A046854 -id:A046854 - cp's OEIS frontend

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

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

A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 7, 43, 265, 1633, 10063, 62011, 382129, 2354785, 14510839, 89419819, 551029753, 3395598337, 20924619775, 128943316987, 794584521697, 4896450447169, 30173287204711, 185936173675435, 1145790329257321

Offset: 0

Olivier Gérard

Row m=6 of A135597.
a(n) = term (1,1) in the 2 X 2 matrix [1,2; 3,5]^n. - Gary W. Adamson, May 30 2008
a(n)/a(n-1) tends to sqrt(10) + 3 = 6.16227766... - Gary W. Adamson, May 30 2008
For n >= 1, row sums of triangle for numbers 6^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
Z[sqrt(10)] is not a unique factorization domain, since, for example, 6 = 2 * 3 = (-1)(2 - sqrt(10))(2 + sqrt(10)) = (4 - sqrt(10))(4 + sqrt(10)). However, the latter two factorizations are not distinct, because 3 + sqrt(10) is a unit in Z[sqrt(10)], and (2 - sqrt(10))(-3 - sqrt(10)) = 4 + sqrt(10). In fact, (2 - sqrt(10))(-3 - sqrt(10))^n gives an algebraic integer b + a(n) * sqrt(10) which, when multiplied by its associate (and by -1 when n is even) is equal to 6. - Alonso del Arte, Mar 15 2014
For n >= 1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,5,6} containing no subwords 00, 11, 22, 33, 44, 55. - Milan Janjic, Jan 31 2015
a(n+1) equals the number of sequences over the alphabet {0,1,2,3,4,5,6} of length n such that no two consecutive terms differ by 4. - David Nacin, May 31 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. 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 (6,1).

Cf. A049310, A055830.

[n le 2 select 1 else 6*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=6*a[n-1]+a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-5*x)/(1-6*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec((1-5*x)/(1-6*x-x^2)) \\ G. C. Greubel, Dec 19 2017

a(n) = Sum_{k=0..n} 5^k * A055830(n, k). - Philippe Deléham, Oct 18 2006
G.f.: (1-5*x)/(1-6*x-x^2). - Philippe Deléham, Nov 20 2008
For n >= 2, a(n) = F_n(6) + F_(n+1)(6), 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)*6^k. - R. J. Mathar, Feb 14 2024

A055244 Number of certain stackings of n+1 squares on a double staircase.

Original entry on oeis.org

1, 1, 3, 6, 12, 23, 43, 79, 143, 256, 454, 799, 1397, 2429, 4203, 7242, 12432, 21271, 36287, 61739, 104791, 177476, 299978, 506111, 852457, 1433593, 2407443, 4037454, 6762708, 11314391, 18909139, 31569799, 52657247, 87751624

Offset: 0

Wolfdieter Lang, May 10 2000

a(n)= G_{n+1} of Turban reference eq.(3.9).
Equals A046854 * [1,2,3,...]. - Gary W. Adamson, Dec 23 2008
(1 + x + 3x^2 + 6x^3 + ...) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + ...) * (1 + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + ...). -Gary W. Adamson, Jul 27 2010
Column 1 of A194540. - R. H. Hardin, Aug 28 2011

L. Turban, Lattice animals on a staircase and Fibonacci numbers, J.Phys. A 33 (2000) 2587-2595.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Arnold Adelberg and Tamás Lengyel, New results on the 2-adic valuation of the central Stirling numbers S(2k, k), Harvey Mudd College (2024).
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A001629, A046854, A055245.

a:= n-> (Matrix([[1,-1,2,-4]]). Matrix(4, (i,j)-> if (i=j-1) then 1 elif j=1 then [2,1,-2,-1][i] else 0 fi)^(n))[1,1] ; seq (a(n), n=0..33); # Alois P. Heinz, Aug 05 2008

a[0] = a[1] = 1; a[n_] := a[n] = (((n-4)*n-6)*a[n-2] + ((n-5)*n-11)*a[n-1]) / ((n-6)*n-1); Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 11 2014 *)
CoefficientList[Series[(1 - x + x^3)/(1 - x - x^2)^2, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 13 2014 *)
LinearRecurrence[{2,1,-2,-1},{1,1,3,6},60] (* Harvey P. Dale, Jul 13 2022 *)

G.f.: (1-x+x^3)/(1-x-x^2)^2. (from Turban reference eq.(3.3) with t=1).
a(n) = ((n+5)*F(n+1)+(2*n-3)*F(n))/5 with F(n)=A000045(n) (Fibonacci numbers) (from Turban reference eq.(3.9)).
a(n) = A001629(n+1) + F(n-1). - Gary W. Adamson, Jul 27 2007
a(n) = (((n-4)*n-6)*a(n-2) + ((n-5)*n-11)*a(n-1)) / ((n-6)*n-1). - Jean-François Alcover, Mar 11 2014

A248750 Decimal expansion of limit of the imaginary 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

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

Offset: 0

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.

