12); A006327 (Kn13); A167616 (Kn14); A180671 (Kn15); A180672 (Kn16); A180673 (Kn17); A180674 (Kn18); A052952 (Kn21 & Kn22 & Kn23 & Fi2 & Ze2); A001906 (Kn3 &Fi1 & Ze3); A004695 (Ca2 & Ze4); A001076 (Ca3 & Ze1); A080239 (Gi2); A081016 (Gi3). - Johannes W. Meijer, Sep 22 2010

Flat(List([1..15], n -> List([1..n], k -> Fibonacci(k)))); # G. C. Greubel, Jul 13 2019

a104763 n k = a104763_tabl !! (n-1) !! (k-1)
a104763_row n = a104763_tabl !! (n-1)
a104763_tabl = map (flip take $ tail a000045_list) [1..]
-- Reinhard Zumkeller, Aug 15 2013

[Fibonacci(k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jul 13 2019

Table[Fibonacci[k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jul 13 2019 *)

for(n=1,15, for(k=1,n, print1(fibonacci(k), ", "))) \\ G. C. Greubel, Jul 13 2019

[[fibonacci(k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Jul 13 2019

F(1) through F(n) starting from the left in n-th row.
T(n,k) = A000045(k), 1<=k<=n. - R. J. Mathar, May 02 2008
a(n) = A000045(m), where m= n-t(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 13 2012
G.f.: (x*y)/((x-1)*(x^2*y^2+x*y-1)). - Vladimir Kruchinin, Jun 21 2025

Edited by R. J. Mathar, May 02 2008

Extended by R. J. Mathar, Aug 27 2008

A140455 13-Fibonacci sequence.

Original entry on oeis.org

0, 1, 13, 170, 2223, 29069, 380120, 4970629, 64998297, 849948490, 11114328667, 145336221161, 1900485203760, 24851643870041, 324971855514293, 4249485765555850, 55568286807740343, 726637214266180309

Offset: 0

R. J. Mathar, Jul 22 2008

The k-Fibonacci sequences for k=2..12 are A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A041041, A049666, A041061. This here is k=13. k=14 is A041085, k=16 A041113, k=18 A041145, k=20 A041181, k=22 A041221.
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 13's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,13} avoiding runs of zeros of odd length. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Apr 21 2023: (Start)
Also called the 13-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 13 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon and Angel Plaza, The k-Fibonacci sequence and Pascal 2-triangle, Chaos, Solit. Fract. 33 (2007) 38-49.
Tanya Khovanova, Recursive sequences.
Index entries for linear recurrences with constant coefficients, signature (13,1).

Row n=13 of A073133, A172236 and A352361 and column k=13 of A157103.

F := proc(n,k) coeftayl( x/(1-k*x-x^2),x=0,n) ; end: for n from 0 to 20 do printf("%d,",F(n,13)) ; od:

LinearRecurrence[{13, 1}, {0, 1}, 30] (* Vincenzo Librandi, Nov 17 2012 *)

[lucas_number1(n,13,-1) for n in range(0, 18)] # Zerinvary Lajos, Apr 29 2009

O.g.f.: x/(1-13*x-x^2).
a(n) = 13*a(n-1) + a(n-2).
a(n-r)*a(n+r) - a(n)^2 = (-1)^(n+1-r)*a(r)^2.
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n,2i+1)*13^(n-1-2*i)*(13^2+4)^i/2^(n-1).
a(n) = ((13+sqrt(173))^n - (13-sqrt(173))^n)/(2^n*sqrt(173)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2*n) = 13*A097844(n), a(2*n+1) = A098244(n).
a(3*n+1) = A041319(5*n), a(3*n+2) = A041319(5*n+3), a(3*n+3) = 2*A041319(5*n+4).
Limit_{k->oo} a(n+k)/a(k) = (A088316(n) + A140455(n)*sqrt(173))/2.
Limit_{n->oo} A088316(n)/A140455(n) = sqrt(173). (End)

A048875 Generalized Pellian with second term of 6.

