A015449 -id:A015449 - cp's OEIS frontend

A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

1, 1, 6, 16, 56, 176, 576, 1856, 6016, 19456, 62976, 203776, 659456, 2134016, 6905856, 22347776, 72318976, 234029056, 757334016, 2450784256, 7930904576, 25664946176, 83053510656, 268766806016, 869747654656, 2814562533376, 9108115685376, 29474481504256

Offset: 0

Paul Barry, May 10 2003

Inverse binomial transform of A001077. Binomial transform of expansion of cosh(sqrt(5)*x) (1,0,5,0,25,...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 5 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(5). - Cino Hilliard, Sep 25 2005
Numerators of fractions in the approximation of the square root of 5 satisfying: a(n) = (a(n-1)+c)/(a(n-1)+1), with c=5 and a(1)=1. For denominators see A063727. - Mark Dols, Jul 24 2009
Equals right border of triangle A143969. (1, 6, 16, 56, ...) = row sums of triangle A143969 and INVERT transform of (1, 5, 5, 5, ...). - Gary W. Adamson, Sep 06 2008
a(n) is the number of compositions of n when there are 1 type of 1 and 5 types of other natural numbers. - Milan Janjic, Aug 13 2010
From Gary W. Adamson, Jul 30 2016: (Start)
The sequence is case N=1 in an infinite set obtained by taking powers of the 2 X 2 matrix M = [(1,5); (1,N)], then extracting the upper left terms. The infinite set begins:
N=1 (A084057):  1, 6, 16,  56,  176,   576,   1856, ...
N=2 (A108306):  1, 6, 21,  81,  306,  1161,   4401, ...
N=3 (A164549):  1, 6, 26, 116,  516,  2296,  10216, ...
N=4 (A015449):  1, 6, 31, 161,  836,  4341,  22541, ...
N=5 (A000400):  1, 6, 36, 216, 1296,  7776,  46656, ...
N=6 (A049685):  1, 6, 41, 281, 1926, 13201,  90481, ...
N=7 (.......):  1, 6, 46, 356, 2756, 21336, 222712, ...
...
Sequences in the above set can be obtained by taking INVERT transforms of the following:
N=1 INVERT transform of (1, 5,  5,  5,   5,    5, ...
N=2 ..."......"......". (1, 5, 10, 20,  40,   80, ...
N=3 ..."......"......". (1, 5, 15, 45, 135,  405, ...
N=4 ..."......"......". (1, 5, 20, 80, 320, 1280, ...
...
with the pattern (1, 5, N*5, (N^2)*5, (N^3)*5, ...
It appears that the sequence generated from powers (n>0) of the matrix P = [(1,a); (1,b)], (a,b > 0), then extracting the upper left terms, is equal to the INVERT transform of the sequence starting: (1, a, b*a, (b^2)*a, (b^3)*a, ...). (End)

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A046717, A002533, A098158, A143969, A269992.
a(n) = A087131(n)/2.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A108306, A164549, A015449, A000400, A049685

I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 31 2016

f[n_] := Simplify[((1 + Sqrt[5])^n + (1 - Sqrt[5])^n)/2]; Array[f, 28, 0] (* Or *)
LinearRecurrence[{2, 4}, {1, 1}, 28] (* Robert G. Wilson v, Sep 18 2013 *)
RecurrenceTable[{a[1] == 1, a[2] == 1, a[n] == 2 a[n-1] + 4 a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Jul 31 2016 *)
Table[2^(n-1) LucasL[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 19 2016 *)

lucas(n)=fibonacci(n-1)+fibonacci(n+1)
a(n)=lucas(n)/2*2^n \\ Charles R Greathouse IV, Sep 18 2013

from sage.combinat.sloane_functions import recur_gen2b; it = recur_gen2b(1,1,2,4, lambda n: 0); [next(it) for i in range(1,26)] # Zerinvary Lajos, Jul 09 2008

[lucas_number2(n,2,-4)/2 for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

a(n) = ((1+sqrt(5))^n + (1-sqrt(5))^n)/2.
G.f.: (1-x) / (1-2*x-4*x^2).
E.g.f.: exp(x) * cosh(sqrt(5)*x).
a(2n+1) = 2*a(n)*a(n+1) - (-4)^n. - Mario Catalani (mario.catalani(AT)unito.it), Jun 13 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*5^k . - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} A098158(n,k)*5^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = 2^(n-1)*A000032(n). - Mark Dols, Jul 24 2009
If p(1)=1, and p(i)=5 for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for i<=j, A(i,j):=-1, (i=j+1), and A(i,j):=0 otherwise, then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(5*k-1)/(x*(5*k+4) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n) = A063727(n) - A063272(n-1). - R. J. Mathar, Jun 06 2019
a(n) = 1 + 5*A014335(n). - R. J. Mathar, Jun 06 2019
Sum_{n>=1} 1/a(n) = A269992. - Amiram Eldar, Feb 01 2021

A055830 Triangle T read by rows: diagonal differences of triangle A037027.

