A046854 -id:A046854 - cp's OEIS frontend

A122765 Triangle read by rows: Let p(k, x) = x*p(k-1, x) - p(k-2, x). Then T(k,x) = dp(k,x)/dx.

1, -1, 2, -2, -2, 3, 2, -6, -3, 4, 3, 6, -12, -4, 5, -3, 12, 12, -20, -5, 6, -4, -12, 30, 20, -30, -6, 7, 4, -20, -30, 60, 30, -42, -7, 8, 5, 20, -60, -60, 105, 42, -56, -8, 9, -5, 30, 60, -140, -105, 168, 56, -72, -9, 10

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 22 2006

Based on the coefficients of derivatives of the polynomials in A130777.

			Triangle begins as:
   1;
  -1,   2;
  -2,  -2,   3;
   2,  -6,  -3,   4;
   3,   6, -12,  -4,   5;
  -3,  12,  12, -20,  -5,   6;
  -4, -12,  30,  20, -30,  -6,   7;
   4, -20, -30,  60,  30, -42,  -7,   8;
   5,  20, -60, -60, 105,  42, -56,  -8,  9;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.

Cf. A000217, A026741, A046854, A066170, A076118, A122766, A130777, A165202.

A122765:= func< n,k | k*(-1)^Binomial(n-k+1, 2)*Binomial(Floor((n+k)/2), k) >;
[A122765(n,k): k in [1..n], n in [1..14]]; // G. C. Greubel, Dec 30 2022

(* First program *)
p[0,x]=1; p[1,x]=x-1; p[k_,x_]:= p[k, x]= x*p[k-1,x] -p[k-2,x]; a = Table[Expand[p[n, x]], {n, 0, 10}]; Table[CoefficientList[D[a[[n]], x], x], {n, 2, 10}]//Flatten
(* Second program *)
T[n_, k_]:= k*(-1)^Binomial[n-k+1,2]*Binomial[Floor[(n+k)/2], k];
Table[T[n, k], {n,14}, {k,n}]//Flatten (* G. C. Greubel, Dec 30 2022 *)

tpol(n) = if (n<=0, 1, if (n==1, x-1, x*tpol(n-1) -tpol(n-2)));
lista(nn) = {for(n=0, nn, pol = deriv(tpol(n)); for (k=0, poldegree(pol), print1(polcoeff(pol, k), ", ");););} \\ Michel Marcus, Feb 07 2014

def A122765(n, k): return k*(-1)^binomial(n-k+1, 2)*binomial(((n+k)//2), k)
flatten( [[A122765(n,k) for k in range(1,n+1)] for n in range(1,15)] ) # G. C. Greubel, Dec 30 2022

From G. C. Greubel, Dec 30 2022: (Start)
T(n, k) = coefficient [x^k]( p(n, x) ), where p(n,x) = (2/(x^2-4))*((n+1)*chebyshev_T(n+1,x/2) -n*chebyshev_T(n,x/2) - (x/2)*(chebyshev_U(n,x/2) - chebyshev_U(n-1,x/2))).
T(n, k) = k*(-1)^binomial(n-k+1, 2)*binomial(floor((n+k)/2), k).
T(n, n) = n.
T(n, n-1) = -(n-1).
T(n, n-2) = -2*A000217(n-2).
T(n, n-3) = 2*A000217(n-3).
T(n, 1) = (-1)^binomial(n, 2)*floor((n+1)/2).
T(n, 2) = 2*(-1)^binomial(n-1, 2)*binomial(floor((n+2)/2), 2).
Sum_{k=1..n} T(n, k) = A076118(n).
Sum_{k=1..n} (-1)^k*T(n, k) = (-1)^(n-1)*A165202(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = [n=1] - [n=2].
Sum_{k=1..floor((n+1)/2)} (-1)^k*T(n-k+1, k) = (-1)^binomial(n+1, 2)*b(n), where b(n) = 4^floor(n/4)*A026741(n/2) if n is even and b(n) = 4^floor((n-1)/4)*A026741((n-1)/4) if n is odd. (End)

Name corrected and more terms from Michel Marcus, Feb 07 2014

A073044 Triangle read by rows: T(n,k) (n >= 1, n-1 >= k >= 0) = number of n-sequences of 0's and 1's with no pair of adjacent 0's and exactly k pairs of adjacent 1's.