Original entry on oeis.org

1, 6, 25, 106, 449, 1902, 8057, 34130, 144577, 612438, 2594329, 10989754, 46553345, 197203134, 835365881, 3538666658, 14990032513, 63498796710, 268985219353, 1139439674122, 4826743915841, 20446415337486, 86612405265785, 366896036400626, 1554196550868289

Offset: 0

Barry E. Williams

			G.f. = 1 + 6*x + 25*x^2 + 106*x^3 + 449*x^4 + 1902*x^5 + 8057*x^6 + 34130*x^7 + ...

T. D. Noe, Table of n, a(n) for n = 0..200
M. Bicknell, A Primer on the Pell Sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Pellian Representations, Fib. Quart. Vol. 10, No. 5, (1972), pp. 449-488.
Tanya Khovanova, Recursive Sequences
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 (4,1).

Cf. A015448, A001076, A001077, A033887, A134418, A097924, A374439.

with(combinat): a:=n->2*fibonacci(n-1,4)+fibonacci(n,4): seq(a(n), n=1..17); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{4,1},{1,6},40] (* Harvey P. Dale, Nov 30 2011 *)
a[ n_] := (4 I ChebyshevT[ n + 1, -2 I] - 3 ChebyshevT[ n, -2 I]) I^n / 5; (* Michael Somos, Feb 23 2014 *)
a[ n_] := If[ n < 0, SeriesCoefficient[ (1 + 6 x) / (1 + 4 x - x^2), {x, 0, -n}], SeriesCoefficient[ (1 + 2 x) / (1 - 4 x - x^2), {x, 0, n}]]; (* Michael Somos, Feb 23 2014 *)

a[0]:1$ a[1]:6$ a[n]:=4*a[n-1]+a[n-2]$ makelist(a[n], n, 0, 30); /* Martin Ettl, Nov 03 2012 */

{a(n) = ( 4*I*polchebyshev( n+1, 1, -2*I) - 3*polchebyshev( n, 1, -2*I)) * I^n / 5}; /* Michael Somos, Feb 23 2014 */

{a(n) = if( n<0, polcoeff( (1 + 6*x) / (1 + 4*x - x^2) + x * O(x^-n), -n), polcoeff( (1 + 2*x) / (1 - 4*x - x^2) + x * O(x^n), n))}; \\ Michael Somos, Feb 23 2014

a(n) = ((4+sqrt(5))*(2+sqrt(5))^n - (4-sqrt(5))*(2-sqrt(5))^n)*sqrt(5)/2.
a(n) = 4a(n-1) + a(n-2); a(0)=1, a(1)=6.
Binomial transform of A134418: (1, 5, 14, 48, 152, ...). - Gary W. Adamson, Nov 23 2007
G.f.: (1+2*x)/(1-4*x-x^2). - Philippe Deléham, Nov 03 2008
a(-1 - n) = (-1)^n * A097924(n) for all n in Z. - Michael Somos, Feb 23 2014
a(n) = A001076(n+1) + 2*A001076(n). - R. J. Mathar, Sep 11 2019
a(n) = 4^n*Sum_{k=0..n} A374439(n, k)*(1/4)^k. - Peter Luschny, Jul 26 2024

Corrected by T. D. Noe, Nov 07 2006

A154597 a(n) = 15*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 15, 226, 3405, 51301, 772920, 11645101, 175449435, 2643386626, 39826248825, 600037119001, 9040383033840, 136205782626601, 2052127122432855, 30918112619119426, 465823816409224245, 7018275358757483101, 105739954197771470760, 1593117588325329544501

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009

Limit_{n -> infinity} a(n)/a(n-1) = (15 + sqrt(229))/2. - Klaus Brockhaus, Oct 07 2009
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 15's along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n - 1 on alphabet {0,1,...,15} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Apr 30 2023: (Start)
Also called the 15-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 15 kinds of squares available. (End)

G. C. Greubel, Table of n, a(n) for n = 0..840 (a(0) = 0 added by Jianing Song)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Tanya Khovanova, Recursive sequences. [From _Johannes W. Meijer_, Jun 12 2010]
Index entries for linear recurrences with constant coefficients, signature (15,1).