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 3, 3, 1, 0, 5, 7, 4, 1, 0, 8, 15, 12, 5, 1, 0, 13, 30, 31, 18, 6, 1, 0, 21, 58, 73, 54, 25, 7, 1, 0, 34, 109, 162, 145, 85, 33, 8, 1, 0, 55, 201, 344, 361, 255, 125, 42, 9, 1, 0, 89, 365, 707, 850, 701, 413, 175, 52, 10, 1, 0, 144, 655, 1416, 1918, 1806, 1239, 630, 236, 63, 11, 1, 0

Offset: 0

Clark Kimberling, May 28 2000

Or, coefficients of a generalized Lucas-Pell polynomial read by rows. - Philippe Deléham, Nov 05 2006

Equals A046854(shifted) * Pascal's triangle; where A046854 is shifted down one row and "1" inserted at (0,0). - Gary W. Adamson, Dec 24 2008

			Triangle begins:
   1
   1,   0
   2,   1,   0
   3,   3,   1,   0
   5,   7,   4,   1,   0
   8,  15,  12,   5,   1,   0
  13,  30,  31,  18,   6,   1,  0
  21,  58,  73,  54,  25,   7,  1, 0
  34, 109, 162, 145,  85,  33,  8, 1, 0
  55, 201, 344, 361, 255, 125, 42, 9, 1, 0
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002) 328-338, Example 3D.

Left-hand columns include A000045, A023610.
Right-hand columns include A055831, A055832, A055833, A055834, A055835, A055836, A055837, A055838, A055839, A055840.
Row sums: A001333 (numerators of continued fraction convergents to sqrt(2)).
Cf. A122075 (another version).
Cf. A046854. - Gary W. Adamson, Dec 24 2008

function T(n,k)
  if k lt 0 or k gt n then return 0;
  elif k eq 0 then return Fibonacci(n+1);
  elif n eq 1 and k eq 1 then return 0;
  else return T(n-1,k-1) + T(n-1,k) + T(n-2,k);
  end if; return T; end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 21 2020

with(combinat);
T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=0 then fibonacci(n+1)
    elif n=1 and k=1 then 0
    else T(n-1, k-1) + T(n-1, k) + T(n-2, k)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..12); # G. C. Greubel, Jan 21 2020

T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, Fibonacci[n+1], If[n==1 && k==1, 0, T[n-1, k-1] + T[n-1, k] + T[n-2, k]]]]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Dec 19 2017 *)

T(n,k) = if(k<0 || k>n, 0, if(k==0, fibonacci(n+1), if(n==1 && k==1, 0, T(n-1, k-1) + T(n-1, k) + T(n-2, k) )));
for(n=0,12, for(k=0, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jan 21 2020

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==0): return fibonacci(n+1)
    elif (n==1 and k==1): return 0
    else: return T(n-1, k-1) + T(n-1, k) + T(n-2, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 21 2020

G.f.: (1-y*z) / (1-y*(1+y+z)).
T(i, j) = R(i-j, j), where R(0, 0)=1, R(0, j)=0 for j >= 1, R(1, j)=1 for j >= 0, R(i, j) = Sum_{k=0..j} (R(i-2, k) + R(i-1, k)) for i >= 1, j >= 1.
Sum_{k=0..n} x^k*T(n,k) = A039834(n-2), A000012(n), A000045(n+1), A001333(n), A003688(n), A015448(n), A015449(n), A015451(n), A015453(n), A015454(n), A015455(n), A015456(n), A015457(n) for x= -2,-1,0,1,2,3,4,5,6,7,8,9,10. - Philippe Deléham, Oct 22 2006
Sum_{k=0..floor(n/2)} T(n-k,k) = A011782(n). - Philippe Deléham, Oct 22 2006
Triangle T(n,k), 0 <= k <= n, given by [1, 1, -1, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2006
T(n,0) = Fibonacci(n+1) = A000045(n+1). Sum_{k=0..n} T(n,k) = A001333(n). T(n,k)=0 if k > n or if k < 0, T(0,0)=1, T(1,1)=0, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k). - Philippe Deléham, Nov 05 2006

Edited by Ralf Stephan, Jan 12 2005

A052961 Expansion of (1 - 3x) / (1 - 5x + 3*x^2).