Original entry on oeis.org

2, 2, 1, 2, 2, 1, 2, 3, 2, 1, 2, 4, 4, 2, 1, 2, 5, 6, 5, 2, 1, 2, 6, 9, 8, 6, 2, 1, 2, 7, 12, 14, 10, 7, 2, 1, 2, 8, 16, 20, 20, 12, 8, 2, 1, 2, 9, 20, 30, 30, 27, 14, 9, 2, 1, 2, 10, 25, 40, 50, 42, 35, 16, 10, 2, 1, 2, 11, 30, 55, 70, 77, 56, 44, 18, 11, 2, 1, 2, 12, 36, 70, 105, 112, 112, 72

Offset: 1

Roger Cuculière, Aug 24 2002

T(n,k) is the number of domino tilings of 2 X (n+1) rectangles that have n+2-k perimeter dominoes. - Bridget Tenner, Oct 14 2019

			T(5,2)=4 because the sequences of length 5 with 2 pairs 11 are 11101, 11011,10111, 01110. Also the 2 X (5+1) rectangle has 4 domino tilings with 5+2-2 perimeter dominoes. - _Bridget Tenner_, Oct 14 2019
Triangle starts:
  2;
  2, 1;
  2, 2, 1;
  2, 3, 2, 1;
  2, 4, 4, 2, 1;

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 67-68).
I. Goulden and D. Jackson, Combinatorial Enumeration, John Wiley and Sons, 1983, page 76.

B. E. Tenner, Tiling-based models of perimeter and area, arXiv:1811.00082 [math.CO], 2018.

Row sums are the Fibonacci numbers (A000045).
Cf. A046854.
Weighted row sums 2*T(n,n) + 3*T(n,n-1) + 4*T(n,n-2) + ... give A320947. - Bridget Tenner, Oct 14 2019

G:=z*(2+2*z-t*z)/(1-t*z-z^2):Gser:=simplify(series(G,z=0,17)):for n from 1 to 15 do P[n]:=sort(coeff(Gser,z^n)) od:for n from 1 to 13 do seq(coeff(t*P[n],t^k),k=1..n) od;# yields sequence in triangular form

nn = 15; f[list_] := Select[list, # > 0 &]; Map[f, Drop[CoefficientList[Series[(1 + x) (1 + x - y x)/(1 - y x - x^2), {x, 0, nn}], {x,y}], 1]] //Flatten (* Geoffrey Critzer, Mar 05 2012 *)

T(n,k) = binomial((n+k-1)\2,k) + binomial((n+k-2)\2,k) \\ Charles R Greathouse IV, Jun 07 2016

Recurrence: T(n, k) = T(n-1, k-1) + T(n-2, k).
G.f.: G(t, z) = z*(2+2*z-t*z)/(1-t*z-z^2). - Emeric Deutsch, Feb 01 2005
T(n,k) = binomial(floor((n+k-1)/2),k) + binomial(floor((n+k-2)/2),k). - Jeremy Dover, Jun 07 2016
T(n,k) = A046854(n-1,k) + A046854(n-2,k), where A046854 is extended so that A046854(-1,0) = 1. - Jeremy Dover, Jun 07 2016

More terms from Emeric Deutsch, Feb 01 2005

A131301 Regular triangle read by rows: T(n,k) = 3*binomial(floor((n+k)/2),k)-2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 7, 1, 1, 1, 7, 7, 10, 1, 1, 1, 7, 16, 10, 13, 1, 1, 1, 10, 16, 28, 13, 16, 1, 1, 1, 10, 28, 28, 43, 16, 19, 1, 1, 1, 13, 28, 58, 43, 61, 19, 22, 1, 1, 1, 13, 43, 58, 103, 61, 82, 22, 25, 1, 1, 1, 16, 43, 103, 103, 166, 82

Offset: 0

Gary W. Adamson, Jun 27 2007

Row sums = A131300: (1, 2, 3, 7, 14, 27, 49, 86, ...). Reversed triangle = A131299.

			First few rows of the triangle:
  1;
  1,  1;
  1,  1,  1;
  1,  4,  1,  1;
  1,  4,  7,  1,  1;
  1,  7,  7, 10,  1,  1;
  1,  7, 16, 10, 13,  1,  1;
  ...

