A168561 -id:A168561 - cp's OEIS frontend

A123018 Triangle read by rows: row n gives the coefficients of x^k (0 <= k <= n) in the expansion of Sum_{j=0..n} A320508(n,j)x^j(1 - x)^(n - j).

1, 1, -2, 1, -2, 2, 1, -2, 1, -1, 1, -2, 0, 2, 0, 1, -2, -1, 5, -4, 0, 1, -2, -2, 8, -7, 2, 1, 1, -2, -3, 11, -9, 0, 3, -2, 1, -2, -4, 14, -10, -6, 12, -6, 2, 1, -2, -5, 17, -10, -16, 27, -15, 3, -1, 1, -2, -6, 20, -9, -30, 47, -24, 0, 4, 0, 1, -2, -7, 23, -7

Offset: 0

Roger L. Bagula and Gary W. Adamson, Sep 24 2006

The n-th row consists of the coefficients in the expansion of (-x)^n - (1 - x)*(((1 - x - sqrt(1 + 2*x - 3*x^2))/2)^n - ((1 - x + sqrt(1 + 2*x - 3*x^2))/2)^n)/sqrt(1 + 2*x - 3*x^2). - Franck Maminirina Ramaharo, Oct 13 2018

			Triangle begins:
     1;
     1, -2;
     1, -2,  2;
     1, -2,  1, -1;
     1, -2,  0,  2,   0;
     1, -2, -1,  5,  -4,   0;
     1, -2, -2,  8,  -7,   2,  1;
     1, -2, -3, 11,  -9,   0,  3,  -2;
     1, -2, -4, 14, -10,  -6, 12,  -6,   2;
     1, -2, -5, 17, -10, -16, 27, -15,   3, -1;
     1, -2, -6, 20,  -9, -30, 47, -24,   0,  4,  0;
     1, -2, -7, 23,  -7, -48, 71, -28, -18, 22, -8, 0;
     ....

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Fibonacci Polynomial

Row sums: A033999.
Cf. A049310, A168561, A320508.
Cf. A122753, A123019, A123021, A123027, A123199, A123202, A123217, A123221.

P[x_, n_]:= Sum[Binomial[n-k-1, k]*x^k*(1-x)^(n-k), {k, 0, n}];
Table[Coefficient[P[x, n], x, k], {n,0,12}, {k,0,n}]//Flatten (* Franck Maminirina Ramaharo, Oct 14 2018 *)

P(x, n) := sum(binomial(n - k - 1, k)*x^k*(1 - x)^(n - k), k, 0, n)$
create_list(ratcoef(expand(P(x, n)), x, k), n, 0, 12, k, 0, n); /* Franck Maminirina Ramaharo, Oct 14 2018 */

def p(n,x): return sum( binomial(n-j-1, j)*x^j*(1-x)^(n-j) for j in (0..n) )
def T(n): return ( p(n,x) ).full_simplify().coefficients(sparse=False)
flatten([T(n) for n in (0..12)]) # G. C. Greubel, Jul 15 2021

From Franck Maminirina Ramaharo, Oct 13 2018: (Start)
G.f.: 1/((1 + x*y)*(1 - y + x*y - x*y^2 + x^2*y^2)).
E.g.f.: exp(-x*y) - (exp(y*(1 - x - sqrt(1 + 2*x - 3*x^2))/2) - exp(y*(1 - x + sqrt(1 + 2*x - 3*x^2))/2))*(1 - x)/sqrt(1 + 2*x - 3*x^2). (End)

Edited by N. J. A. Sloane, May 26 2007

Edited by Franck Maminirina Ramaharo, Oct 14 2018

A162515 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = (U^n - L^n)/d, where U = (x + d)/2, L = (x - d)/2, d = sqrt(x^2 + 4).