Original entry on oeis.org

1, 2, 7, 29, 124, 533, 2293, 9866, 42451, 182657, 785932, 3381689, 14550649, 62608178, 269388943, 1159120181, 4987434076, 21459809837, 92336746957, 397304305274, 1709511285499, 7355643511673, 31649683701868, 136181487974321, 585958388766001, 2521247479907042

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

a(n) is the number of tilings of a 2 X n rectangle using integer dimension tiles at least one of whose dimensions is 1, so allowable dimensions are 1 X 1, 1 X 2, 1 X 3, 1 X 4, ..., and 2 X 1. - David Callan, Aug 27 2014
a(n+1) counts closed walks on K_2 containing one loop on the index vertex and four loops on the other vertex. Equivalently the (1,1)entry of A^(n+1) where the adjacency matrix of digraph is A=(1,1;1,4). - _David Neil McGrath, Nov 05 2014
A production matrix for the sequence is M =
  1, 1, 0, 0, 0, 0, 0, ...
  1, 0, 4, 0, 0, 0, 0, ...
  1, 0, 0, 4, 0, 0, 0, ...
  1, 0, 0, 0, 4, 0, 0, ...
  1, 0, 0, 0, 0, 4, 0, ...
  1, 0, 0, 0, 0, 0, 4, ...
  ...
Take powers of M and extract the upper left term, getting the sequence starting (1, 1, 2, 7, 29, 124, ...). - Gary W. Adamson, Jul 22 2016
From Gary W. Adamson, Jul 29 2016: (Start)
The sequence is N=1 in an infinite set obtained from matrix powers of [(1,N); (1,4)], extracting the upper left terms.
The infinite set begins:
N=1  (A052961):  1,  2,  7,  29   124,  533,   2293, ...
N=2  (A052984):  1,  3, 13,  59,  269, 1227,   5597, ...
N=3  (A004253):  1,  4, 19,  91,  436, 2089,  10009, ...
N=4  (A000351):  1,  5, 25, 125,  625, 3125,  15625, ...
N=5  (A015449):  1,  6, 31, 161,  836, 4341,  22541, ...
N=6  (A124610):  1,  7, 37, 199, 1069, 5743,  30853, ...
N=7  (A111363):  1,  8, 43, 239, 1324, 7337,  40653, ...
N=8  (A123270):  1,  9, 49, 281, 1601, 9129,  52049, ...
N=9  (A188168):  1, 10, 55, 325, 1900, 11125, 65125, ...
N=10 (A092164):  1, 11, 61, 371, 2221, 13331, 79981, ...
... (End)

Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1032
Kai Liang, Solving tiling enumeration problems by tensor network contractions, arXiv:2503.17698, March 2025.
Index entries for linear recurrences with constant coefficients, signature (5,-3).

Column k=2 of A254414.

Cf. A000351, A004253, A015449, A052984, A092164, A111363, A123270, A124610, A188168.

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

I:=[1,2]; [n le 2 select I[n] else 5*Self(n-1)-3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 17 2014

spec:= [S,{S = Sequence(Union(Prod(Sequence(Union(Z,Z,Z)),Z),Z))}, unlabeled ]: seq(combstruct[count ](spec,size = n), n = 0..20);
seq(coeff(series((1-3*x)/(1-5*x+3*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 23 2019

CoefficientList[Series[(1-3x)/(1-5x+3x^2),{x,0,30}],x] (* or *) LinearRecurrence[{5,-3},{1,2},30] (* Harvey P. Dale, Nov 23 2013 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-5*x+3*x^2)) \\ G. C. Greubel, Oct 23 2019

def A052961_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-3*x)/(1-5*x+3*x^2)).list()
A052961_list(30) # G. C. Greubel, Oct 23 2019