Nathaniel Johnston, Rows 0..100, flattened

Cf. A046854, A000012, A131299, A131300.

for n from 0 to 6 do seq(3*binomial(floor((n+k)/2),k)-2,k=0..n); od; # Nathaniel Johnston, Jun 29 2011

t[n_, k_] := 3 Binomial[Floor[(n + k)/2], k] - 2; Table[t[n, k], {n, 11}, {k, 0, n}] // Flatten
(* to view triangle: Table[t[n, k], {n, 5}, {k, 0, n}] // TableForm *) (* Robert G. Wilson v, Feb 28 2015 *)

3*A046854 - 2*A000012 as infinite lower triangular matrices (former name).

T(n,k) = 3*binomial(floor((n+k)/2),k)-2. - Nathaniel Johnston, Jun 29 2011

A248749 Decimal expansion of limit of the real part of f(1+i,n), where f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1).

Original entry on oeis.org

1, 5, 2, 9, 0, 8, 5, 5, 1, 3, 6, 3, 5, 7, 4, 6, 1, 2, 5, 1, 6, 0, 9, 9, 0, 5, 2, 3, 7, 9, 0, 2, 2, 5, 2, 1, 0, 6, 1, 9, 3, 6, 5, 0, 4, 9, 8, 3, 8, 9, 0, 9, 7, 4, 3, 1, 4, 0, 7, 7, 1, 1, 7, 6, 3, 2, 0, 2, 3, 9, 8, 1, 1, 5, 7, 9, 1, 8, 9, 4, 6, 2, 7, 7, 1, 1

Offset: 1

Clark Kimberling, Oct 13 2014

See A046854 for a triangle of coefficients of the numerators and denominators of f(x,n). Note that the limit of f(1,n) is the golden ratio.

			limit = 1.52908551363574612516099052379022521061936504983890974314077117...
n   f(x,n)                                 Re(f(1+i,n))  Im(f(1+i,n))
0   1                                      1             0
1   1 + x                                  2             1
2   (1 + x + x^2)/(1 + x)                  7/5           4/5
3   (1 + 2*x + x^2 + x^3)/(1 + x + x^2)    20/13         9/13
Re(f(1+i,10)) = 815/533 = 1.529080...
Im(f(1+i,10)) = 396/533 = 0.742964...

Cf. A248750, A046854.

evalf((1+sqrt(2+sqrt(5)))/2, 120); # Vaclav Kotesovec, Oct 19 2014

$RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1; t = Table[Factor[f[x, n]], {n, 1, 12}]; u = t /. x -> I + 1; {Re[u], Im[u]}
{N[Re[u], 12], N[Im[u], 12]}
t = Table[Factor[f[x, n]], {n, 1, 300}]; u = t /. x -> I + 1;
r1 = N[Re[u][[300]], 130]
r2 = N[Im[u][[300]], 130]
d1 = RealDigits[r1]  (* A248749 *)
d2 = RealDigits[r2]  (* A248750 *)

Equals (1+sqrt(2+sqrt(5)))/2. - Vaclav Kotesovec, Oct 19 2014

A274228 Triangle read by rows: T(n,k) (n>=3, 0<=k<=n-3) = number of n-sequences of 0's and 1's with exactly one pair of adjacent 0's and exactly k pairs of adjacent 1's.

Original entry on oeis.org

2, 3, 2, 4, 4, 2, 5, 8, 5, 2, 6, 12, 12, 6, 2, 7, 18, 21, 16, 7, 2, 8, 24, 36, 32, 20, 8, 2, 9, 32, 54, 60, 45, 24, 9, 2, 10, 40, 80, 100, 90, 60, 28, 10, 2, 11, 50, 110, 160, 165, 126, 77, 32, 11, 2, 12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2, 13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2

Offset: 3

Jeremy Dover, Jun 14 2016

			n=3 => 100, 001 -> T(3,0) = 2.
n=4 => 0010, 0100, 1001 -> T(4,0) = 3; 0011, 1100 -> T(4,1) = 2.
Triangle starts:
2,
3, 2,
4, 4, 2,
5, 8, 5, 2,
6, 12, 12, 6, 2,
7, 18, 21, 16, 7, 2,
8, 24, 36, 32, 20, 8, 2,
9, 32, 54, 60, 45, 24, 9, 2,
10, 40, 80, 100, 90, 60, 28, 10, 2,
11, 50, 110, 160, 165, 126, 77, 32, 11, 2,
12, 60, 150, 240, 280, 252, 168, 96, 36, 12, 2,
13, 72, 195, 350, 455, 448, 364, 216, 117, 40, 13, 2,
...