			0.742934135878322839091431937947266281096242992001186505475869206219...
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...

Index entries for algebraic numbers, degree 4.

Cf. A248749, A046854, A248751, A248752, A001622. A300072.

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

$RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
z = 20; (* For more accuracy, increase z *)
f[x_, n_] := x + 1/f[x, n - 1];
f[x_, 1] = 1; t = Table[Factor[f[x, n]], {n, 1, z}];
u = t /. x -> I + 1; t = Table[Factor[f[x, n]], {n, 1, z}]; u = t /. x -> I + 1;
r1 = N[Re[u][[z]], 130]
r2 = N[Im[u][[z]], 130]
d1 = RealDigits[r1]  (*A248749*)
d2 = RealDigits[r2]  (*A248750*)

Equals (1+sqrt(sqrt(5)-2))/2. - Vaclav Kotesovec, Oct 19 2014
From Wolfdieter Lang, Mar 02 2018: (Start)
Equals (1 + (2 - phi)*sqrt(phi))/2, with phi = A001622.
Equals (1/10)*y*(1 - (1/50)*y^2) with y = -A300072. (End)

A106195 Riordan array (1/(1-2x), x(1-x)/(1-2*x)).

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 8, 8, 4, 1, 16, 20, 13, 5, 1, 32, 48, 38, 19, 6, 1, 64, 112, 104, 63, 26, 7, 1, 128, 256, 272, 192, 96, 34, 8, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1024, 2816, 4096, 4048, 2972, 1683, 743, 253, 64, 11

Offset: 0

Gary W. Adamson, Apr 24 2005; Paul Barry, May 21 2006

Extract antidiagonals from the product P * A, where P = the infinite lower triangular Pascal's triangle matrix; and A = the Pascal's triangle array:
  1, 1,  1,  1, ...
  1, 2,  3,  4, ...
  1, 3,  6, 10, ...
  1, 4, 10, 20, ...
  ...
Row sums are Fibonacci(2n+2). Diagonal sums are A006054(n+2). Row sums of inverse are A105523. Product of Pascal triangle A007318 and A046854.
A106195 with an appended column of ones = A055587. Alternatively, k-th column (k=0, 1, 2) is the binomial transform of bin(n, k).
T(n,k) is the number of ideals in the fence Z(2n) having k elements of rank 1. - Emanuele Munarini, Mar 22 2011
Subtriangle of the triangle given by (0, 2, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012

			Triangle begins
   1;
   2,   1;
   4,   3,   1;
   8,   8,   4,  1;
  16,  20,  13,  5,  1;
  32,  48,  38, 19,  6, 1;
  64, 112, 104, 63, 26, 7, 1;
(0, 2, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, ...) begins :
  1;
  0,  1;
  0,  2,   1;
  0,  4,   3,   1;
  0,  8,   8,   4,  1;
  0, 16,  20,  13,  5,  1;
  0, 32,  48,  38, 19,  6, 1;
  0, 64, 112, 104, 63, 26, 7, 1. - _Philippe Deléham_, Mar 22 2012

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
E. Munarini, N. Zagaglia Salvi, On the Rank Polynomial of the Lattice of Order Ideals of Fences and Crowns, Discrete Mathematics 259 (2002), 163-177.

Column 0 = 1, 2, 4...; (binomial transform of 1, 1, 1...); column 1 = 1, 3, 8, 20...(binomial transform of 1, 2, 3...); column 2: 1, 4, 13, 38...= binomial transform of bin(n, 2): 1, 3, 6...