G.f.: (1-3*x)/(1-5*x+3*x^2).
a(n) = 5*a(n-1) - 3*a(n-2), with a(0) = 1, a(1) = 2.
a(n) = Sum_{alpha=RootOf(1-5*z+3*z^2)} (-1 + 9*alpha)*alpha^(-1-n)/13.
E.g.f.: (1 + sqrt(13) + (sqrt(13)-1) * exp(sqrt(13)*x)) / (2*sqrt(13) * exp(((sqrt(13)-5)*x)/2)). - Vaclav Kotesovec, Feb 16 2015
a(n) = A116415(n) - 3*A116415(n-1). - R. J. Mathar, Feb 27 2019

A135597 Square array read by antidiagonals: row m (m >= 1) satisfies b(0) = b(1) = 1; b(n) = m*b(n-1) + b(n-2).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 5, 1, 1, 5, 13, 17, 8, 1, 1, 6, 21, 43, 41, 13, 1, 1, 7, 31, 89, 142, 99, 21, 1, 1, 8, 43, 161, 377, 469, 239, 34, 1, 1, 9, 57, 265, 836, 1597, 1549, 577, 55, 1, 1, 10, 73, 407, 1633, 4341, 6765, 5116, 1393, 89, 1, 1, 11, 91, 593, 2906

Offset: 1

N. J. A. Sloane, Mar 02 2008

For n > 1, the number of independent vertex sets in the graph K_m X P_{n-1}. For example, in K_3 X P_1 there are 4 independent vertex sets. - Andrew Howroyd, May 23 2017

			Array begins:
========================================================
m\n| 0 1 2  3   4    5     6      7       8        9
---|----------------------------------------------------
1  | 1 1 2  3   5    8    13     21      34       55 ...
2  | 1 1 3  7  17   41    99    239     577     1393 ...
3  | 1 1 4 13  43  142   469   1549    5116    16897 ...
4  | 1 1 5 21  89  377  1597   6765   28657   121393 ...
5  | 1 1 6 31 161  836  4341  22541  117046   607771 ...
6  | 1 1 7 43 265 1633 10063  62011  382129  2354785 ...
7  | 1 1 8 57 407 2906 20749 148149 1057792  7552693 ...
8  | 1 1 9 73 593 4817 39129 317849 2581921 20973217 ...
...

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Cf. A121875, A000045, A287376.

Rows 2-11 are A001333, A003688, A015448, A015449, A015451, A015453-A015457.

A135597 := proc(m,c) coeftayl( (m*x-x-1)/(x^2+m*x-1),x=0,c) ; end: for d from 1 to 15 do for c from 0 to d-1 do printf("%d,",A135597(d-c,c)) ; od: od: # R. J. Mathar, Apr 21 2008

a[, 0] = a[, 1] = 1; a[m_, n_] := m*a[m, n-1] + a[m, n-2]; Table[a[m-n+1, n], {m, 0, 11}, {n, 0, m}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)

O.g.f. row m: (mx-x-1)/(x^2+mx-1). - R. J. Mathar, Apr 21 2008

More terms from R. J. Mathar, Apr 21 2008

A152187 a(n) = 3a(n-1) + 5a(n-2), with 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, 56920427728295

Offset: 0

Philippe Deléham, Nov 28 2008

Unsigned version of A152185.
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 and 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.
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 central square to A179606.
This sequence belongs to a family of sequences with g.f. (1+2*x)/(1 - 3*x - k*x^2). Red king sequences that are members of this family are A007483 (k=2), A108981 (k=4), A152187 (k=5; this sequence), A154964 (k=6), A179602 (k=7) and A179598 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A036563 (k=-2), A054486 (k=-1), A084244 (k=0), A108300 (k=1) and A000351 (k=10).
Inverse binomial transform of A015449 (without the first leading 1).
(End)

Index entries for linear recurrences with constant coefficients, signature (3, 5).

Cf. A015523, A072263, A072264, A179606, A197189.

LinearRecurrence[{3,5},{1,5},40] (* Harvey P. Dale, May 03 2013 *)

G.f.: (1+2*x)/(1 - 3*x - 5*x^2).
Lim_{k->infinity} a(n+k)/a(k) = (A072263(n) + A015523(n)*sqrt(29))/2. - Johannes W. Meijer, Aug 01 2010
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

A180028 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + 3x)/(1 - 6x - 3*x^2).

