A001077 -id:A001077 - cp's OEIS frontend

A213893 Fixed points of a sequence h(n) defined by the minimum number of 4's in the relation n*[n,4,4,...,4,n] = [x,...,x] between simple continued fractions.

3, 7, 43, 67, 103, 127, 163, 223, 283, 367, 463, 487, 523, 547, 607, 643, 727, 787, 823, 883, 907, 1063, 1123, 1303, 1327, 1423, 1447, 1543, 1567, 1627, 1663, 1723, 1747, 1783, 1867, 1987, 2083, 2143, 2203, 2287, 2347, 2383, 2467, 2683, 2707, 2767, 2803, 2887

Offset: 1

Art DuPre, Jun 23 2012

nonn

In a variant of A213891, multiply n by a number with simple continued fraction [n,4,4,...,4,n] and increase the number of 4's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2*[2,4,2] = [4,2,4],
3*[3,4,4,4,3] = [9,1,2,2,2,1,9],
4*[4,4,4] = [16,1,16],
5*[5,4,4,4,4,5] = [26,5,1,1,5,26].
The number of 4's needed defines the sequence h(n) = 1, 3, 1, 4, 3, 7, 3, 3, 9, ... (n>=2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences(sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n) = 4*f(n-1) + f(n-2), A001076, A001077, A015448, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 4*f(n-1) + f(n-2).
The above sequence h() is recorded as A262214. - M. F. Hasler, Sep 15 2015

Cf. A000057, A213891, A213892, A213894 - A213899, A261311; A213358.

Cf. A213648, A262212 - A262220, A213900, A262211.

f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[4, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)

{a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,4), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A226956 a(0)=a(1)=1, a(n+2) = a(n+1) + a(n) - A128834(n).

Original entry on oeis.org

1, 1, 2, 2, 3, 5, 9, 15, 24, 38, 61, 99, 161, 261, 422, 682, 1103, 1785, 2889, 4675, 7564, 12238, 19801, 32039, 51841, 83881, 135722, 219602, 355323, 574925, 930249, 1505175, 2435424, 3940598, 6376021, 10316619, 16692641, 27009261, 43701902, 70711162, 114413063, 185124225, 299537289

Offset: 0

Paul Curtz, Jun 24 2013

a(n) and differences:
    1,  1, 2, 2, 3, 5, 9, 15, 24, 38, ...             a(n)
    0,  1, 0, 1, 2, 4, 6,  9, 14, 23, 38, ...         b(n)
    1, -1, 1, 1, 2, 2, 3,  5,  9, 15, 24, 38, ...     a(n-2)
   -2,  2, 0, 1, 0, 1, 2,  4,  6,  9, 14, 23, 38, ... b(n-2)
    4, -2, 1,-1, 1, 1, 2,  2,  3,  5,  9, ...         a(n-4)
   -6,  3,-2, 2, 0, 1, 0,  1,  2,  4,  6,  9, ...     b(n-4)
    9, -5, 4,-2, 1,-1, 1,  1,  2,  2,  3,  5, 9, ...  a(n-6)
  -14,  9,-6, 3,-2, 2, 0,  1   0,  1,  2, ...         b(n-6)
   23,-15, 9,-5, 4,-2, 1, -1,  1,  1,  2,  2, ...     a(n-8)
a(n)-b(n+1) = period 6: repeat 0, 1, 1, 0, -1, -1 = A128834(n).
Diagonals with the same number give 1, 2, 9, 38, ... = A001077(n).
Second column: the (n+2)-th term is identical to a(n+1) signed.
a(n+1) is identical to its twice shifted inverse binomial transform signed.
a(n+1)/a(n) tends to A001622 (the golden ratio) as n -> infinity.

			a(0) = a(1) = 1.
a(2) = a(3) = 2.
a(4) = 2*a(3) - a(2) + a(0) = 4-2+1 = 3.
a(5) = 6-2+1 = 5.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1)

Cf. Diagonals in A024490.