Row n=15 of A073133, A172236 and A352361 and column k=15 of A157103.
First bisection is A098247.
Cf. A166125 (decimal expansion of sqrt(229)), A166126 (decimal expansion of (15 + sqrt(229))/2).
Cf. also A041427, A090301, A098245.
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), this sequence (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-229); S:=[ ((15+r)^n-(15-r)^n)/(2^n*r): n in [1..17] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jan 12 2009

[n le 2 select n-1 else 15*Self(n-1) +Self(n-2): n in [1..30]]; // G. C. Greubel, Sep 20 2024

LinearRecurrence[{15,1}, {0,1}, 31] (* Vladimir Joseph Stephan Orlovsky, Oct 27 2009 *)
CoefficientList[Series[x/(1-15*x-x^2), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)

my(x='x+O('x^50)); concat([0], Vec(x/(1-15*x-x^2))) \\ G. C. Greubel, Apr 16 2017

def A154597(n): return (-i)^(n-1)*chebyshev_U(n-1, 15*i/2)
[A154597(n) for n in range(31)] # G. C. Greubel, Sep 20 2024

G.f.: x/(1 - 15*x - x^2). - Klaus Brockhaus, Jan 12 2009, corrected Oct 07 2009
a(n) = ((15 + sqrt(229))^n - (15 - sqrt(229))^n)/(2^n*sqrt(229)).
From Johannes W. Meijer, Jun 12 2010: (Start)
Limit_{k -> infinity} a(n+k)/a(k) = (A090301(n) + a(n)*sqrt(229))/2.
Limit_{n -> infinity} A090301(n)/a(n) = sqrt(229).
a(2n+1) = 15*A098245(n-1).
a(3n+1) = A041427(5n), a(3n+2) = A041427(5n+3), a(3n+3) = 2*A041427(5n+4). (End)
E.g.f.: (2/sqrt(229))*exp(15*x/2)*sinh(sqrt(229)*x/2). - G. C. Greubel, Sep 20 2024

Extended beyond a(7) by Klaus Brockhaus and Philippe Deléham, Jan 12 2009
Name from Philippe Deléham, Jan 12 2009
Edited by Klaus Brockhaus, Oct 07 2009
Missing a(0) added by Jianing Song, Jan 29 2019

A110679 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.

Original entry on oeis.org

1, 2, 11, 44, 189, 798, 3383, 14328, 60697, 257114, 1089155, 4613732, 19544085, 82790070, 350704367, 1485607536, 6293134513, 26658145586, 112925716859, 478361013020, 2026369768941, 8583840088782, 36361730124071, 154030760585064, 652484772464329

Offset: 0

Creighton Dement, Aug 02 2005

2tesseq[A*B*cyc(A)] (see program code) gives an alternative formula for A110528.

a(n) is the number of tilings of a 2 X n rectangle by using 1 X 1 squares, dominoes and right trominoes. - Roberto Tauraso, Mar 21 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3,5,1).

Cf. A110528, A110680, A001076, A110526, A110526, A033887.

[(Fibonacci(3*n+2) +(-1)^n)/2: n in [0..30]]; // G. C. Greubel, Apr 19 2019

seriestolist(series((-1+x)/((x+1)*(x^2+4*x-1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: -1jesseq[A*B*cyc(A)] with A = - 'j + 'k - 'ii' - 'ij' - 'ik' and B = - .5'i - .5i' - .5'ii' + .5'jj' - .5'kk' + .5'jk' + .5'kj' - .5e

a[n_] := (Fibonacci[3*n+2] + (-1)^n)/2; a /@ Range[0, 22] (* Giovanni Resta, Mar 21 2017 *)