Original entry on oeis.org

1, 9, 57, 369, 2385, 15417, 99657, 644193, 4164129, 26917353, 173996505, 1124731089, 7270376049, 46996449561, 303789825513, 1963728301761, 12693739287105, 82053620627913, 530402941628793, 3428578511656497

Offset: 0

Johannes W. Meijer, Aug 09 2010; edited Jun 21 2013

The a(n) represent the number of n-move routes of a fairy chess piece starting in the center square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen.
On a 3 X 3 chessboard there are 2^9 = 512 ways to explode with fury on the center square (off the center square the piece behaves like a normal queen). The red queen is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program and A180140. For the center square the 512 red queens lead to 17 red queen sequences, see the overview of red queen sequences and the crossreferences.
The sequence above corresponds to just one red queen vector, i.e., A[5] = [111 111 111] vector. The other squares lead for this vector to A090018.
This sequence belongs to a family of sequences with g.f. (1+k*x)/(1 - 6*x - k*x^2). The members of this family that are red queen sequences are A180028 (k=3; this sequence), A180029 (k=2), A015451 (k=1), A000400 (k=0), A001653 (k=-1), A180034 (k=-2), A084120 (k=-3), A154626 (k=-4) and A000012 (k=-5). Other members of this family are A123362 (k=5), 6*A030192(k=-6).
Inverse binomial transform of A107903.

Gary Chartrand, Introductory Graph Theory, pp. 217-221, 1984.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Johannes W. Meijer, The red queen sequences.
Wikipedia, Alice in Wonderland (2010 film).
Index entries for linear recurrences with constant coefficients, signature (6, 3).

Cf. A180140 (berserker sequences)
Cf. A180032 (Corner and side squares).
Cf. Red queen sequences center square [decimal value A[5]]: A180028 [511], A180029 [255], A180031 [495], A015451 [127], A152240 [239], A000400 [63], A057088 [47], A001653 [31], A122690 [15], A180034 [23], A180036 [7], A084120 [19], A180038 [3], A154626 [17], A015449 [1], A000012 [16], A000007 [0].

I:=[1,9]; [n le 2 select I[n] else 6*Self(n-1)+3*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 15 2011

nmax:=19; m:=5; A[1]:=[0,1,1,1,1,0,1,0,1]: A[2]:=[1,0,1,1,1,1,0,1,0]: A[3]:=[1,1,0,0,1,1,1,0,1]: A[4]:=[1,1,0,0,1,1,1,1,0]: A[5]:=[1,1,1,1,1,1,1,1,1]: A[6]:=[0,1,1,1,1,0,0,1,1]: A[7]:=[1,0,1,1,1,0,0,1,1]: A[8]:=[0,1,0,1,1,1,1,0,1]: A[9]:=[1,0,1,0,1,1,1,1,0]: 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[{6,3},{1,9},50] (* Vincenzo Librandi, Nov 15 2011 *)

G.f.: (1+3*x)/(1 - 6*x - 3*x^2).
a(n) = 6*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 9.
a(n) = ((1-A)*A^(-n-1) + (1-B)*B^(-n-1))/4 with A=(-1+2*sqrt(3)/3) and B=(-1-2*sqrt(3)/3).
Lim_{k->infinity} a(n+k)/a(k) = (-1)^(n-1)*A108411(n+1)/(A041017(n-1)*sqrt(12) - A041016(n-1)) for n >= 1.

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).

Original entry on oeis.org

1, 6, 37, 227, 1394, 8559, 52553, 322678, 1981261, 12165051, 74694082, 458625767, 2815987409, 17290317414, 106163498933, 651849716563, 4002393075346, 24574913392671, 150891318490777, 926480986202582, 5688644160448349

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move routes of a fairy chess piece starting in a given corner or side square (m = 1, 3, 7, 9; 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a white chess queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen.
On a 3 X 3 chessboard there are 2^9 = 512 ways to explode with fury on the central square (we assume here that a red queen might behave like a white queen). The red queen is represented by the A[5] vector in the fifth row of the adjacency matrix A, see the Maple program. For the corner and side squares the 512 red queens lead to 17 red queen sequences, see the cross-references for the complete set.
The sequence above corresponds to 8 red queen vectors, i.e., A[5] vectors, with decimal values 239, 367, 431, 463, 487, 491, 493 and 494. The central square leads for these vectors to A152240.
This sequence belongs to a family of sequences with g.f. (1+x)/(1 - 5*x - k*x^2). The members of this family that are red queen sequences are A180030 (k=8), A180032 (k=7; this sequence), A000400 (k=6), A180033 (k=5), A126501 (k=4), A180035 (k=3), A180037 (k=2) A015449 (k=1) and A003948 (k=0). Other members of this family are A030221 (k=-1), A109114 (k=-3), A020989 (k=-4), A166060 (k=-6).
Inverse binomial transform of A054413.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Wikipedia, Alice in Wonderland (2010 film)
Index entries for linear recurrences with constant coefficients, signature (5, 7).

