A099453 -id:A099453 - cp's OEIS frontend

A099454 A Chebyshev transform of A099453 associated to the knot 8_12.

1, 7, 37, 175, 792, 3521, 15539, 68369, 300431, 1319472, 5793745, 25437727, 111681277, 490315231, 2152620360, 9450575729, 41490490763, 182153978153, 799702876895, 3510901281888, 15413758929889, 67670362004791, 297090274041301, 1304302623454159, 5726223576745848

Offset: 0

Paul Barry, Oct 16 2004

The denominator is a parameterization of the Alexander polynomial for the knot 8_12. The g.f. is the image of the g.f. of A099453 under the Chebyshev transform A(x)->(1/(1+x^2))A(x/(1+x^2)).

Stefano Spezia, Table of n, a(n) for n = 0..1500
Dror Bar-Natan, The Rolfsen Knot Table.
Index entries for linear recurrences with constant coefficients, signature (7,-13,7,-1).

Cf. A099453, A099455.

CoefficientList[Series[(1+x^2)/(1-7x+13x^2-7x^3+x^4),{x,0,24}],x] (* Stefano Spezia, May 13 2024 *)

G.f.: (1+x^2)/(1-7x+13x^2-7x^3+x^4).
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*Sum{j=0..n-2*k} C(n-2*k-j, j)*(-11)^j*7^(n-2*k-2*j).
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*A099453(n-2*k).
a(n) = Sum_{k=0..n} binomial((n+k)/2, k)*(-1)^((n-k)/2)*(1+(-1)^(n+k))*A099453(k)/2.
a(n) = Sum_{k=0..n} A099455(n-k)*binomial(1, k/2)*(1+(-1)^k)/2.

A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)

Original entry on oeis.org

1, 0, 2, 1, 6, 7, 22, 36, 89, 168, 377, 756, 1630, 3353, 7110, 14783, 31130, 65016, 136513, 285648, 599041, 1254456, 2629418, 5508097, 11542854, 24183271, 50674318, 106173180, 222470009, 466131960, 976694489, 2046447180, 4287928678, 8984443769, 18825088134

Offset: 1

Clark Kimberling, Jun 26 2011

Polynomial reduction: an introduction
...
We begin with an example.  Suppose that p(x) is a polynomial, so that p(x)=(x^2)t(x)+r(x) for some polynomials t(x) and r(x), where r(x) has degree 0 or 1.  Replace x^2 by x+1 to get (x+1)t(x)+r(x), which is (x^2)u(x)+v(x) for some u(x) and v(x), where v(x) has degree 0 or 1.  Continuing in this manner results in a fixed polynomial w(x) of degree 0 or 1.  If p(x)=x^n, then w(x)=x*F(n)+F(n-1), where F=A000045, the sequence of Fibonacci numbers.
In order to generalize, write d(g) for the degree of an arbitrary polynomial g(x), and suppose that p, q, s are polynomials satisfying d(s)s in this manner until reaching w such that d(w)s.
The coefficients of (reduction of p by q->s) comprise a vector of length d(q)-1, so that a sequence p(n,x) of polynomials begets a sequence of vectors, such as (F(n), F(n-1)) in the above example.  We are interested in the component sequences (e.g., F(n-1) and F(n)) for various choices of p(n,x).
Following are examples of reduction by x^2->x+1:
n-th Fibonacci p(x) -> A192232+x*A112576
n-th cyclotomic p(x) -> A192233+x*A051258
n-th 1st-kind Chebyshev p(x) -> A192234+x*A071101
n-th 2nd-kind Chebyshev p(x) -> A192235+x*A192236
x(x+1)(x+2)...(x+n-1) -> A192238+x*A192239
(x+1)^n -> A001519+x*A001906
(x^2+x+1)^n -> A154626+x*A087635
(x+2)^n -> A020876+x*A030191
(x+3)^n -> A192240+x*A099453
...
Suppose that b=(b(0), b(1),...) is a sequence, and let p(n,x)=b(0)+b(1)x+b(2)x^2+...+b(n)x^n.  We define (reduction of sequence b by q->s) to be the vector given by (reduction of p(n,x) by q->s), with components in the order of powers, from 0 up to d(q)-1.  For k=0,1,...,d(q)-1, we then have the "k-sequence of (reduction of sequence b by q->s)".  Continuing the example, if b is the sequence given by b(k)=1 if k=n and b(k)=0 otherwise, then the 0-sequence of (reduction of b by x^2->x+1) is (F(n-1)), and the 1-sequence is (F(n)).
...
For selected sequences b, here are the 0-sequences and 1-sequences of (reduction of b by x^2->x+1):
b=A000045, Fibonacci sequence (1,1,2,3,5,8,...) yields
   0-sequence A166536 and 1-sequence A064831.
b=(1,A000045)=(1,1,1,2,3,5,8,...) yields
   0-sequence A166516 and 1-sequence A001654.
b=A000027, natural number sequence (1,2,3,4,...) yields
  0-sequence A190062 and 1-sequence A122491.
b=A000032, Lucas sequence (1,3,4,7,11,...) yields
  0-sequence A192243 and 1-sequence A192068.
b=A000217, triangular sequence (1,3,6,10,...) yields
  0-sequence A192244 and 1-sequence A192245.
b=A000290, squares sequence (1,4,9,16,...) yields
  0-sequence A192254 and 1-sequence A192255.
More examples:  A192245-A192257.
...
More comments:
(1) If s(n,x)=(reduction of x^n by q->s) and
    p(x)=p(0)x^n+p(1)x^(n-1)+...+p(n)x^0, then
    (reduction of p by q->s)=p(0)s(n,x)+p(1)s(n-1,x)
    +...+p(n-1)s(1,x)+p(n)s(0,x).  See A192744.
(2) For any polynomial p(x), let P(x)=(reduction of p(x)
    by q->s).  Then P(r)=p(r) for each zero r of
    q(x)-s(x).  In particular, if q(x)=x^2 and s(x)=x+1,
    then P(r)=p(r) if r=(1+sqrt(5))/2 (golden ratio) or
    r=(1-sqrt(5))/2.