I:=[1, 1, 2, 2]; [n le 4 select I[n] else 2*Self(n-1) - Self(n-2) + Self(n-4): n in [1..30]]; // G. C. Greubel, Jan 15 2018

a[n_] := (LucasL[n] + {0, 1, 1, 0, -1, -1}[[Mod[n, 6] + 1]])/2; Table[a[n], {n, 0, 42}] (* Jean-François Alcover, Jun 28 2013, after R. J. Mathar *)
LinearRecurrence[{2,-1,0,1}, {1,1,2,2}, 30] (* G. C. Greubel, Jan 15 2018 *)

x='x+O('x^30); Vec((x-1)*(1+x^2)/((x^2+x-1)*(x^2-x+1))) \\ G. C. Greubel, Jan 15 2018

a(n+6) - a(n-6) = 20*A000045(n).
a(n) = 2*a(n-1) - a(n-2) + a(n-4).
a(n) = 3*a(n-3) + 5*a(n-6) + a(n-9) (plus many similar by telescoping the fundamental recurrence).
a(n+3) - a(n-3) = 2*A000032(n).
G.f.: (x-1)*(1+x^2) / ( (x^2+x-1)*(x^2-x+1) ). - R. J. Mathar, Jun 26 2013
2*a(n) = A000032(n) + A010892(n-1). - R. J. Mathar, Jun 26 2013
a(n+5) = a(n+4) + a(n+2) + A108014(n).
a(2n+1) + A226447(2n+2) = 2*A182895(n).
a(n) - a(n-2) = 0,2,1,1,1,3,6,... = abs(A111734(n-2)).

A328695 Rectangular array R read by descending antidiagonals: divide to each even term of the Wythoff array (A035513) by 2, and delete all others.

Original entry on oeis.org

1, 4, 2, 17, 9, 3, 72, 38, 5, 12, 305, 161, 8, 51, 6, 1292, 682, 13, 216, 10, 7, 5473, 2889, 21, 915, 16, 30, 14, 23184, 12238, 34, 3876, 26, 127, 59, 25, 98209, 51841, 55, 16419, 42, 538, 250, 106, 11, 416020, 219602, 89, 69552, 68, 2279, 1059, 449, 18, 33

Offset: 1

Clark Kimberling, Oct 26 2019

Every positive integer occurs exactly once in R, and every row of R is a linear recurrence sequence.  The appearance of a sequence s(r) below means that corresponding row of R is the same as s(r) except possibly for one or more initial terms of s(r).
Row 1 of R: A001076
Row 2 of R: A001077
Row 3 of R: A000045
Row 4 of R: A115179
Row 5 of R: A006355
Row 6 of R: A097924
Row 8 of R: A048875
Row 9 of R: A000032

			Row 1 of the Wythoff array is (1,2,3,5,8,13,21,34,55,89,144,...), so that row 1 of R is (1,4,17,72,...).
_______________
Northwest corner of R:
   1   4   17   72  305  1292   5473
   2   9   38  161  682  2889  12238
   3   5    8   13   21    34     55
  12  51  216  915 3876 16419  69552
   6  10   16   26   42    68    110
   7  30  127  538 2279  9654  40895

Cf. A035513, A001076, A001077, A000045, A115179, A006355, A097924, A048875, A000032, A328696, A328697.

w[n_, k_] := Fibonacci[k + 1] Floor[n*GoldenRatio] + (n - 1) Fibonacci[k];
Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten;
q[n_, k_] := If[Mod[w[n, k], 2] == 0, w[n, k]/2, 0];
t[n_] := Union[Table[q[n, k], {k, 1, 50}]];
u[n_] := If[First[t[n]] == 0, Rest[t[n]], t[n]]
Table[u[n], {n, 1, 10}] (* A328695 array *)
v[n_, k_] := u[n][[k]];
Table[v[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* A328695 sequence *)

A048877 a(n) = 4*a(n-1) + a(n-2); a(0)=1, a(1)=8.

