A162516 -id:A162516 - cp's OEIS frontend

A162517 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x + d)^n - (x - d)^n)/(2*d), where d = sqrt(x+4).

0, 1, 2, 0, 3, 1, 4, 4, 4, 16, 0, 5, 10, 41, 8, 16, 6, 20, 86, 48, 96, 0, 7, 35, 161, 169, 348, 48, 64, 8, 56, 280, 456, 992, 384, 512, 0, 9, 84, 462, 1044, 2449, 1744, 2400, 256, 256, 10, 120, 732, 2136, 5482, 5920, 8640, 2560, 2560, 0, 11, 165, 1122, 4026, 11407, 16721, 26420, 14240, 14720, 1280, 1024

Offset: 1

Clark Kimberling, Jul 05 2009

			First six rows:
  0
  1
  2...0
  3...1...4
  4...4...16...0
  5...10..41...8...16

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

Cf. A162514, A162515, A162516.

Cf. A002605, A063727.

m:=12;
Q:= func< n,x | ((x+Sqrt(x+4))^n - (x-Sqrt(x+4))^n)/(2*Sqrt(x+4)) >;
R:=PowerSeriesRing(Rationals(), m+1);
T:= func< n,k | Coefficient(R!( Q(n, x) ), n-k) >;
[0] cat [T(n,k): k in [1..n], n in [1..m]]; // G. C. Greubel, Jul 09 2023

Q[n_, x_]:= Q[n, x]= ((x+Sqrt[x+4])^n -(x-Sqrt[x+4])^n)/(2*Sqrt[x+4]);
T[n_, k_]:= Coefficient[Series[P[n,x], {x,0,n-k+1}], x, n-k];
Join[{0}, Table[T[n,k], {n,12}, {k,n}]//Flatten] (* G. C. Greubel, Jul 09 2023 *)

def Q(n,x): return ((x+sqrt(x+4))^n - (x-sqrt(x+4))^n)/(2*sqrt(x+4))
def T(n,k):
    P. = PowerSeriesRing(QQ)
    return P( Q(n,x) ).list()[n-k]
[0]+flatten([[T(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Jul 09 2023

Q(n,x) = (P(n+1, x) - x*P(n,x))/(x+4), where P(n, x) is the n-th polynomial of A162516.
Q(n, x) also has the recurrence Q(n, x) = 2*x*Q(n-1, x) - (x^2 - x - 4)*Q(n-2, x).
From G. C. Greubel, Jul 09 2023: (Start)
T(n, k) = [x^(n-k)](((x+sqrt(x+4))^n -(x-sqrt(x+4))^n)/(2*sqrt(x+4))).
Sum_{k=1..n-1} T(n, k) = A063727(n-2), n >= 2.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = A002605(n-1). (End)

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

A162514 Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.

Original entry on oeis.org

2, 1, 0, 1, 0, 2, 1, 0, 3, 0, 1, 0, 4, 0, 2, 1, 0, 5, 0, 5, 0, 1, 0, 6, 0, 9, 0, 2, 1, 0, 7, 0, 14, 0, 7, 0, 1, 0, 8, 0, 20, 0, 16, 0, 2, 1, 0, 9, 0, 27, 0, 30, 0, 9, 0, 1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2, 1, 0, 11, 0, 44, 0, 77, 0, 55, 0, 11, 0, 1, 0, 12, 0, 54, 0, 112, 0, 105, 0, 36, 0, 2, 1, 0, 13, 0

Offset: 0

Clark Kimberling, Jul 05 2009

For a signed version of this triangle corresponding to the row reversed version of the triangle A127672 see A244422. - Wolfdieter Lang, Aug 07 2014

The row reversed triangle is A114525. - Paolo Bonzini, Jun 23 2016

			Triangle begins
   2;  == 2
   1, 0;  == x + 0
   1, 0,  2;  == x^2 + 2
   1, 0,  3, 0;  == x^3 + 3*x + 0
   1, 0,  4, 0,  2;
   1, 0,  5, 0,  5, 0;
   1, 0,  6, 0,  9, 0,  2;
   1, 0,  7, 0, 14, 0,  7, 0;
   1, 0,  8, 0, 20, 0, 16, 0,  2;
   1, 0,  9, 0, 27, 0, 30, 0,  9, 0;
   1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2;
   ...
From _Wolfdieter Lang_, Aug 07 2014: (Start)
The row polynomials R(n, x) are:
  R(0, x) = 2, R(1, x) = 1 =   x*P(1,1/x),  R(2, x) = 1 + 2*x^2 = x^2*P(2,1/x), R(3, x) = 1 + 3*x^2 = x^3*P(3,1/x), ...
(End)

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

Cf. A000032, A114525, A162515, A162516, A162517.