Cf. A001792, A001906, A002620, A007318, A029653, A049612, A055587, A078812, A208341.

a106195 n k = a106195_tabl !! n !! k
a106195_row n = a106195_tabl !! n
a106195_tabl = [1] : [2, 1] : f [1] [2, 1] where
   f us vs = ws : f vs ws where
     ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
                      ([0] ++ us ++ [0])
-- Reinhard Zumkeller, Dec 16 2013

[ (&+[Binomial(n-k, n-j)*Binomial(j, k): j in [0..n]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Mar 15 2020

T := (n, k) -> hypergeom([-n+k, k+1],[1],-1):
seq(lprint(seq(simplify(T(n, k)), k=0..n)), n=0..7); # Peter Luschny, May 20 2015

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + (x + 1) v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207605 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A106195 *)
(* Clark Kimberling, Feb 19 2012 *)
Table[Hypergeometric2F1[-n+k, k+1, 1, -1], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 15 2020 *)

create_list(sum(binomial(i,k)*binomial(n-k,n-i),i,0,n),n,0,8,k,0,n); /* Emanuele Munarini, Mar 22 2011 */

from sympy import Poly, symbols
x = symbols('x')
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

from mpmath import hyp2f1, nprint
def T(n, k): return hyp2f1(k - n, k + 1, 1, -1)
for n in range(13): nprint([int(T(n, k)) for k in range(n + 1)]) # Indranil Ghosh, May 28 2017, after formula from Peter Luschny

[[sum(binomial(n-k,n-j)*binomial(j,k) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 15 2020

T(n,k) = Sum_{j=0..n} C(n-k,n-j)*C(j,k).
From Emanuele Munarini, Mar 22 2011: (Start)
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-k,i)*2^(n-k-i).
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-i,k)*(-1)^i*2^(n-k-i).
Recurrence: T(n+2,k+1) = 2*T(n+1,k+1)+T(n+1,k)-T(n,k). (End)
From Clark Kimberling, Feb 19 2012: (Start)
Define u(n,x) = u(n-1,x)+v(n-1,x), v(n,x) = u(n-1,x)+(x+1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1. Then v matches A106195 and u matches A207605. (End)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k-1). - Philippe Deléham, Mar 22 2012
T(n+k,k) is the coefficient of x^n y^k in 1/(1-2x-y+xy). - Ira M. Gessel, Oct 30 2012
T(n, k) = A208341(n+1,n-k+1), k = 0..n. - Reinhard Zumkeller, Dec 16 2013
T(n, k) = hypergeometric_2F1(-n+k, k+1, 1 , -1). - Peter Luschny, May 20 2015
G.f. 1/(1-2*x+x^2*y-x*y). - R. J. Mathar, Aug 11 2015
Sum_{k=0..n} T(n, k) = Fibonacci(2*n+2) = A088305(n+1). - G. C. Greubel, Mar 15 2020

Edited by N. J. A. Sloane, Apr 09 2007, merging two sequences submitted independently by Gary W. Adamson and Paul Barry

A187660 Triangle read by rows: T(n,k) = (-1)^(floor(3k/2))binomial(floor((n+k)/2),k), 0 <= k <= n.

Original entry on oeis.org

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

Offset: 0

L. Edson Jeffery, Mar 12 2011

Conjecture: (i) Let n > 1 and N=2*n+1. Row n of T gives the coefficients of the characteristic polynomial p_N(x)=Sum_{k=0..n} T(n,k)*x^(n-k) of the n X n Danzer matrix D_{N,n-1} = {{0,...,0,1}, {0,...,0,1,1}, ..., {0,1,...,1}, {1,...,1}}. (ii) Let S_0(t)=1, S_1(t)=t and S_r(t)=t*S_(r-1)(t)-S_(r-2)(t), r > 1 (cf. A049310). Then p_N(x)=0 has solutions w_{N,j}=S_(n-1)(phi_{N,j}), where phi_{N,j}=2*(-1)^(j+1)*cos(j*Pi/N), j = 1..n. - L. Edson Jeffery, Dec 18 2011

			Triangle begins:
  1;
  1,  -1;
  1,  -1,  -1;
  1,  -2,  -1,   1;
  1,  -2,  -3,   1,   1;
  1,  -3,  -3,   4,   1,  -1;
  1,  -3,  -6,   4,   5,  -1,  -1;
  1,  -4,  -6,  10,   5,  -6,  -1,   1;
  1,  -4, -10,  10,  15,  -6,  -7,   1,   1;
  1,  -5, -10,  20,  15, -21,  -7,   8,   1,  -1;
  1,  -5, -15,  20,  35, -21, -28,   8,   9,  -1,  -1;
  1,  -6, -15,  35,  35, -56, -28,  36,   9, -10,  -1,   1;

L. E. Jeffery, Danzer matrices
Guoce Xin and Yueming Zhong, Proving some conjectures on Kekulé numbers for certain benzenoids by using Chebyshev polynomials, arXiv:2201.02376 [math.CO], 2022.