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 3, 0, 1, 1, 0, 4, 0, 3, 0, 1, 0, 5, 0, 6, 0, 1, 1, 0, 6, 0, 10, 0, 4, 0, 1, 0, 7, 0, 15, 0, 10, 0, 1, 1, 0, 8, 0, 21, 0, 20, 0, 5, 0, 1, 0, 9, 0, 28, 0, 35, 0, 15, 0, 1, 1, 0, 10, 0, 36, 0, 56, 0, 35, 0, 6, 0, 1, 0, 11, 0, 45, 0, 84, 0, 70, 0, 21, 0, 1

Offset: 0

Clark Kimberling, Jul 05 2009

Row sums 0,1,1,2,3,5,... are the Fibonacci numbers, A000045.
Note that the coefficients are given in decreasing order. - M. F. Hasler, Dec 07 2011
Essentially a mirror image of A168561. - Philippe Deléham, Dec 08 2013

			Polynomial expansion:
  0;
  1;
  x;
  x^2 + 1;
  x^3 + 2*x;
  x^4 + 3*x^2 + 1;
First rows:
  0;
  1;
  1, 0;
  1, 0, 1;
  1, 0, 2, 0;
  1, 0, 3, 0, 1;
  1, 0, 4, 0, 3, 0;
Row 6 matches P(6,x)=x^5 + 4*x^3 + 3*x.

G. C. Greubel, Rows n = 0..101 of triangle
T. Copeland, Addendum to Elliptic Lie Triad

Cf. A000045, A049310, A053119, A162514, A162516, A162517.

T:= function(n,k)
    if (k mod 2)=0 then return Binomial(n- k/2, k/2);
    else return 0;
    fi; end;
Concatenation([0], Flat(List([0..15], n-> List([0..n], k-> T(n,k) ))) ); # G. C. Greubel, Jan 01 2020

function T(n,k)
  if (k mod 2) eq 0 then return Round( Gamma(n-k/2+1)/(Gamma(k/2+1)*Gamma(n-k+1)));
  else return 0;
  end if; return T; end function;
[0] cat [T(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jan 01 2020

0, seq(seq(`if`(`mod`(k,2)=0, binomial(n-k/2, k/2), 0), k = 0..n), n = 0..15); # G. C. Greubel, Jan 01 2020

Join[{0}, Table[If[EvenQ[k], Binomial[n-k/2, k/2], 0], {n,0,15}, {k,0,n} ]//Flatten] (* G. C. Greubel, Jan 01 2020 *)

row(n,d=sqrt(1+x^2/4+O(x^n))) = Vec(if(n,Pol(((x/2+d)^n-(x/2-d)^n)/d)>>1)) \\ M. F. Hasler, Dec 07 2011, edited Jul 05 2021

@CachedFunction
def T(n, k):
    if (k%2==0): return binomial(n-k/2, k/2)
    else: return 0
[0]+flatten([[T(n, k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jan 01 2020

P(n,x) = x*P(n-1, x) + P(n-2, x), where P(0,x)=0 and P(1,x)=1.

T(n,k) = T(n-1, k) + T(n-2, k-2) for n>=2. - Philippe Deléham, Dec 08 2013

A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 4, 4, 8, 1, 4, 12, 8, 16, 1, 6, 12, 32, 16, 32, 1, 6, 24, 32, 80, 32, 64, 1, 8, 24, 80, 80, 192, 64, 128, 1, 8, 40, 80, 240, 192, 448, 128, 256, 1, 10, 40, 160, 240, 672, 448, 1024, 256, 512, 1, 10, 60, 160, 560, 672, 1792, 1024, 2304, 512, 1024, 1, 12, 60, 280, 560, 1792, 1792, 4608, 2304, 5120, 1024, 2048