Vec((1 - x) / ((1 + x)*(1 - 4*x - x^2)) + O(x^30)) \\ Colin Barker, Mar 21 2017

{a(n) = -(-1)^n * (fibonacci(-2 - 3*n)\2)}; /* Michael Somos, Mar 26 2017 */

[(fibonacci(3*n+2) +(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

Program "FAMP" finds: 2*(-1^(n+1)) = A110528(n) - A001076(n+1) - 2*a(n). Program "Superseeker" finds: a(n) = A110526(n+1) - A110526(n); a(n) + a(n+1) = A033887(n+1).
a(n) = (-1)^n*Sum_{k=0..n} (-1)^k*Fibonacci(3*k+1). - Gary Detlefs, Jan 22 2013
a(n) = (Fibonacci(3*n+2)+(-1)^n)/2. - Roberto Tauraso, Mar 21 2017
From Colin Barker, Mar 21 2017: (Start)
G.f.: (1 - x) / ((1 + x)*(1 - 4*x - x^2)).
a(n) = 3*a(n-1) + 5*a(n-2) + a(n-3) for n>2.
(End)
a(n) = -(-1)^n * A049651(-1 - n) for all n in Z. - Michael Somos, Mar 26 2017
a(2*n) = A254627(2*n+1); a(2*n+1) = A077259(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
2*a(n) = A015448(n+1)+(-1)^n. - R. J. Mathar, Oct 03 2021

Typo in program code fixed by Creighton Dement, Dec 11 2009

A167808 Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.

Original entry on oeis.org

0, 1, 1, 1, 3, 5, 4, 13, 21, 17, 55, 89, 72, 233, 377, 305, 987, 1597, 1292, 4181, 6765, 5473, 17711, 28657, 23184, 75025, 121393, 98209, 317811, 514229, 416020, 1346269, 2178309, 1762289, 5702887, 9227465, 7465176, 24157817, 39088169, 31622993

Offset: 0

Reinhard Zumkeller, Nov 12 2009

Define a sequence c(n) by c(0)=0, c(1)=1; thereafter c(n) = (c(n-2)*c(n-1)-1)/(c(n-2)+c(n-1)+2). Then it appears that (apart from signs), a(n) is the denominator of c(n). - Jonas Holmvall, Jun 21 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,1).

Cf. A000045, A130196 (denominator).

The a(2*n) appear in A179135. - Johannes W. Meijer, Jul 01 2010

a:=[0,1,1,1,3,5];; for n in [7..40] do a[n]:=4*a[n-3]+a[n-6]; od; a; # Muniru A Asiru, Oct 16 2018

nmax:=39; x(0):=0: x(1):=1/2:for n from 2 to nmax do x(n):=x(n-1)+x(n-2) od: for n from 0 to nmax do a(n):= numer(x(n)) od: seq(a(n),n=0..nmax); # Johannes W. Meijer, Jul 01 2010
with(combinat):f:=n->fibonacci(n):L:=n->f(n)+2*f(n-1):seq(numer(f(n)/L(n)), n=0..39); # Gary Detlefs, Dec 11 2010

f[n_]:=Numerator[Fibonacci[n]/Fibonacci[n+3]];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)
Numerator[LinearRecurrence[{1,1},{0,1/2},40]] (* Harvey P. Dale, Aug 08 2014 *)
CoefficientList[Series[-x (1 + x + x^2 - x^3 + x^4)/((x^2 + x - 1) (x^4 - x^3 + 2 x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 08 2014 *)
LinearRecurrence[{0, 0, 4, 0, 0, 1},{0, 1, 1, 1, 3, 5},40] (* Ray Chandler, Aug 03 2015 *)
a[n_]:=If[Mod[n,3]==0, Fibonacci[n]/2, Fibonacci[n]]; Array[a, 40, 0] (* Stefano Spezia, Oct 16 2018 *)