Signed version of A046854.
Absolute values of a(n) form a reflected version of A065941, which is considered the main entry.
Cf. A046854, A066170, A130777, A267482.

A187660 := proc(n,k): (-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k) end: seq(seq(A187660(n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

t[n_, k_] := (-1)^Floor[3 k/2] Binomial[Floor[(n + k)/2], k]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] (* L. Edson Jeffery, Oct 20 2017 *)

T(n,k) = (-1)^n*A066170(n,k).
abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) = abs(A130777(n,k)).
abs(T(n,k)) = A065941(n,n-k) = abs(A108299(n,n-k)).

Edited and corrected by L. Edson Jeffery, Oct 20 2017

A267482 Triangle of coefficients of Gaussian polynomials [2n+1,1]_q represented as finite sum of terms (1+q^2)^k*q^(g-k), where k = 0,1,...,g with g=n.

Original entry on oeis.org

1, 1, 1, -1, 1, 1, -1, -2, 1, 1, 1, -2, -3, 1, 1, 1, 3, -3, -4, 1, 1, -1, 3, 6, -4, -5, 1, 1, -1, -4, 6, 10, -5, -6, 1, 1, 1, -4, -10, 10, 15, -6, -7, 1, 1, 1, 5, -10, -20, 15, 21, -7, -8, 1, 1

Offset: 0

Stephen O'Sullivan, Jan 15 2016

The entry a(n,k), n >= 0, k = 0,1,...,g, where g=n, of this irregular triangle is the coefficient of (1+q^2)^k*q^(g-k) in the representation of the Gaussian polynomial [2n+1,1]q = Sum{k=0..g) a(n,k)*(1+q^2)^k*q^(g-k).
The sequence arises in the formal derivation of the stability polynomial B(x) = Sum_{i=0..N} d_i T(iM,x) of rank N, and degree L, where T(iM,x) denotes the Chebyshev polynomial of the first kind of degree iM. The coefficients d_i are determined by order conditions on the stability polynomial.
Conjecture: More generally, the Gaussian polynomial [2*n+m+1-(m mod 2),m]q = Sum{k=0..g(m;n)} a(m;n,k)*(1+q^2)^k*q^(g(m;n)-k), for m >= 0, n >= 0, where g(m;n) = m*n if m is odd and (2*n+1)*m/2 if m is even, and the tabf array entries a(m;n,k) are the coefficients of the g.f. for the row n polynomials G(m;n,x) = (d^m/dt^m)G(m;n,t,x)/m!|{t=0}, with G(m;n,t,x) = (1+t)*Product{k=1..n+(m - m (mod 2))/2}(1 + t^2 + 2*t*T(k,x/2) (Chebyshev's T-polynomials). Hence a(m;n,k) = [x^k]G(m;n,x), for k=0..g(m;n). The present entry is the instance m = 2. (Thanks to Wolfdieter Lang for clarifying the text on the general prescription of a(m;n,k).)
Signed version of A046854, A130777.
Conjecture: row n is U(n, x/2) + U(n-1, x/2) where U is the sequence of Chebyshev polynomials of the second kind. - Thomas Baruchel, Jun 03 2018 [For a proof see the following comment.]
From Wolfdieter Lang, Oct 19 2019: (Start)
The row polynomial R(n, x) = Sum_{k=0..n} a(n, k)*x^k = [2*n+1]_q / q^n with the q-number [2*n+1]_q := (1 - q^n)/(1 - q), which for q = 1 becomes 2*n+1, and x = x(q) = q + q^(-1). See the simplified Name and the first comment. In terms of Chebyshev S polynomials (A049310) this q-number is written as [2*n+1]_q = q^n*S(2*n, q^(1/2) + q^(-1/2)), hence R(n, x) = S(2*n, sqrt(2+x)) = S(n, x) + S(n-1, x) (which proves the conjecture of the previous comment).
For the o.g.f. of R(n, x) see the formula section.
My motivation for looking at this sequence came from the Brändli and Beyne paper's recurrence for the polynomial P_m(s) which coincides with R(n, x), with m -> n and s -> x. (End)
A294099(n, k) = Sum_{j=0..k} n^j * T(n, j) for all n, k in Z. - Michael Somos, Jun 19 2023

			Triangle begins:
   1;
   1,   1;
  -1,   1,   1;
  -1,  -2,   1,   1;
   1,  -2,  -3,   1,   1;
   1,   3,  -3,  -4,   1,   1;
  -1,   3,   6,  -4,  -5,   1,   1;
  -1,  -4,   6,  10,  -5,  -6,   1,   1;
   1,  -4, -10,  10,  15,  -6,  -7,   1,   1;
   1,   5, -10, -20,  15,  21,  -7,  -8,   1,   1;