Offset: 0

Clark Kimberling, Mar 05 2009

Conjecture 1. If n>1 is even then F(n,x) has no real roots.
Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,
and if n>4 then 0 < -r < n.
Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]
Cayley (1876) states "We, in fact, find  1 + sin u = 1 + x,  1 - sin 3u = (1 + x)(1 - 2x)^2,  1 + sin 5u = (1 + x)(1 + 2x - 4x^2)^2,  1 - sin 7u = (1 + x)(1 - 4x - 4x^2 + 8x^3)^2, &c.". - Michael Somos, Jun 19 2012
Appears to be the unsigned row reverse of A180870 and A228565. - Peter Bala, Feb 17 2014
From Wolfdieter Lang, Jul 29 2014: (Start)
This triangle is the Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). For Riordan triangles see the W. Lang link 'Sheffer a-and z-sequences' under A006232, also for references. The o.g.f. given by Peter Bala in the formula section refers to the row reversed triangle. The usual information on this triangle, like o.g.f. for the columns, the row sums, the alternating row sums, the recurrences using A- and Z-sequences, etc. follows from this Riordan property. The Riordan proof follows from the given o.g.f. by Peter Bala, call it Grev(x,z), by row reversion: G(x,z) = Grev(1/x,x*z) = (1+z)/(1- 2*x*z - z^2) = G(z)*(1/(1 - x*F(z))) with G(z) = (1+z)/(1-z^2) and F(z) = z*2/(1-z^2). See A244419 for the discussion for a signed version of this triangle.
(End)

			Rows 0 to 8:
1
1 2
1 2 4
1 4 4 8
1 4 12 8 16
1 6 12 32 16 32
1 6 24 32 80 32 64
1 8 24 80 80 192 64 128
1 8 40 80 240 192 448 128 256
(Row 8) = (1, 4*2, 10*4, 10*8, 15*16, 6*32, 7*64, 1*128, 1*256).
First few polynomials:
F(0,x)=1, F(1,x)=x+2, F(2,x)=x^2+2*x+4, F(3,x)=x^3+4*x^2+4*x+8.
The row polynomials R(n,x) start: 1, 1 + 2*x = x*F(1,1/x), 1 + 2*x + 4^x^2 = x^2*F(2,1/x), ...  - _Wolfdieter Lang_, Jul 29 2014

A. Cayley, On an Expression for 1 +- sin(2p+1)u in Terms of sin u, Messenger of Mathematics, 5 (1876), pp. 7-8 = Mathematical Papers Vol. 10, n. 630, pp. 1-2.

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

Cf. A140070, A140071, A180870, A228565, A078057, A123335, A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980.

T[n_, 0]:= 1; T[n_, n_]:= 2^n; T[n_, k_]:= T[n, k] = T[n-1, k] + (1 + (-1)^(n-k))*T[n-1, k-1]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] (* G. C. Greubel, Sep 24 2018 *)

t(n,k) = if(k==0, 1, if(k==n, 2^n, t(n-1,k) + (1+(-1)^(n-k))*t(n-1,k-1)));
for(n=0,15, for(k=0,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Sep 24 2018

Count the top row as row 0 and let C(n,k) denote the usual binomial
coefficient. For row 2n, define p(0)=C(n,0), p(1)=C(n,1), p(2)=C(n+1,2),
p(3)=C(n+1,3), p(4)=C(n+2,4), p(5)=c(n+2,5),..., until reaching two final
1's: p(2n-1)=C(2n-1,2n-1) and p(2n)=C(2n,2n). Then the k-th number in row
2n is p(k)*2^k. For row 2n+1, define q(0)=C(n,0), q(1)=C(n+1,1),
q(2)=C(n+1,2), q(3)=C(n+2,3),..., until reaching q(2n+1)=C(2n+1,2n+1).
Then the k-th number in row 2n+1 is q(k)*2^k.
From Peter Bala, Jan 17 2014: (Start)
Working with an offset of 0, the o.g.f. is (1 + x*z)/(1 - 2*z - x^2*z^2) = 1 + (x + 2)*z + (x^2 + 2*x + 4)*z^2 + ....
Recurrence equation: T(n,k) = 2*T(n-1,k-1) + T(n-2,k-2) with T(n,0) = 1.
The polynomials G(n,x) defined by G(0,x) = 1 and G(n,x) = x*F(n-1,x) for n >= 1 satisfy G(n,x) = (x + 1)*G(n-1,x) - G(n-1,-x). Cf. A140070 and A140071. (End)
From Wolfdieter Lang, Jul 29 2014: (Start)
O.g.f. for the row polynomials (rising powers of x) R(n,x) = x^n*F(n,1/x): (1+z)/(1 - 2*x*z - z^2). Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). See a comment above.
Recurrence for the row polynomials R(n,x) = (1+x)*R(n-1,x) - (-1)^n*x*R(n-1,-x), n >= 1, R(0,x) = 1.
R(n,x) = Ftilde(n,2*x) + Ftilde(n-1,2*x) with the monic Fibonacci polynomials Ftilde(n,x) given in A168561.
Recurrence for the triangle: R(n,m) = R(n-1,m) + (1 + (-1)^(n-m))*R(n-1,m-1), n >= m >= 1, R(n,m) = 0 if n < m, R(n,0) = 1.
O.g.f. column sequences ((1+x)/(1-x^2))*(2*x/(1-x^2))^m, m >= 0. See A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980, ...
Row sums A078057. Alternating row sums A123335.
(End)