a(n) = (a(n-1)*A131534(n) + a(n-2)*A131534(n+2))/A131534(n+1) for n > 1.
a(3*n) = A001076(n) = (a(3*n-1) + a(3*n-2))/2;
a(3*n+1) = A033887(n) = 2*a(3*n-1) + a(3*n-2);
a(3*n+2) = A015448(n+1) = a(3*n-1) + 2*a(3*n-2).
From Johannes W. Meijer, Jul 01 2010: (Start)
a(2*n) = A001906(n)/A131534(n+1) for n >= 0 and a(2*n+1) = A179131(n)/5 for n >= 1.
a(6*n+2) - 2*a(6*n) = A134493(n);
2*a(6*n+1) - a(6*n+3) = A023039(n);
2*a(6*n+4) - a(6*n+2) = A134497(n);
a(6*n+5) - 2*a(6*n+3) = A103134(n);
2*a(6*n+4) - a(6*n+6) = A075796(n).
(End)
From Gary Detlefs, Dec 11 2010: (Start)
a(n) = numerator(A000045(n)/A000032(n)).
If n mod 3 = 0 then a(n) = Fibonacci(n)/2, else a(n)= Fibonacci(n). (End)
G.f.: -x*(1 + x + x^2 - x^3 + x^4) / ( (x^2 + x - 1)*(x^4 - x^3 + 2*x^2 + x + 1) ). - R. J. Mathar, Mar 08 2011
a(n) = 4*a(n-3) + a(n-6). - Muniru A Asiru, Oct 16 2018

Typo in title corrected by Johannes W. Meijer, Jun 26 2010

A180141 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + x - 2x^2)/(1 - 3x - 6*x^2).

Original entry on oeis.org

1, 4, 16, 72, 312, 1368, 5976, 26136, 114264, 499608, 2184408, 9550872, 41759064, 182582424, 798301656, 3490399512, 15261008472, 66725422488, 291742318296, 1275579489816, 5577192379224, 24385054076568, 106618316505048

Offset: 0

Johannes W. Meijer, Aug 13 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner square (m = 1, 3, 7 or 9) on a 3 X 3 chessboard. This fairy chess piece behaves like a rook on the eight side and corner squares but on the central square the rook goes berserk and turns into a berserker, see A180140.
On a 3 X 3 chessboard there are 2^9 = 512 ways to go berserk on the central square (we assume here that a berserker might behave like a rook). The berserker is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program. For the corner squares the 512 berserkers lead to 42 berserker sequences, see the cross-references for some examples.
The sequence above corresponds to just one A[5] vectors with decimal value 495. This vector leads for the side squares to 4*A154964 (for n >= 1 with a(0) = 1) and for the central square to 2*A180141 (for n >= 1 with a(0)=1).
This sequence belongs to a family of sequences with g.f. (1 + x + k*x^2)/(1 - 3*x + (k-4)*x^2), see A123620.

Index entries for linear recurrences with constant coefficients, signature (3, 6).

Cf. A180140 (side squares) and A180147 (central square).

Cf. Berserker sequences corner squares [numerical value A[5]]: 4*A055099 [0, with leading 1 added], A180143 [16], 4*A001353 [17, n>=1 and a(0)=1], A123620 [3], 2*A018916 [19, with leading 1 added], A000302 [15], 4*A179606 [111, with leading 1 added], A089979 [343], 4*A001076 [95, n>=1 and a(0)=1], A180145 [191], A180141 [495, this sequence], 4*A090017 [383, n>=1 and a(0)=1].

with(LinearAlgebra): nmax:=22; m:=1; A[5]:= [1,1,1,1,0,1,1,1,1]: A:= Matrix([[0,1,1,1,0,0,1,0,0], [1,0,1,0,1,0,0,1,0], [1,1,0,0,0,1,0,0,1], [1,0,0,0,1,1,1,0,0], A[5], [0,0,1,1,1,0,0,0,1], [1,0,0,1,0,0,0,1,1], [0,1,0,0,1,0,1,0,1], [0,0,1,0,0,1,1,1,0]]): 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);

LinearRecurrence[{3, 6}, {1, 4, 16}, 23] (* Jean-François Alcover, Jan 18 2025 *)

G.f.: (1 + x - 2*x^2)/(1 - 3*x - 6*x^2).
a(n) = 4*a(n-1) - 2*a(n-3) with a(0)=2, a(1)=8 and a(2)=31.
a(n) = 3*a(n-1) + 6*a(n-2) for n >= 3 with a(0)=1, a(1)=4 and a(2)=16.
a(n) = (6+2*A)*A^(-n-1)/33 + (6+2*B)*B^(-n-1)/33 with A=(-3-sqrt(33))/12 and B=(-3+sqrt(33))/12 for n >= 1 with a(0)=1.