Stephen O'Sullivan, Table of n, a(n) for n = 0..495
Gerold Brändli and Tim Beyne, Modified Congruence Modulo n with Half the Amount of Residues, arXiv:1504.02757 [math.NT], 2016-2017. Definition 6.for polynomials P_m(s).
Stephen O'Sullivan, A class of high-order Runge-Kutta-Chebyshev stability polynomials, Journal of Computational Physics, 300 (2015), 665-678.
Wikipedia, Gaussian binomial coefficients.
Index entries for sequences related to Chebyshev polynomials.

Cf. A049310, A267120, A267483, A267484, A267485, A267486, A294099.

Cf. A046854, A066170, A130777, A187660 all signed versions.

A267482 := proc (n, k) local y: y := expand(subs(t = 0, diff((1+t)*product(1+t^2+2*t*ChebyshevT(i, x/2), i = 1 .. n),t))): if k = 0 then subs(x = 0, y) else subs(x = 0, diff(y, x$k)/k!) end if: end proc: seq(seq(A267482(n, k), k = 0 .. n), n = 0 .. 20);

row[n_] := D[(1+t)*Product[1+t^2+2*t*ChebyshevT[i, x/2], {i, 1, n}], t] /. t -> 0 // CoefficientList[#, x]&; Table[row[n], {n, 0, 20}] // Flatten (* Jean-François Alcover, Jan 16 2016 *)

T(n,k) = (-1)^((n-k)\2)*binomial((n+k)\2, k); \\François Marques, Sep 28 2021

G.f. for row polynomial: G(n,x) = (d^2/dt^2)((1+t)*Product_{i=1..n+1}(1+t^2+2t*T(i,x/2)))|_{t=0}.
From Wolfdieter Lang, Oct 19 2019: (Start)
Row polynomial R(n, x) = S(2*n, sqrt(2+x)) = S(n, x) + S(n-1, x) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*(2 + x)^(n-k), for n >= 0. (See the Thomas Baruchel conjecture and the proof above.) For the S(n, x) coefficients see A049310.
R(n, x) = Sum_{j=0} (-1)^e(n,j)*binomial(e(n,j) + j, j)*x^j*, with e(n,j) := floor((n-j)/2). See eq. (12) of the Brändli and Beyne paper.
G.f. for row polynomials R(n, x) (that is of the triangle): G(x,z) = (1 + z)/(1 - x*z + z^2).
Recurrence for R(n, x): R(-1, x)  = -1, R(0, x) = 1, R(n, x) = x*R(n-1, x) - R(n-2, x), for n >= 1. (See the Brändli and Beyne link, polynomials P_m(s) in Definition 6.)
(End)
T(n,k) = (-1)^(floor((n-k)/2))*binomial(floor((n+k)/2), k). - François Marques, Sep 28 2021

A131246 Row sums of triangle A131245.

Original entry on oeis.org

1, 3, 6, 13, 27, 57, 119, 250, 523, 1097, 2297, 4815, 10086, 21137, 44283, 92793, 194419, 407378, 853559, 1788481, 3747361, 7851867, 16451910, 34471669, 72228171, 151339401, 317100335, 664418698, 1392152131

Offset: 0

Gary W. Adamson, Jun 22 2007

nonn

A131244 is a companion sequence.

			a(3) = 13 = sum of row 3 terms of triangle A131245: (5 + 5 + 2 + 1)

Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-1)

Cf. A131243, A131244, A131245, A065941, A046854.

A046854 := proc(n,k) binomial(floor((n+k)/2),k) ; end proc:
A131245 := proc(n,k) local a,j ; a := 0 ; for j from k to n do a := a+ A046854(n,j)*A046854(j,k) ;  end do: a ; end proc:
A131246 := proc(n) add(A131245(n,k),k=0..n) ; end proc:
seq(A131246(n),n=0..50) ; # R. J. Mathar, Jan 29 2011

CoefficientList[Series[-(1+x)(x^2-x-1)/(1-x-3x^2+x^3+x^4),{x,0,30}],x] (* or *) LinearRecurrence[{1,3,-1,-1},{1,3,6,13},30] (* Harvey P. Dale, Sep 07 2013 *)

G.f. -(1+x)*(x^2-x-1)/ ( 1-x-3*x^2+x^3+x^4 ). - R. J. Mathar, Jan 29 2011

a(0)=1, a(1)=3, a(2)=6, a(3)=13, a(n)=a(n-1)+3*a(n-2)-a(n-3)-a(n-4). - Harvey P. Dale, Sep 07 2013

A015453 Generalized Fibonacci numbers.