Offset corrected to 0. Cf.s added, keyword easy added by Wolfdieter Lang, Jul 29 2014

A374439 Triangle read by rows: the coefficients of the Lucas-Fibonacci polynomials. T(n, k) = T(n - 1, k) + T(n - 2, k - 2) with initial values T(n, k) = k + 1 for k < 2.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 3, 4, 1, 1, 2, 4, 6, 3, 2, 1, 2, 5, 8, 6, 6, 1, 1, 2, 6, 10, 10, 12, 4, 2, 1, 2, 7, 12, 15, 20, 10, 8, 1, 1, 2, 8, 14, 21, 30, 20, 20, 5, 2, 1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1, 1, 2, 10, 18, 36, 56, 56, 70, 35, 30, 6, 2

Offset: 0

Peter Luschny, Jul 22 2024

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.
A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.
The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).
Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.
The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 1]
  [ 3] [1, 2, 2,  2]
  [ 4] [1, 2, 3,  4,  1]
  [ 5] [1, 2, 4,  6,  3,  2]
  [ 6] [1, 2, 5,  8,  6,  6,  1]
  [ 7] [1, 2, 6, 10, 10, 12,  4,  2]
  [ 8] [1, 2, 7, 12, 15, 20, 10,  8,  1]
  [ 9] [1, 2, 8, 14, 21, 30, 20, 20,  5,  2]
  [10] [1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1]
.
Table of interpolated sequences:
  |  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |
  |  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|
  |    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |
  |  0 |        -1         |     1   |       1     |       1    |
  |  1 |         1         |     3   |       0     |       4    |
  |  2 |         0         |     4   |       1     |       9    |
  |  3 |         1         |     7   |       2     |      22    |
  |  4 |         1         |    11   |       5     |      53    |
  |  5 |         2         |    18   |      12     |     128    |
  |  6 |         3         |    29   |      29     |     309    |
  |  7 |         5         |    47   |      70     |     746    |
  |  8 |         8         |    76   |     169     |    1801    |
  |  9 |        13         |   123   |     408     |    4348    |

Paolo Xausa, Rows n = 0..150 of the triangle, flattened
Peter Luschny, Illustrating the polynomials.

Triangles related to Lucas polynomials: A034807, A114525, A122075, A061896, A352362.
Triangles related to Fibonacci polynomials: A162515, A053119, A168561, A049310, A374441.
Sums include: A000204 (Lucas numbers, row), A000045 & A212804 (even sums, Fibonacci numbers), A006355 (odd sums), A039834 (alternating sign row).
Type m^n*P(n, 1/m): A000129 & A048654 (Pell, m=2), A108300 & A003688 (m=3), A001077 & A048875 (m=4).
Adding and subtracting the values in a row of the table (plus halving the values obtained in this way): A022087, A055389, A118658, A052542, A163271, A371596, A324969, A212804, A077985, A069306, A215928.
Columns include: A040000 (k=1), A000027 (k=2), A005843 (k=3), A000217 (k=4), A002378 (k=5).
Diagonals include: A000034 (k=n), A029578 (k=n-1), abs(A131259) (k=n-2).
Cf. A029578 (subdiagonal), A124038 (row reversed triangle, signed).
Cf. A039961, A065941 (qStirling2), A103631, A108299, A131259. A133607, A194005, A333143, A374438 (m=3).