Cf. A180028 (Central square).

Cf. Red queen sequences corner and side squares [decimal value A[5]]: A090018 [511], A135030 [255], A180030 [495], A005668 [127], A180032 [239], A000400 [63], A180033 [47], A001109 [31], A126501 [15], A154244 [23], A180035 [7], A138395 [19], A180037 [3], A084326 [17], A015449 [1], A003463 [16], A003948 [0].

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=20; m:=1; A[5]:= [1,1,1,1,0,1,1,1,0]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): 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[{5,7},{1,6},40] (* Vincenzo Librandi, Nov 15 2011 *)
CoefficientList[Series[(1+x)/(1-5x-7x^2),{x,0,30}],x] (* Harvey P. Dale, Apr 04 2024 *)

G.f.: (1+x)/(1 - 5*x - 7*x^2).
a(n) = 5*a(n-1) + 7*a(n-2) with a(0) = 1 and a(1) = 6.
a(n) = ((7+9*A)*A^(-n-1) + (7+9*B)*B^(-n-1))/53 with A = (-5+sqrt(53))/14 and B = (-5-sqrt(53))/14.

A013946 Least d for which the number with continued fraction [n,n,n,n...] is in Q(sqrt(d)).

Original entry on oeis.org

5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 5, 37, 173, 2, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 5, 226, 965, 257, 1093, 290, 1229, 13, 1373, 362, 61, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677, 2813, 730, 3029, 785, 3253

Offset: 1

Clark Kimberling

nonn

Square roots of a(n) are found in the limiting ratios of A000045, A001333, A003688, A015448, A015449, A015451 and so on. I.e., the limiting ratios are the golden ratio, silver mean, bronze ratio and so on. - Mats Granvik, Oct 20 2010

Clark Kimberling, Table of n, a(n) for n = 1..1000
Robin James Spivey, Close encounters of the golden and silver ratios, Notes on Number Theory and Discrete Mathematics (2019) Vol. 25, No. 3, 170-184.
Ofer Yifrach-Stav, Fast and Private Pool Testing and Contributions to Experimental Mathematics, Doctoral thesis, École normale supérieure (Paris, France), HAL Science [math.cs] 2024, Art. No. tel-04513104. See p. 104.

a(n) = 2 is equivalent to "n is in the sequence A077444", a(n) = 5 is equivalent to "n is in the sequence A002878".

z = 5000; u = Table[{p, e} = Transpose[FactorInteger[n]];
Times @@ (p^Mod[e, 2]), {n, z}]; Table[u[[n^2 + 4]], {n, 1, Sqrt[z - 4]}]  (* Clark Kimberling, Jul 20 2015, based on T. D. Noe's program at A007913 *)

A013946(n)=core(n^2+4)  \\ M. F. Hasler, Dec 08 2010

a(n) = A007913(n^2+4). - David W. Wilson, Dec 08 2010

More terms from David W. Wilson

A164581 a(n) = 5*a(n - 1) + a(n - 2), with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 57, 296, 1537, 7981, 41442, 215191, 1117397, 5802176, 30128277, 156443561, 812346082, 4218173971, 21903215937, 113734253656, 590574484217, 3066606674741, 15923607857922, 82684645964351, 429346837679677, 2229418834362736, 11576441009493357

Offset: 0

Vincenzo Librandi, Aug 17 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,1).

Cf. A000032, A001077, A052924, A078343, A179237, A180148, A329723.