Original entry on oeis.org

1, 8, 33, 140, 593, 2512, 10641, 45076, 190945, 808856, 3426369, 14514332, 61483697, 260449120, 1103280177, 4673569828, 19797559489, 83863807784, 355252790625, 1504874970284, 6374752671761

Offset: 0

Barry E. Williams

Generalized Pellian with second term of 8.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A015448, A001076, A001077, A033887.

a048877 n = a048877_list !! n
a048877_list = 1 : 8 : zipWith (+) a048877_list (map (* 4) $ tail a048877_list)
-- Reinhard Zumkeller, May 01 2013

with(combinat): a:=n->4*fibonacci(n-1,4)+fibonacci(n,4): seq(a(n), n=1..16); # Zerinvary Lajos, Apr 04 2008

CoefficientList[Series[(1+4x)/(1-4x-x^2),{x,0,20}],x]  (* Harvey P. Dale, Mar 30 2011 *)
LinearRecurrence[{4,1},{1,8},30] (* Harvey P. Dale, Nov 03 2013 *)

a(n) = ((6+sqrt(5))*(2+sqrt(5))^n - (6-sqrt(5))*(2-sqrt(5))^n )/(2*sqrt(5)).
G.f.: (1+4*x)/(1-4*x-x^2). - Philippe Deléham, Nov 03 2008
a(n)=4*a(n-1) + a(n-2); a(0)=1, a(1)=8.

A048879 Generalized Pellian with second term of 10.

Original entry on oeis.org

1, 10, 41, 174, 737, 3122, 13225, 56022, 237313, 1005274, 4258409, 18038910, 76414049, 323695106, 1371194473, 5808472998, 24605086465, 104228818858, 441520361897, 1870310266446, 7922761427681, 33561355977170, 142168185336361, 602234097322614

Offset: 0

Barry E. Williams

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (4,1)

Cf. A015448, A001077, A001076, A033887.

a048879 n = a048879_list !! n
a048879_list = 1 : 10 : zipWith (+)
                        a048879_list (map (* 4) $ tail a048879_list)
-- Reinhard Zumkeller, Mar 03 2014

with(combinat): a:=n->6*fibonacci(n-1,4)+fibonacci(n,4): seq(a(n), n=1..16); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{4,1},{1,10},30] (* Harvey P. Dale, Jul 18 2011 *)

a(n) = ((8+sqrt(5))*(2+sqrt(5))^n - (8-sqrt(5))*(2-sqrt(5))^n)2*sqrt(5).
From Philippe Deléham, Nov 03 2008: (Start)
a(n) = 4*a(n-1) + a(n-2); a(0)=1, a(1)=10.
G.f.: (1+6*x)/(1-4*x-x^2). (End)
For n >= 1, a(n) equals the denominator of the continued fraction [4, 4, ..., 4, 10] (with n copies of 4). The numerator of that continued fraction is a(n+1). - ZhenShu Luan, Aug 05 2019

More terms from Harvey P. Dale, Jul 18 2011

A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

Original entry on oeis.org

3, 21, 72, 329, 1353, 5796, 24447, 103729, 439128, 1860621, 7880997, 33385604, 141421803, 599075421, 2537719272, 10749959329, 45537545553, 192900159396, 817138154247, 3461452823129, 14662949371128, 62113250430021

Offset: 1

Gary W. Adamson, Jun 12 2003

a(n+1)/a(n) -> (phi)^3 = ((1 + sqrt(5))/2)^3 = 4.236067...

			a(4) = Lucas(4)*Lucas(8) = 7*47 = 329.

Vincenzo Librandi, Table of n, a(n) for n = 1..160
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7
Index entries for linear recurrences with constant coefficients, signature (3, 6, -3, -1).

Third row of array A028412.