function T(n,k) // T = A374439
  if k lt 0 or k gt n then return 0;
  elif k le 1 then return k+1;
  else return T(n-1,k) + T(n-2,k-2);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 23 2025

A374439 := (n, k) -> ifelse(k::odd, 2, 1)*binomial(n - irem(k, 2) - iquo(k, 2), iquo(k, 2)):
# Alternative, using the function qStirling2 from A333143:
T := (n, k) -> 2^irem(k, 2)*qStirling2(n, k, -1):
seq(seq(T(n, k), k = 0..n), n = 0..10);

A374439[n_, k_] := (# + 1)*Binomial[n - (k + #)/2, (k - #)/2] & [Mod[k, 2]];
Table[A374439[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Paolo Xausa, Jul 24 2024 *)

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k > n: return 0
    if k < 2: return k + 1
    return T(n - 1, k) + T(n - 2, k - 2)

from math import comb as binomial
def T(n: int, k: int) -> int:
    o = k & 1
    return binomial(n - o - (k - o) // 2, (k - o) // 2) << o

def P(n, x):
    if n < 0: return P(n, x)
    return sum(T(n, k)*x**k for k in range(n + 1))
def sgn(x: int) -> int: return (x > 0) - (x < 0)
# Table of interpolated sequences
print("|  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |")
print("|  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|")
print("|    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |")
f = "| {0:2d} | {1:9d}         |  {2:4d}   |   {3:5d}     |    {4:4d}    |"
for n in range(10): print(f.format(n, -P(n, -1), P(n, 1), int(2**n*P(n, -1/2)), int(2**n*P(n, 1/2))))

from sage.combinat.q_analogues import q_stirling_number2
def A374439(n,k): return (-1)^((k+1)//2)*2^(k%2)*q_stirling_number2(n+1, k+1, -1)
print(flatten([[A374439(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 23 2025

T(n, k) = 2^k' * binomial(n - k' - (k - k') / 2, (k - k') / 2) where k' = 1 if k is odd and otherwise 0.
T(n, k) = (1 + (k mod 2))*qStirling2(n, k, -1), see A333143.
2^n*P(n, -1/2) = A000129(n - 1), Pell numbers, P(-1) = 1.
2^n*P(n, 1/2) = A048654(n), dual Pell numbers.
T(2*n, n) = (1/2)*(-1)^n*( (1+(-1)^n)*A005809(n/2) - 2*(1-(-1)^n)*A045721((n-1)/2) ). - G. C. Greubel, Jan 23 2025

A180148 a(n) = 3*a(n-1) + a(n-2) with a(0)=2 and a(1)=5.

Original entry on oeis.org

2, 5, 17, 56, 185, 611, 2018, 6665, 22013, 72704, 240125, 793079, 2619362, 8651165, 28572857, 94369736, 311682065, 1029415931, 3399929858, 11229205505, 37087546373, 122491844624, 404563080245, 1336181085359, 4413106336322, 14575500094325, 48139606619297

Offset: 0

Johannes W. Meijer, Aug 13 2010

Inverse binomial transform of A052961 (without the leading 1).

For n >= 1, also the number of matchings in the n-alkane graph. - Eric W. Weisstein, Jul 14 2021

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Alkane Graph
Eric Weisstein's World of Mathematics, Matching
Eric Weisstein's World of Mathematics, Maximum Independent Edge Set
Index entries for linear recurrences with constant coefficients, signature (3,1).