[ n le 2 select (n) else 5*Self(n-1)+Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Sep 12 2013

LinearRecurrence[{5, 1}, {1, 2}, 40] (* or *) Rest[CoefficientList[Series [x (1 - 3 x) / (1 - 5 x - x^2), {x, 0, 40}], x]] (* Harvey P. Dale, May 02 2011 *)

Vec((1-3*x)/(1-5*x-x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

a(n) = 5*a(n-1)+a(n-2) = A052918(n)-3*A052918(n-1).
G.f.: (1-3*x)/(1-5*x-x^2).
a(n) = A052918(n) + A015449(n). - R. J. Mathar, Jul 06 2012
a(n) = (2^(-1-n)*((5-sqrt(29))^n*(1+sqrt(29))+(-1+sqrt(29))*(5+sqrt(29))^n))/sqrt(29). - Colin Barker, Oct 13 2015
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*5^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*5^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = A052918(n) -3*A052918(n-1). - R. J. Mathar, Feb 14 2024
From Peter Bala, Jul 08 2025: (Start)
The following series telescope:
Sum_{n >= 1} 1/(a(n) - 7*(-1)^n/a(n)) = 3/10, since 1/(a(n) - 7*(-1)^n/a(n)) = b(n) - b(n+1), where b(n) = (1/5) * (a(n) + a(n-1)) / (a(n)*a(n-1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 7*(-1)^n/a(n)) = 1/10, since 1/(a(n) - 7*(-1)^n/a(n)) = c(n) + c(n+1), where c(n) = (1/5) * (a(n) - a(n-1)) / (a(n)*a(n-1)). (End)

A180031 Number of n-move paths on a 3 X 3 chessboard of a queen starting or ending in the central square.

Original entry on oeis.org

1, 8, 48, 304, 1904, 11952, 74992, 470576, 2952816, 18528688, 116265968, 729559344, 4577924464, 28726097072, 180253881072, 1131078181936, 7097421958256, 44535735246768, 279458051899888, 1753576141473584

Offset: 0

Johannes W. Meijer, Aug 09 2010

The a(n) represent the number of n-move paths of a chess queen starting or ending in the central square (m = 5) on a 3 X 3 chessboard. The other squares lead to A180030.
To determine the a(n) we can either sum the components of the column vector A^n[k,m], with A the adjacency matrix of the queen's graph, or we can sum the components of the row vector A^n[m,k], see the Maple program.
Closely related with this sequence are the red queen sequences, see A180028 and A180032.
This sequence belongs to a family of sequences with g.f. (1+k*x)/(1 - 5*x - (k+5)*x^2). The members of this family that are red queen sequences are A180031 (k=3; this sequence), A152240 (k=2), A000400 (k=1), A057088 (k=0), A122690 (k=-1), A180036 (k=-2), A180038 (k=-3), A015449 (k=-4) and A000007 (k=-5). Other members of this family are A030221 (k= -6), 3*A109114 (k=-8), 4*A020989 (k=-9), 6*A166060 (k=-11).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (5, 8).

I:=[1,8]; [n le 2 select I[n] else 5*Self(n-1)+8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

with(LinearAlgebra): nmax:=19; m:=5; A[5]:= [1,1,1,1,0,1,1,1,1]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): 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[{5,8},{1,8},50] (* Vincenzo Librandi, Nov 15 2011 *)

G.f.: (1+3*x)/(1 - 5*x - 8*x^2).
a(n) = 5*a(n-1) + 8*a(n-2) with a(0) = 1 and a(1) = 8.
a(n) = ((A+11)*A^(-n-1) + (B+11)*B^(-n-1))/57 with A = (-5+sqrt(57))/16 and B = (-5-sqrt(57))/16.

cp's OEIS Frontend

A084057 a(n) = 2*a(n-1) + 4*a(n-2), a(0)=1, a(1)=1.

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A055830 Triangle T read by rows: diagonal differences of triangle A037027.

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A052961 Expansion of (1 - 3*x) / (1 - 5*x + 3*x^2).

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A135597 Square array read by antidiagonals: row m (m >= 1) satisfies b(0) = b(1) = 1; b(n) = m*b(n-1) + b(n-2).

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A152187 a(n) = 3*a(n-1) + 5*a(n-2), with a(0)=1, a(1)=5.

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A180028 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + 3*x)/(1 - 6*x - 3*x^2).

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A084057 a(n) = 2a(n-1) + 4a(n-2), a(0)=1, a(1)=1.

A052961 Expansion of (1 - 3x) / (1 - 5x + 3*x^2).

A152187 a(n) = 3a(n-1) + 5a(n-2), with a(0)=1, a(1)=5.

A180028 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + 3x)/(1 - 6x - 3*x^2).

A180032 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1+x)/(1-5x-7x^2).