Row sums give A001629.
Cf. A073044.
Columns of table:
  T(n,0)=A000027(n-1)
  T(n,1)=A007590(n-1)
  T(n,2)=A080838(n-1)
  T(n,3)=A032091(n)

Table[(k + 1) (Binomial[Floor[(n + k - 2)/2], k + 1] + Binomial[Floor[(n + k - 3)/2], k + 1]) + 2 Binomial[Floor[(n + k - 3)/2], k], {n, 3, 14}, {k, 0, n - 3}] // Flatten (* Michael De Vlieger, Jun 16 2016 *)

T(n,k) = (k+1)*(binomial((n+k-2)\2,k+1)+binomial((n+k-3)\2,k+1))+2*binomial((n+k-3)\2,k); \\ Michel Marcus, Jun 17 2016

T(n,k) = (k+1)*(binomial(floor((n+k-2)/2),k+1)+binomial(floor((n+k-3)/2),k+1))+2*binomial(floor((n+k-3)/2),k).
T(n,k) = (k+1)*A073044(n-2,k+1) + 2*A046854(n-3,k).
T(n,k) = A274742(n,k)+A274742(n-1,k)+A046854(n-3,k).

A122766 Triangle read by rows: let p(n, x) = x*p(n-1, x) - p(n-2, x), then T(n, x) = d^2/dx^2 (p(n, x)).

Original entry on oeis.org

2, -2, 6, -6, -6, 12, 6, -24, -12, 20, 12, 24, -60, -20, 30, -12, 60, 60, -120, -30, 42, -20, -60, 180, 120, -210, -42, 56, 20, -120, -180, 420, 210, -336, -56, 72, 30, 120, -420, -420, 840, 336, -504, -72, 90, -30, 210, 420, -1120, -840, 1512, 504, -720, -90, 110

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 22 2006

			Triangle begins as:
    2;
   -2,    6;
   -6,    6,   12;
    6,  -24,  -12,   20;
   12,   24,  -60,  -20,   30;
   12,   60,   60, -120,  -30,   42;
  -20,  -60,  180,  120, -210,  -42,  56;
   20, -120, -180,  420,  210, -336, -56,  72;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.

Cf. A000217, A046854, A066170, A122765, A130777.

A122766:= func< n,k | 2*(-1)^Binomial(n-k+1, 2)*Binomial(k+1,2)*Binomial(Floor((n+k+2)/2), k+1) >;
[A122766(n,k): k in [1..n], n in [1..14]]; // G. C. Greubel, Dec 31 2022

(* First program *)
p[0, x]=1; p[1, x]=x-1; p[k_, x_]:= p[k, x]= x*p[k-1, x] -p[k-2, x]; b = Table[Expand[p[n,x]], {n,0,15}]; Table[CoefficientList[D[b[[n]], {x,2}], x], {n,2,14}]//Flatten
(* Second program *)
T[n_, k_]:= 2*(-1)^Binomial[n-k+1,2]*Binomial[k+1,2]*Binomial[Floor[(n +k+2)/2], k+1]; Table[T[n,k], {n,14}, {k,n}]//Flatten (* G. C. Greubel, Dec 31 2022 *)

tpol(n) = if (n <= 0, 1, if (n == 1, x -1, x*tpol(n-1) - tpol(n-2)));
lista(nn) = {for(n=0, nn, pol = deriv(deriv(tpol(n))); for (k=0, poldegree(pol), print1(polcoeff(pol, k), ", ");););} \\ Michel Marcus, Feb 07 2014

def A122766(n, k): return 2*(-1)^binomial(n-k+1,2)*binomial(k+1,2)*binomial(((n+k+2)//2), k+1)
flatten([[A122766(n, k) for k in range(1, n+1)] for n in range(1, 15)]) # G. C. Greubel, Dec 31 2022

From G. C. Greubel, Dec 31 2022: (Start)
T(n, k) = 2*(-1)^binomial(n-k+1, 2)*binomial(k+1,2)*binomial(floor((n+k +2)/2), k+1).
T(n, 1) = 2*(-1)^binomial(n,2)*binomial(floor((n+3)/2), 2)
T(n, n) = 2*A000217(n).
Sum_{k=1..n} T(n, k) = 2*A104555(n).
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = 2*([n=1] - [n=2]). (End)