Cf. A003688, A052906.
Appears in A180142.
Cf. A000602 (more information on n-alkanes).

a:= n-> (<<0|1>, <1|3>>^n. <<2, 5>>)[1,1]:
seq(a(n), n=0..27);  # Alois P. Heinz, Jul 14 2021

LinearRecurrence[{3, 1}, {5, 7}, 20] (* Eric W. Weisstein, Jul 14 2021 *)
CoefficientList[Series[(2 - x)/(1 - 3 x - x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Jul 14 2021 *)

a(n)=([0,1;1,3]^n*[2;5])[1,1] \\ Charles R Greathouse IV, Oct 13 2016

G.f.: (2-x)/(1-3*x-x^2).
a(n) = 3*a(n-1) + a(n-2) with a(0)=2 and a(1)=5.
a(n) = ((4+7*A)*A^(-n-1) + (4+7*B)*B^(-n-1))/13 with A = (-3+sqrt(13))/2 and B = (-3-sqrt(13))/2.
Lim_{k->infinity} a(n+k)/a(k) = (-1)^n*2/(A006497(n) - A006190(n)*sqrt(13)).
a(n) = 2 * Sum_{k=0..n-2} A168561(n-2,k)*3^k + 5 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 2*A006190(n+1) - A006190(n). - R. J. Mathar, Feb 14 2024

A073028 a(n) = max{ C(n,0), C(n-1,1), C(n-2,2), ..., C(n-n,n) }.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 6, 10, 15, 21, 35, 56, 84, 126, 210, 330, 495, 792, 1287, 2002, 3003, 5005, 8008, 12376, 19448, 31824, 50388, 77520, 125970, 203490, 319770, 497420, 817190, 1307504, 2042975, 3268760, 5311735, 8436285, 13123110, 21474180, 34597290

Offset: 0

Miklos Kristof, Aug 22 2002

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

a(n-1) is the max coefficient in n-th Fibonacci polynomial (the polynomial F_0(x) is constant zero, and is not included in this sequence). - Vladimir Reshetnikov, Oct 09 2016

			For n = 6, C(6,0) = 1, C(5,1) = 5, C(4,2) = 6, C(3,3) = 1. These binomial coefficients are the coefficients in the Fibonacci polynomial F_7(x) = x^6 + 5*x^4 + 6*x^2 + 1. The max coefficient is 6, so a(6) = 6.

Peter Boros (borospet(AT)freemail.hu): Lectures on Fibonacci's World at the SOTERIA Foundation, 1999.

Charles R Greathouse IV, Table of n, a(n) for n = 0..4793
Benjamin Aram Berendsohn, László Kozma, and Dániel Marx, Finding and counting permutations via CSPs, arXiv:1908.04673 [cs.DS], 2019.
Charles Bouillaguet, Boolean Polynomial Evaluation for the Masses, LIP6 Laboratory, Sorbonne Université (Paris, France) Cryptology ePrint Archive (2022) No. 1412.
S. M. Tanny and M. Zuker, On a unimodal sequence of binomial coefficients, Discrete Math. 9 (1974), 79-89.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Cf. A060065, A277282, A168561.

Table[Max[CoefficientList[Fibonacci[n + 1, x], x]], {n, 1, 30}] (* Vladimir Reshetnikov, Oct 07 2016 *)

a(n)=my(k=(5*n-sqrtint(5*n^2+10*n+9)+6)\10); binomial(n-k,k) \\ Charles R Greathouse IV, Sep 22 2016

a(n) = binomial(n-A060065(n), A060065(n)). - Vladeta Jovovic, Jun 16 2004

a(n) ~ 5^(1/4) * phi^(n+1) / sqrt(2*Pi*n), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 09 2016

a(0) = 1 prepended by Vladimir Reshetnikov, Oct 09 2016

A179237 a(0) = 1, a(1) = 2; a(n+1) = 6*a(n) + a(n-1) for n>1.

Original entry on oeis.org

1, 2, 13, 80, 493, 3038, 18721, 115364, 710905, 4380794, 26995669, 166354808, 1025124517, 6317101910, 38927735977, 239883517772, 1478228842609, 9109256573426, 56133768283165, 345911866272416, 2131604965917661, 13135541661778382, 80944854936587953

Offset: 0

Gary W. Adamson, Jul 04 2010

a(n)/a(n-1) converges to 1/(sqrt(10) - 3) = 6.16227766017... = A176398.

			a(5) = 3038 = 6*a(5) + a(4) = 6*493 + 80.
a(5) = term (1,1) in M^5 where M^5 = [3038, 4215, 4215, 5848].