			The first four Fibonacci polynomials and their reductions by x^2->x+1 are shown here:
F1(x)=1 -> 1 + 0x
F2(x)=x -> 0 + 1x
F3(x)=x^2+1 -> 2+1x
F4(x)=x^3+2x -> 1+4x
F5(x)=x^4+3x^2+1 -> (x+1)^2+3(x+1)+1 -> 6+6x.
From these, read A192232=(1,0,1,1,6,...) and A112576=(0,1,1,4,6,...).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-1).

Cf. A168561, A192233 - A192240, A192744.

Cf. A206296, A265398, A265399, A265752, A265753.

q[x_] := x + 1;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),  x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[FixedPoint[Expand[#1 /. reductionRules] &, Fibonacci[n, x]], {n, 1, 40}];
Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]
  (* A192232 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
(* A112576 *)
(* Peter J. C. Moses, Jun 25 2011 *)
LinearRecurrence[{1, 3, -1, -1}, {1, 0, 2, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

Vec((1-x-x^2)/(1-x-3*x^2+x^3+x^4)+O(x^99)) \\ Charles R Greathouse IV, Jan 08 2013

Empirical G.f.: -x*(x^2+x-1)/(x^4+x^3-3*x^2-x+1). - Colin Barker, Sep 11 2012
The above formula is correct. - Charles R Greathouse IV, Jan 08 2013
a(n) = A265752(A206296(n)). - Antti Karttunen, Dec 15 2015
a(n) = A112576(n) -A112576(n-1) -A112576(n-2). - R. J. Mathar, Dec 16 2015

Example corrected by Clark Kimberling, Dec 18 2017

A094440 Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 8, 6, 4, 5, 15, 20, 10, 5, 8, 30, 45, 40, 15, 6, 13, 56, 105, 105, 70, 21, 7, 21, 104, 224, 280, 210, 112, 28, 8, 34, 189, 468, 672, 630, 378, 168, 36, 9, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 89, 605, 1870, 3465, 4290, 3696, 2310, 990, 330, 55, 11

Offset: 1

Clark Kimberling, May 03 2004

Row sums yield the even-subscripted Fibonacci numbers (A001906).

Row n shows the coefficients of the numerator of the n-th derivative of c(n)/(x^2+x-1), where c(n) = ((-1)^(n + 1))/n!; see the Mathematica program. - Clark Kimberling, Oct 22 2019

			Triangle starts:
   1;
   1,  2;
   2,  3,   3;
   3,  8,   6,   4;
   5, 15,  20,  10,  5;
   8, 30,  45,  40, 15,  6;
  13, 56, 105, 105, 70, 21, 7;
  ...
T(4,3) = F(2)*C(4,2) = 1*6 = 6.