Original entry on oeis.org

1, 1, 8, 57, 407, 2906, 20749, 148149, 1057792, 7552693, 53926643, 385039194, 2749201001, 19629446201, 140155324408, 1000716717057, 7145172343807, 51016923123706, 364263634209749, 2600862362591949, 18570300172353392

Offset: 0

Olivier Gérard

Row m=7 of A135597.
For n >= 1, row sums of triangle for numbers 7^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 13 2012
For n >= 1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,5,6,7} containing no subwords ii, (i=0,1,...,6). - Milan Janjic, Jan 31 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 18.
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 (7,1)

Cf. A054413, A055830.

Row m=7 of A135597.

[n le 2 select 1 else 7*Self(n-1) + Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{7, 1}, {1, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
CoefficientList[Series[(1-6*x)/(1-7*x-x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)

my(x='x+O('x^30)); Vec((1-6*x)/(1-7*x-x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number1(n+1, 7, -1) - 6*lucas_number1(n, 7, -1) for n in (0..30)] # G. C. Greubel, Dec 24 2021

a(n) = 7*a(n-1) + a(n-2).
a(n) = Sum_{k=0..n} 6^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
G.f.: (1-6*x)/(1-7*x-x^2). - Philippe Deléham, Nov 20 2008
For n >= 2, a(n) = F_(n)(7) + F_(n+1)(7), 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) = A054413(n) - 6*A054413(n-1). - R. J. Mathar, Jul 06 2012
a(n) = Sum_{k=0..n} A046854(n-1,k)*7^k. - R. J. Mathar, Feb 14 2024

A030111 Triangular array in which k-th entry in n-th row is C([ (n+k)/2 ],k) (1<=k<=n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 3, 3, 4, 1, 1, 3, 6, 4, 5, 1, 1, 4, 6, 10, 5, 6, 1, 1, 4, 10, 10, 15, 6, 7, 1, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 6, 21, 35, 70, 56, 84, 36, 45, 10, 11, 1, 1, 7, 21, 56, 70, 126, 84, 120, 45, 55, 11, 12, 1

Offset: 1

Jacques Haubrich (jhaubrich(AT)freeler.nl)

Same as A046854, but missing the initial column of ones.

Riordan array (1/((1-x)(1-x^2)),x/(1-x^2)). Diagonal sums are A052551. - Paul Barry, Sep 30 2006

			1;
1 1;
2 1 1;
2 3 1 1;
3 3 4 1 1;
3 6 4 5 1 1;
...

Indranil Ghosh, Rows 0..125, flattened

Cf. A066170.

Flatten[Table[Binomial[Floor[(n+k)/2],k],{n,20},{k,n}]] (* Harvey P. Dale, Jun 03 2014 *)

{T(n, k) = binomial((n+k)\2, k)}; /* Michael Somos, Jul 23 1999 */

printp(matrix(8,8,n,k,binomial((n+k)\2,k)))

for(n=1,7, for(k=1,n,print1(binomial((n+k)\2,k)); if(k==n,print1("; ")); print1(" ")))

G.f.: 1 / (1 - x - xy - x^2 + x^2y + x^3). - Ralf Stephan, Feb 13 2005

Sum(k=1, n, T(n, k)) = F(n+2)-1 where F(n) is the n-th Fibonacci number. - Benoit Cloitre, Oct 07 2002

Description corrected by Michael Somos, Jul 23 1999

Corrected and extended by Harvey P. Dale, Jun 03 2014

cp's OEIS Frontend

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

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A015451 a(n) = 6*a(n-1) + a(n-2) for n > 1, with a(0) = a(1) = 1.

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A055244 Number of certain stackings of n+1 squares on a double staircase.

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A248750 Decimal expansion of limit of the imaginary 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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A106195 Riordan array (1/(1-2*x), x*(1-x)/(1-2*x)).

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A187660 Triangle read by rows: T(n,k) = (-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k), 0 <= k <= n.

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

A106195 Riordan array (1/(1-2x), x(1-x)/(1-2*x)).

A187660 Triangle read by rows: T(n,k) = (-1)^(floor(3k/2))binomial(floor((n+k)/2),k), 0 <= k <= n.