Edited by N. J. A. Sloane, Oct 01 2006

Name corrected and more terms from Michel Marcus, Feb 07 2014

A131238 Triangle read by rows: T(n,k) = 2*binomial(n,k) - binomial(floor((n+k)/2), k) (0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 4, 5, 1, 1, 6, 9, 7, 1, 1, 7, 17, 16, 9, 1, 1, 9, 24, 36, 25, 11, 1, 1, 10, 36, 60, 65, 36, 13, 1, 1, 12, 46, 102, 125, 106, 49, 15, 1, 1, 13, 62, 148, 237, 231, 161, 64, 17, 1, 1, 15, 75, 220, 385, 483, 392, 232, 81, 19, 1, 1, 16, 95, 295, 625, 868, 896, 624, 321, 100, 21, 1

Offset: 0

Gary W. Adamson, Jun 21 2007

Row sums = A027934: (1, 2, 5, 11, 24, 51, 107, ...).

A131239 = 3*A007318 - 2*A046854.

			First few rows of the triangle:
  1;
  1,  1;
  1,  3,  1;
  1,  4,  5,  1;
  1,  6,  9,  7,  1;
  1,  7, 17, 16,  9,  1;
  1,  9, 24, 36, 25, 11,  1;
  1, 10, 36, 60, 65, 36, 13, 1;
  ...

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A027934, A131239, A007318, A046854.

B:=Binomial;; Flat(List([0..12], n-> List([0..n], k-> 2*B(n,k) - B(Int((n+k)/2), k) ))); # G. C. Greubel, Jul 12 2019

B:=Binomial; [2*B(n,k) - B(Floor((n+k)/2), k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 12 2019

T := proc (n, k) options operator, arrow; 2*binomial(n, k)-binomial(floor((1/2)*n+(1/2)*k), k) end proc: for n from 0 to 9 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form - Emeric Deutsch, Jul 09 2007

With[{B = Binomial}, Table[2*B[n, k] - B[Floor[(n+k)/2], k], {n,0,12}, {k,0,n}]]//Flatten (* G. C. Greubel, Jul 12 2019 *)

b=binomial; T(n,k) = 2*b(n,k) - b((n+k)\2, k);
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 12 2019

b=binomial; [[2*b(n,k) - b(floor((n+k)/2), k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jul 12 2019

T(n,k) = 2*A007318(n,k) - A046854(n,k) as infinite lower triangular matrices, where A007318 = Pascal's triangle and A046854 = Pascal's triangle with repeats, by columns.

More terms added by G. C. Greubel, Jul 12 2019

A134513 Triangle read by rows: T(n, k) = binomial(ceiling((n+k)/2), floor((n-k)/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 3, 3, 4, 4, 1, 1, 1, 6, 6, 5, 5, 1, 1, 4, 4, 10, 10, 6, 6, 1, 1, 1, 10, 10, 15, 15, 7, 7, 1, 1, 5, 5, 20, 20, 21, 21, 8, 8, 1, 1, 1, 15, 15, 35, 35, 28, 28, 9, 9, 1, 1, 6, 6, 35, 35, 56, 56, 36, 36, 10, 10, 1, 1

Offset: 0

Gary W. Adamson, Oct 28 2007

Old name: abs(A049310 * A097806).
Equivalently, T(n,k) = A168561(n,k) + A168561(n,k+1).
Row sums = A062114: (1, 2, 3, 6, 9, 16, 25, 42, 67, ...).
Triangle A046854 = abs(A097806 * A049310).

			First few rows of the triangle:
  1;
  1,  1;
  1,  1,  1;
  2,  2,  1,  1;
  1,  3,  3,  1,  1;
  3,  3,  4,  4,  1,  1;
  1,  6,  6,  5,  5,  1,  1;
  4,  4, 10, 10,  6,  6,  1,  1;
  1, 10, 10, 15, 15,  7,  7,  1,  1;
  ...

Cf. A046854, A049310, A062114, A097806.

abs(A049310 * A097806) as infinite lower triangular matrices.

Better definition, offset changed to 0, and more terms from Jinyuan Wang, Jan 25 2025

A122173 Expansion of -x * (x^5+x^4-15x^3+19x^2-8x+1) / (x^6-12x^5+34x^4-30x^3+6x^2+3x-1).