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,1).

I:=[1,2]; [n le 2 select I[n] else 6*Self(n-1)+Self(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 13 2015

CoefficientList[Series[(-1 + 4 x)/(-1 + 6 x + x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Oct 13 2015 *)

Vec((-1+4*x)/(-1+6*x+x^2) + O(x^40)) \\ Colin Barker, Oct 13 2015

Let M = the 2x2 matrix [2,3; 3,4]. a(n) = term (1,1) in M^n.
G.f.: ( -1+4*x ) / ( -1+6*x+x^2 ). a(n) = A005668(n) + A015451(n). - R. J. Mathar, Jul 06 2012
a(n) = ((3-sqrt(10))^n*(1+sqrt(10))+(-1+sqrt(10))*(3+sqrt(10))^n)/(2*sqrt(10)). - Colin Barker, Oct 13 2015
a(n) = Sum_{k=0..n-2} A168561(n-2,k)*6^k + 2 * Sum_{k=0..n-1} A168561(n-1,k)*6^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = A005668(n+1) - 4*A005668(n). - R. J. Mathar, Feb 14 2024

Corrected by R. J. Mathar, Jul 06 2012

A097837 Chebyshev polynomials S(n,51) + S(n-1,51) with Diophantine property.

Original entry on oeis.org

1, 52, 2651, 135149, 6889948, 351252199, 17906972201, 912904330052, 46540213860451, 2372638002552949, 120957997916339948, 6166485255730784399, 314369790044353664401, 16026692807006306100052, 817046963367277257438251

Offset: 0

Wolfdieter Lang, Sep 10 2004

(7*a(n))^2 - 53*b(n)^2 = -4 with b(n)=A097838(n) gives all positive solutions of this Pell equation.

			All positive solutions of Pell equation x^2 - 53*y^2 = -4 are (7=7*1,1), (364=7*52,50), (18557=7*2651,2549), (946043=7*135149,129949), ...

Indranil Ghosh, Table of n, a(n) for n = 0..584
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (51,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A097783, A097834, A097836, A097840.

a:=[1,52];; for n in [3..30] do a[n]:=51*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-51*x+x^2) )); // G. C. Greubel, Jan 12 2019

LinearRecurrence[{51,-1}, {1,52}, 30] (* G. C. Greubel, Jan 12 2019 *)

my(x='x+O('x^30)); Vec((1+x)/(1-51*x+x^2)) \\ G. C. Greubel, Jan 12 2019

((1+x)/(1-51*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019

a(n) = S(n, 51) + S(n-1, 51) = S(2*n, sqrt(53)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 51)=A097836(n).
a(n) = (-2/7)*i*((-1)^n)*T(2*n+1, 7*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-51*x+x^2).
a(n) = 51*a(n-1) - a(n-2); a(0)=1, a(1)=52. - Philippe Deléham, Nov 18 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/7)*(7/2 o 7/2 o ... o 7/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 52, 0, 2651, 0, 135149, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 51*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -49, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,7) + (1/7)*( 1 + (-1)^(n+1) )*F(n+1,7), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 14*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*x^n.
Exp( Sum_{n >= 1} (-14)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 14*A054413(n)*(-x)^n. (End)

A097840 Chebyshev polynomials S(n,83) + S(n-1,83) with Diophantine property.

Original entry on oeis.org

1, 84, 6971, 578509, 48009276, 3984191399, 330639876841, 27439125586404, 2277116783794691, 188973253929372949, 15682502959354160076, 1301458772372465913359, 108005395603955316648721

Offset: 0

Wolfdieter Lang, Sep 10 2004

(9*a(n))^2 - 85*b(n)^2 = -4 with b(n)=A097841(n) give all positive solutions of this Pell equation.

			All positive solutions of Pell equation x^2 - 85*y^2 = -4 are (9=9*1,1), (756=9*84,82), (62739=9*6971,6805), (5206581=9*578509,564733), ...