G. C. Greubel, Rows n = 1..100 of triangle, flattened
M. Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

Cf. A000032, A000045, A001906, A030191, A081574, A094435, A132148.
Cf. A094435, A094436, A094437, A094438, A094439, A094441, A094442, A094443, A094444.
Cf. A367208, A367209, A367210, A367211.

Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k-1)* Fibonacci(n-k+1) ))); # G. C. Greubel, Oct 30 2019

/* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Aug 15 2017

with(combinat): T:=(n,k)->binomial(n,k-1)*fibonacci(n+1-k): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # Emeric Deutsch

Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* Harvey P. Dale, Sep 14 2016 *)
(* Next program outputs polynomials having coefficients T(n,k) *)
g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[1/(1 - x - x^2), {x, n}]]]
Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* Clark Kimberling, Oct 22 2019 *)

T(n,k) = binomial(n,k-1)*fibonacci(n-k+1);
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 30 2019

[[binomial(n,k-1)*fibonacci(n-k+1) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Oct 30 2019

From Peter Bala, Aug 17 2007: (Start)
With an offset of 0, the row polynomials F(n,x) = Sum_{k = 0..n} C(n,k)* Fibonacci(n-k)*x^k satisfy F(n,x)*L(n,x) = F(2*n,x), where L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k.
Other identities and formulas include:
F(n+1,x)^2 - F(n,x)*F(n+2,x) = (x^2 + x - 1)^n;
Sum_{k = 0..n} C(n,k)*F(n-k,x)*L(k,x) = (2^n)*F(n,x);
F(n,2*x) = Sum_{k = 0..n} C(n,k)*F(n-k,x)*x^k;
F(n,3*x) = Sum_{k = 0..n} C(n,k)*F(n-k,2*x)*x^k, etc.
The sequence {F(n,r)}n>=1 gives the r-th binomial transform of the Fibonacci numbers: r = 1 gives A001906, r = 2 gives A030191, r = 3 gives A099453, r = 4 gives A081574, r = 5 gives A081575.
F(n,1/phi) = (-1)^(n-1)*F(n,-phi) = sqrt(5)^(n-1) for n >= 1, where phi = (1 + sqrt(5))/2.
The polynomials F(n,-x) satisfy a Riemann hypothesis: the zeros of F(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
G.f.: t/(1 - (2*x + 1)*t + (x^2 + x - 1)*t^2) = t + (1 + 2*x)*t^2 + (2 + 3*x + 3*x^2)*t^3 + (3 + 8*x + 6*x^2 + 4*x^3)*t^4 + ... . (End)
From Peter Bala, Jun 29 2016: (Start)
Working with an offset of 0, the n-th row polynomial F(n,x) = 1/sqrt(5)*( (x + phi)^n - (x - 1/phi)^n ), where phi = (1 + sqrt(5))/2.
d/dx(F(n,x)) = n*F(n-1,x).
F(-n,x) = -F(n,x)/(x^2 + x - 1)^n.
F(n,x - 1) = (-1)^(n-1)*F(n,-x).
F(n,x) is a divisibility sequence of polynomials, that is, if n divides m then F(n,x) divides F(m,x) in the polynomial ring Z[x]. (End)
From G. C. Greubel, Oct 30 2019: (Start)
Sum_{k = 1..n} T(n,k) = Fibonacci(2*n).
Sum_{k = 1..n} (-1)^k * T(n,k) = (-1)^n * Fibonacci(n). (End)
From Clark Kimberling, Oct 30 2019: (Start)
F(n,x) is a strong divisibility sequence of polynomials in Z[x]; that is,
gcd(F(x,h),F(x,k)) = F(x,gcd(h,k)) for h,k >= 1.  Thus, if x is an integer, then F(n,x) is a strong divisibility sequence of integers; e.g., for x=3, we have A099453. (End)
Let p(n) denote the polynomial F(x,n). Then p(n) = k(b^n - c^n), where k = -1/sqrt(5), b = (1/2)(2x + 1 - sqrt(5)), c = (1/2)(2x + 1 + sqrt(5)), and for n >=3, p(n) = u*p(n - 1) + v*p(n - 2), where u = 1 + 2 x, v = 1 - x - x^2. - Clark  Kimberling, Nov 11 2023

Error in expansion of generating function corrected by Peter Bala, Sep 24 2008

A081568 Third binomial transform of Fibonacci(n+1).