Original entry on oeis.org

1, -5, 10, -45, 110, -421, 1148, -4037, 11697, -39250, 117736, -384657, 1177235, -3787218, 11727187, -37389217, 116571621, -369712938, 1157315631, -3659226205, 11481436216, -36237006073, 113856243558, -358967583724, 1128781753801, -3556642214960, 11189229179710

Offset: 1

Gary W. Adamson and Roger L. Bagula, Oct 17 2006

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Peter Steinbach, Golden fields: a case for the heptagon, Math. Mag. Vol. 70, No. 1, Feb. 1997, 22-31.
Index entries for linear recurrences with constant coefficients, signature (3,6,-30,34,-12,1).

Cf. A046854. Cf. A046854. Cf. A007700, A059455. Cf. A065941.

M = {{0, -1, -1, -1, -1, -1}, {-1, 0, -1, -1, -1, 0}, {-1, -1, 0, -1, 0, 0}, {-1, -1, -1, 1, 0, 0}, {-1, -1, 0, 0, 1, 0}, {-1, 0, 0, 0, 0, 1}}; v[1] = {1, 1, 1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1]; a = Table[Floor[v[n][[1]]], {n, 1, 50}]
LinearRecurrence[{3,6,-30,34,-12,1},{1,-5,10,-45,110,-421},30] (* Harvey P. Dale, Mar 16 2025 *)

G.f.: -x*(x^5+x^4-15*x^3+19*x^2-8*x+1)/(x^6-12*x^5+34*x^4-30*x^3+6*x^2+3*x-1). [Colin Barker, Oct 19 2012]

Sequence edited by Joerg Arndt, Colin Barker, Bruno Berselli, Oct 19 2012

A122174 First row sum of the matrix M^n, where M is the 5 X 5 matrix {{0,-1,-1,-1,-1}, {-1,0,-1,-1,0}, {-1,-1,0,0,0}, {-1,-1,0,1,0}, {-1,0,0,0,1}}.

Original entry on oeis.org

1, -4, 6, -24, 41, -145, 273, -886, 1789, -5457, 11605, -33807, 74761, -210366, 479256, -1313465, 3061242, -8222492, 19501429, -51579259, 123983182, -324067194, 787044384, -2038584810, 4990387355, -12836179872, 31614557443, -80883958143, 200146505560, -509959672813

Offset: 0

Gary W. Adamson and Roger L. Bagula, Oct 17 2006

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Peter Steinbach, Golden fields: a case for the heptagon, Math. Mag. Vol. 70, No. 1, Feb. 1997, 22-31.
Index entries for linear recurrences with constant coefficients, signature (2,5,-13,7,-1).

Cf. A046854, A007700, A059455, A065941.

with(linalg): M[1]:=matrix(5,5,[0,-1,-1,-1,-1,-1,0,-1,-1,0,-1,-1,0,0,0,-1,-1,0,1, 0,-1,0,0,0,1]): for n from 2 to 30 do M[n]:=multiply(M[n-1],M[1]) od: 1,seq(M[n][1,1]+M[n][1,2]+M[n][1,3]+M[n][1,4]+M[n][1,5],n=1..30);

M = {{0, -1, -1, -1, -1}, {-1, 0, -1, -1, 0}, {-1, -1, 0, 0, 0}, {-1, -1, 0, 1, 0}, {-1, 0, 0, 0, 1}}; v[1] = {1, 1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1]; a1 = Table[v[n][[1]], {n, 1, 25}]

a(n) = my(m=[0,-1,-1,-1,-1; -1,0,-1,-1,0; -1,-1,0,0,0; -1,-1,0,1,0; -1,0,0,0,1]); vecsum((m^n)[1,]); \\ Michel Marcus, Jun 21 2017

a(n) = 2*a(n-1)+5*a(n-2)-13*a(n-3)+7*a(n-4)-a(n-5); a(0)=1, a(1)=-4, a(2)=6, a(3)=-24, a(4)=41 (follows from the minimal polynomial x^5-2*x^4-5*x^3+13*x^2-7*x+1 of the matrix M).

G.f.: (1-3*x^3+9*x^2-6*x)/(1+x^5-7*x^4+13*x^3-5*x^2-2*x). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009

Edited by N. J. A. Sloane, Oct 29 2006

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A248749 Decimal expansion of limit of the real part of f(1+i,n), where f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1).

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