Table[Reverse[CoefficientList[LucasL[n, x], x]], {n, 0, 12}]//Flatten  (* G. C. Greubel, Nov 05 2018 *)

P(n)=
{
    local(U, L, d, r, x);
    if ( n<0, return(0) );
    x = 'x+O('x^(n+1));
    d=(4 + x^2)^(1/2);
    U=(x+d)/2;  L=(x-d)/2;
    r = U^n+L^n;
    r = truncate(r);
    return( r );
}
for (n=0, 10, print(Vec(P(n))) ); /* show triangle */
/* Joerg Arndt, Jul 24 2011 */

P(n,x) = x*P(n-1,x) + P(n-2,x) for n >= 2, P(0,x) = 2, P(1,x) = x.
From Wolfdieter Lang, Aug 07 2014: (Start)
T(n,m) = [x^(n-m)] P(n,x), m = 0, 1, ..., n and  n >= 0.
G.f. of polynomials P(n,x): (2 - x*z)/(1 - x*z - z^2).
G.f. of row polynomials R(n,x) = Sum_{m=0..n} T(n,m)*x^m:  (2 - z)/(1 - z - (x*z)^2) (rows for P(n,x) reversed).
(End)
For n > 0, T(n,2*m+1) = 0, T(n,2*m) = A034807(n,m). - Paolo Bonzini, Jun 23 2016

Name clarified by Wolfdieter Lang, Aug 07 2014

A163762 Triangle of coefficients of polynomials H(n,x)=(U^n+L^n)/2+(U^n-L^n)/(2d), where U=x+d, L=x-d, d=(x+4)^(1/2).

Original entry on oeis.org

1, 1, 1, 1, 3, 4, 1, 6, 13, 4, 1, 10, 29, 24, 16, 1, 15, 55, 81, 88, 16, 1, 21, 95, 207, 300, 144, 64, 1, 28, 154, 448, 813, 684, 496, 64, 1, 36, 238, 868, 1913, 2352, 2272, 768, 256, 1, 45, 354, 1554, 4077, 6625, 7984, 4704, 2560, 256, 1, 55, 510, 2622, 8061, 16283

Offset: 1

Clark Kimberling, Aug 04 2009

H(n,x)=P(n,x)+Q(n,x), where P and Q are given by A162516, A162517.
H(n,0)=4^Floor(n/2) for n=0,1,2,...
H(n,1)=A063727(n); row sums
(Column 2)=A000217 (triangular numbers)

			First six rows:
1
1...1
1...3...4
1...6..13...4
1..10..29..24..16
1..15..55..81..88..16
Row 6 represents x^5+15*x^4+55*x^3+81*x^2+88*x+16.

A000217, A063727, A162516, A162517.

H(n,x)=2*x*H(n-1,x)-(x^2-x-4)*H(n-2,x), where H(0,x)=1, H(1,x)=x+1.

H(n,x)=(1+1/d)*U^n+(1-1/d)*L^n, where U=x+d, L=x-d, d=(x+4)^(1/2).

A152429 a(n) = (11^n + 5^n)/2.

Original entry on oeis.org

1, 8, 73, 728, 7633, 82088, 893593, 9782648, 107374753, 1179950408, 12973595113, 142680249368, 1569336258673, 17261966423528, 189877968549433, 2088639343496888, 22974941225731393, 252723895719373448

Offset: 0

Philippe Deléham, Dec 03 2008

Binomial transform of A081343.

Inverse binomial transform of A143079.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (16,-55).

Cf. A162516.

List([0..20], n-> (11^n+5^n)/2); # G. C. Greubel, Jan 08 2020

[(11^n+5^n)/2: n in [0..20]]; // Vincenzo Librandi, Jun 01 2011

seq( (11^n+5^n)/2, n=0..20); # G. C. Greubel, Jan 08 2020

LinearRecurrence[{16,-55}, {1,8}, 20] (* G. C. Greubel, Jan 08 2020 *)

vector(21, n, (11^(n-1) + 5^(n-1))/2 ) \\ G. C. Greubel, Jan 08 2020

[(11^n+5^n)/2 for n in (0..20)] # G. C. Greubel, Jan 08 2020

a(n) = 16*a(n-1) - 55*a(n-2), with a(0)=1, a(1)=8.
G.f.: (1-8*x)/(1 - 16*x + 55*x^2).
a(n) = Sum_{k=0..n} A098158(n,k)*8^(2k-n)*9^(n-k).
a(n) = ((8 + sqrt(9))^n + (8 - sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008
E.g.f.: (exp(11*x) + exp(5*x))/2. - G. C. Greubel, Jan 08 2020

cp's OEIS Frontend

A162517 Triangle of coefficients of polynomials defined by Binet form: P(n,x) = ((x + d)^n - (x - d)^n)/(2*d), where d = sqrt(x+4).

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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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A162514 Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.

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A163762 Triangle of coefficients of polynomials H(n,x)=(U^n+L^n)/2+(U^n-L^n)/(2d), where U=x+d, L=x-d, d=(x+4)^(1/2).

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A152429 a(n) = (11^n + 5^n)/2.

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