[Lucas(n)*Lucas(2*n): n in [1..25]]; // Vincenzo Librandi, May 03 2011

Table[Fibonacci[n*4]/Fibonacci[n],{n,50}] (* Vladimir Joseph Stephan Orlovsky, May 02 2011 *)

From Benoit Cloitre, Aug 30 2003: (Start)
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4);
a(n) = Fibonacci(4*n)/Fibonacci(n) = A000045(4*n)/A000045(n). (End)
a(n) = Lucas(3*n) + (-1)^n*Lucas(n).
From R. J. Mathar, Oct 27 2008: (Start)
G.f.: x*(3+12*x-9*x^2-4*x^3)/((1+x-x^2)*(1-4*x-x^2)).
a(n) = A061084(n+1) + 2*A001077(n). (End)
a(n) = (1+phi)^n + (-phi)^n + (2*phi+1)^n + (3-2*phi)^n, phi = (1+sqrt(5))/2. - Gary Detlefs, Dec 09 2012

A162770 a(n) = ((2+sqrt(5))(1+sqrt(5))^n + (2-sqrt(5))(1-sqrt(5))^n)/2.

Original entry on oeis.org

2, 7, 22, 72, 232, 752, 2432, 7872, 25472, 82432, 266752, 863232, 2793472, 9039872, 29253632, 94666752, 306348032, 991363072, 3208118272, 10381688832, 33595850752, 108718456832, 351820316672, 1138514460672, 3684310188032

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jul 13 2009

nonn

Binomial transform of A162963. Inverse binomial transform of A001077 without initial 1.

Michael De Vlieger, Table of n, a(n) for n = 0..1960
Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 35.
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A000032, A001077, A162963.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-5); S:=[ ((2+r)*(1+r)^n+(2-r)*(1-r)^n)/2: n in [0..24] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jul 19 2009

LinearRecurrence[{2,4},{2,7},30] (* Harvey P. Dale, Jan 13 2015 *)

a(n) = 2*a(n-1) + 4*a(n-2) for n > 1; a(0) = 2, a(1) = 7.
G.f.: (2+3*x)/(1-2*x-4*x^2).
a(n) = 2^(n-1) * A000032(n+3). - Diego Rattaggi, Jun 24 2020

Edited and extended beyond a(5) by Klaus Brockhaus, Jul 19 2009

A179604 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2x - 9x^2 - 2*x^3).

Original entry on oeis.org

1, 3, 15, 59, 259, 1079, 4607, 19443, 82507, 349215, 1479879, 6267707, 26552755, 112474631, 476459471, 2018296131, 8549676763, 36216937647, 153417558423, 649886909195, 2752965719491, 11661748738583, 49399962770975

Offset: 0

Johannes W. Meijer, Jul 28 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 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 4 red king vectors, i.e., A[5] vectors, with decimal [binary] values 327 [1,0,1,0,0,0,1,1,1], 333 [1,0,1,0,0,1,1,0,1], 357 [1,0,1,1,0,0,1,0,1] and 453 [1,1,1,0,0,0,1,0,1]. These vectors lead for the side squares to A015448 and for the central square to A179605.

Index entries for linear recurrences with constant coefficients, signature (2,9,2).

Cf. A001076, A001077, A015448, A179605, A179596.

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

LinearRecurrence[{2,9,2},{1,3,15},30] (* or *) CoefficientList[ Series[ (x+1)/(-2 x^3-9 x^2-2 x+1),{x,0,30}],x] (* Harvey P. Dale, Mar 17 2012 *)

G.f.: ( -1-x ) / ( (2*x+1)*(x^2 + 4*x - 1) ).
a(n) = 2*a(n-1) + 9*a(n-2) + 2*a(n-3) with a(0)=1, a(1)=3 and a(2)=15.
a(n) = (20*(-1/2)^(-n) + (5+7*sqrt(5))*A^(-n-1) + (5-7*sqrt(5))*B^(-n-1))/110 with A = (-2+sqrt(5)) and B:= (-2-sqrt(5)).
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n+1)/(A001076(n)*sqrt(5) - A001077(n)).

A179605 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + 3x - 2x^2)/(1 - 2x - 9x^2 - 2*x^3).