Indranil Ghosh, Table of n, a(n) for n = 0..520
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (83, -1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A097783, A097834, A097837, A097839.

a:=[1,84];; for n in [3..20] do a[n]:=83*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-83*x+x^2) )); // G. C. Greubel, Jan 13 2019

CoefficientList[Series[(1+x)/(1-83x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)

my(x='x+O('x^20)); Vec((1+x)/(1-83*x+x^2)) \\ G. C. Greubel, Jan 13 2019

((1+x)/(1-83*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019

a(n) = S(n, 83) + S(n-1, 83) = S(2*n, sqrt(85)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 83) = A097839(n).
a(n) = (-2/9)*i*((-1)^n)*T(2*n+1, 9*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1 - 83*x + x^2).
a(n) = 83*a(n-1) - a(n-2) for n > 1; a(0)=1, a(1)=84. - Philippe Deléham, Nov 18 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/9)*(9/2 o 9/2 o ... o 9/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 84, 0, 6971, 0, 578509, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 83*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -81, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = 1/2*( (-1)^n - 1 )*F(n,9) + 1/9*( 1 + (-1)^(n+1) )*F(n+1,9), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 18*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 18*A099371(n)*x^n.
Exp( Sum_{n >= 1} (-18)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 18*A099371(n)*(-x)^n. (End)

A097834 Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.

Original entry on oeis.org

1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781

Offset: 0

Wolfdieter Lang, Sep 10 2004

(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
From Klaus Purath, Sep 24 2024: (Start)
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)), where (t) is any sequence satisfying t(i) = 28t(i-1) - 28t(i-2) + t(i-3) or t(i) = 27t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0.
a(n) = (Sum_{i .. i+2n} t(i))/t(i+n), where (t) is any recurrence of the form (27,-1) without regard to initial values and including this sequence itself, as long as t(i+n) != 0 for integer i and n >= 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/28, where (t) is any third order recurrence with constant coefficients (28,-28,1) and initial values t(0) = x, t(1) = x + 1, t(2) = x + 29 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/26, where (t) is any third order recurrence with constant coefficients (26,26,-1) and initial values t(0) = x, t(1) = 1 - x, t(2) = x + 27 for any integer x.
a(n) =  (t(i+4n+2) - t(i))/(t(i+2n+2) - t(i+2n)), where (t) is any recurrence of the form (5,1) without regard to initial values, as long as t(i+2n+2) - t(i+2n) != 0 for nonnegative integer i and n. (End)

			All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...

Indranil Ghosh, Table of n, a(n) for n = 0..697 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (27,-1).
Index entries for sequences related to Chebyshev polynomials.

A087130(2*n + 1) = 5 * a(n). - Michael Somos, Nov 01 2008

Cf. A097781, A097783, A097837, A097840.

a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)

{a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */

a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)=A097781(n).
a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-27*x+x^2).
a(n) = - a(-1-n) for all n in Z. - Michael Somos, Nov 01 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*x^n.
Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*(-x)^n.
(End)
a(n) = (a(n-1)*a(n-2) - 783)/a(n-3) for n >= 3. - Klaus Purath, Sep 24 2024

cp's OEIS Frontend

A123018 Triangle read by rows: row n gives the coefficients of x^k (0 <= k <= n) in the expansion of Sum_{j=0..n} A320508(n,j)*x^j*(1 - x)^(n - j).

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A162515 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = (U^n - L^n)/d, where U = (x + d)/2, L = (x - d)/2, d = sqrt(x^2 + 4).

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A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

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A374439 Triangle read by rows: the coefficients of the Lucas-Fibonacci polynomials. T(n, k) = T(n - 1, k) + T(n - 2, k - 2) with initial values T(n, k) = k + 1 for k < 2.

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A180148 a(n) = 3*a(n-1) + a(n-2) with a(0)=2 and a(1)=5.

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A073028 a(n) = max{ C(n,0), C(n-1,1), C(n-2,2), ..., C(n-n,n) }.

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A123018 Triangle read by rows: row n gives the coefficients of x^k (0 <= k <= n) in the expansion of Sum_{j=0..n} A320508(n,j)x^j(1 - x)^(n - j).