A180226 a(n) = 4a(n-1) + 10a(n-2), with a(1)=0 and a(2)=1.

Original entry on oeis.org

0, 1, 4, 26, 144, 836, 4784, 27496, 157824, 906256, 5203264, 29875616, 171535104, 984896576, 5654937344, 32468715136, 186424233984, 1070384087296, 6145778689024, 35286955629056, 202605609406464, 1163291993916416, 6679224069730304, 38349816218085376

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 (4, 10).

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

I:=[0,1]; [n le 2 select I[n] else 4*Self(n-1) + 10*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2018

Join[{a=0,b=1},Table[c=4*b+10*a;a=b;b=c,{n,100}]]
LinearRecurrence[{4,10}, {0,1}, 30] (* G. C. Greubel, Jan 16 2018 *)

x='x+O('x^30); concat([0], Vec(x^2/(1-4*x-10*x^2))) \\ G. C. Greubel, Jan 16 2018

a(n) = ((2+sqrt(14))^(n-1) - (2-sqrt(14))^(n-1))/(2*sqrt(14)). - Rolf Pleisch, May 14 2011

G.f.: x^2/(1-4*x-10*x^2).

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

A181866 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the fourth powers of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

Original entry on oeis.org

1, 2, 58, 200195112, 580000000008023436288643185139644928

Offset: 1

Peter Bala, Nov 29 2010

The calculations for the first few values of the sequence are
... 2^4 = 16 so a(3) = 116/2 = 58
... 58^4 = 11316496 so a(4) = 11611316496/58 = 200195112.
The value of a(6) is calculated in the Example section below.
For similarly defined sequences see A181754 through A181756 and
A181864 through A181870.

			The recurrence relation (2) above gives
a(6) = a(5)^3+10^144*a(4)
= 200 19511 20000 00000 00000 00000 00000 00000 00000 00195 11200
00080 97251 90261 07158 64917 75226 28886 69453 43420 83613 55167
37330 42401 44438 01550 47183 94579 01959 53586 66752.
a(6) has 153 digits.

Cf. A000045, A001076 (F(n,4)), A181754, A181755, A181756, A181864, A181865,

A181867, A181868, A181869, A181870

#A181866
M:=5:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^4,string):
s[2]:=cat(s[1],convert(a[2]^4,string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n]^4,string));
end do:
seq(a[n],n = 1..M);

DEFINITION
a(1) = 1, a(2) = 2, and for n >= 3
(1)... a(n) = concatenate(a(1)^4,a(2)^4,...,a(n-1)^4)/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = a(n+1)^3 + (100^F(n,4))*a(n)
= a(n+1)^3 + (10^F(3*n))*a(n),
where F(n,4) is the Fibonacci polynomial F(n,x) evaluated at x = 4
and where F(n) denotes the n-th Fibonacci number A000045(n).
F(n,4) = A001076(n).

cp's OEIS Frontend

A104763 Triangle read by rows: Fibonacci(1), Fibonacci(2), ..., Fibonacci(n) in row n.

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A140455 13-Fibonacci sequence.

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A048875 Generalized Pellian with second term of 6.

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A154597 a(n) = 15*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.

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A110679 a(n+3) = 3*a(n+2) + 5*a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.

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A167808 Numerator of x(n), where x(n) = x(n-1) + x(n-2) with x(0)=0, x(1)=1/2.

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A110679 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 1, a(1) = 2, a(2) = 11.

A180141 Eight rooks and one berserker on a 3 X 3 chessboard. G.f.: (1 + x - 2x^2)/(1 - 3x - 6*x^2).

A180226 a(n) = 4a(n-1) + 10a(n-2), with a(1)=0 and a(2)=1.

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