Original entry on oeis.org

1, 5, 17, 81, 325, 1413, 5913, 25193, 106429, 451421, 1911089, 8097825, 34298293, 145299189, 615478665, 2607246617, 11044399597, 46784976077, 198184041761, 839521667409, 3556269662821, 15064602415845, 63814675131897

Offset: 0

Johannes W. Meijer, Jul 28 2010

The a(n) represent 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 toes crazy and turns into a red king, see A179596.

The sequence above corresponds to 4 red king vectors, A[5] vectors, with decimal [binary] values 327 [1,0,1,0,0,0,1,1,1], 333 [1,0,1,0,0,1,1,0,1], 357 [1,0,1,1,0,0,1,0,1] and 453 [1,1,1,0,0,0,1,0,1]. These vectors lead for the corner squares to A179604 and for the side squares to A015448.

Index entries for linear recurrences with constant coefficients, signature (2,9,2).

Cf. A001076, A001077, A015448, A179596, A179597 (central square), A179604.

with(LinearAlgebra): nmax:=21; 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]:= [1,0,1,1,0,0,1,0,1]: 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);

G.f.: ( -1 - 3*x + 2*x^2 ) / ( (2*x+1)*(x^2 + 4*x - 1) ).
a(n) = 2*a(n-1) + 9*a(n-2) + 2*a(n-3) with a(0)=1, a(1)=5 and a(2)=17.
a(n) = (-4/11)*(-1/2)^(-n) + ((17+41*A)*A^(-n-1) + (17+41*B)*B^(-n-1))/110 with A = (-2+sqrt(5)) and B =(-2-sqrt(5)).
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n+1)/(A001076(n)*sqrt(5) - A001077(n)).

A089926 a(n) = 12*a(n-1) + a(n-2), a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 73, 882, 10657, 128766, 1555849, 18798954, 227143297, 2744518518, 33161365513, 400680904674, 4841332221601, 58496667563886, 706801342988233, 8540112783422682, 103188154744060417, 1246797969712147686

Offset: 0

Paul Barry, Nov 15 2003

The family of recurrences a(n) = 2*k*a(n-1) + a(n-2), a(0)=1, a(1)=k has solution a(n) = ((k+sqrt(k^2+1))^n + (k-sqrt(k^2+1))^n)/2; a(n) = Sum_{j=0..floor(n/2)} C(n,2k)*(k^2+1)^jk^(n-2j); a(n) = T(n,ki)*(-i)^n; e.g.f. exp(kx)*cosh(sqrt(k^2+1)*x).

Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (12,1).

Cf. A088320, A088317, A005667, A001077.

Essentially the same as A041060.

E.g.f.: exp(6x)*cosh(sqrt(37)x);
a(n) = ((6+sqrt(37))^n + (6-sqrt(37))^n)/2;
a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*37^k*6^(n-2k).
a(n) = T(n, 6i)*(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = -1.
G.f.: (1-6x)/(1-12*x-x^2). - Philippe Deléham, Nov 21 2008

cp's OEIS Frontend

A213893 Fixed points of a sequence h(n) defined by the minimum number of 4's in the relation n*[n,4,4,...,4,n] = [x,...,x] between simple continued fractions.

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A226956 a(0)=a(1)=1, a(n+2) = a(n+1) + a(n) - A128834(n).

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A328695 Rectangular array R read by descending antidiagonals: divide to each even term of the Wythoff array (A035513) by 2, and delete all others.

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A048877 a(n) = 4*a(n-1) + a(n-2); a(0)=1, a(1)=8.

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A048879 Generalized Pellian with second term of 10.

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A083564 a(n) = L(n)*L(2n), where L(n) are the Lucas numbers (A000204).

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A162770 a(n) = ((2+sqrt(5))*(1+sqrt(5))^n + (2-sqrt(5))*(1-sqrt(5))^n)/2.

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A162770 a(n) = ((2+sqrt(5))(1+sqrt(5))^n + (2-sqrt(5))(1-sqrt(5))^n)/2.

A179604 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 2x - 9x^2 - 2*x^3).

A179605 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + 3x - 2x^2)/(1 - 2x - 9x^2 - 2*x^3).