Original entry on oeis.org

1, 4, 17, 75, 338, 1541, 7069, 32532, 149965, 691903, 3193706, 14745009, 68084297, 314394980, 1451837593, 6704518371, 30961415074, 142980203437, 660285858245, 3049218769908, 14081386948661, 65028302171639, 300302858766202, 1386808687475385, 6404329365899473

Offset: 0

Paul Barry, Mar 22 2003

Binomial transform of A081567.
Case k=3 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2) for n >= 2, with a(0) = 1 and a(1) = k + 1.
a(n) = 4^n*a(n;1/4) = Sum_{k=0..n} binomial(n,k) * (-1)^k * F(k-1) * 4^(n-k), which also implies Deléham's formula given below and where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also Witula's et al. papers). - Roman Witula, Jul 12 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (7,-11).

Cf. A000045, A161731 (INVERT transform), A007582 (INVERTi transform), A028387, A081567, A081569 (binomial transform), A094441, A099453.

a:=[1,4];; for n in [3..30] do a[n]:=7*a[n-1]-11*a[n-2]; od; a; # G. C. Greubel, Aug 12 2019

I:=[1, 4]; [n le 2 select I[n] else 7*Self(n-1)-11*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 09 2013

seq(coeff(series((1-3*x)/(1-7*x+11*x^2), x, n+1), x, n), n = 0 .. 30); # G. C. Greubel, Aug 12 2019

CoefficientList[Series[(1-3x)/(1 -7x +11x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 09 2013 *)
LinearRecurrence[{7,-11},{1,4},30] (* Harvey P. Dale, Feb 01 2015 *)

Vec((1-3*x)/(1-7*x+11*x^2) + O(x^30)) \\ Altug Alkan, Dec 10 2015

def A081568_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-3*x)/(1-7*x+11*x^2)).list()
A081568_list(30) # G. C. Greubel, Aug 12 2019

a(n) = 7*a(n-1) - 11*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 4.
a(n) = (1/2 - sqrt(5)/10)*(7/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2)*(sqrt(5)/2 + 7/2)^n = A099453(n) - 3*A099453(n-1).
G.f.: (1 - 3*x)/(1 - 7*x + 11*x^2).
a(n) = Sum_{k=0..n} A094441(n,k)*3^k. - Philippe Deléham, Dec 14 2009
G.f.: Q(0,u)/x - 1/x, where u = x/(1 - 3*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k + 1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
E.g.f.: exp(7*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A081574 Fourth binomial transform of Fibonacci numbers F(n).

Original entry on oeis.org

0, 1, 9, 62, 387, 2305, 13392, 76733, 436149, 2467414, 13919895, 78398189, 441105696, 2480385673, 13942462833, 78354837710, 440286745563, 2473838793577, 13899100976496, 78088971710501, 438717826841085

Offset: 0

Paul Barry, Mar 22 2003

Binomial transform of A099453(n-1):= [0,1,7,38,189,905,...].

Case k=4 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=0, a(1)=1.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
S. Falcon, Iterated Binomial Transforms of the k-Fibonacci Sequence, British Journal of Mathematics & Computer Science, 4 (22): 2014.
J. Pan, Multiple Binomial Transforms and Families of Integer Sequences , J. Int. Seq. 13 (2010), 10.4.2, F^(4).
Index entries for linear recurrences with constant coefficients, signature (9,-19).

Cf. A000045, A081569.

a:=[0,1];; for n in [3..30] do a[n]:=9*a[n-1]-19*a[n-2]; od; a; # G. C. Greubel, Aug 13 2019

[n le 2 select (n-1) else 9*Self(n-1)-19*Self(n-2): n in [1..25]]; // Vincenzo Librandi, Aug 09 2013

seq(coeff(series(x/(1-9*x+19*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Aug 13 2019

Join[{a=0,b=1},Table[c=9*b-19*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2011 *)
LinearRecurrence[{9,-19},{0,1},30] (* Harvey P. Dale, Dec 03 2011 *)
CoefficientList[Series[x/(1 -9x +19x^2), {x, 0,  30}], x] (* Vincenzo Librandi, Aug 09 2013 *)

my(x='x+O('x^30)); Vec(x/(1 - 9*x + 19*x^2)) \\ G. C. Greubel, Aug 13 2019

[lucas_number1(n,9,19) for n in range(0, 21)] # Zerinvary Lajos, Apr 23 2009

a(n) = 9*a(n-1) - 19*a(n-1), a(0)=0, a(1)=1.
a(n) = ((sqrt(5)/2 + 9/2)^n - (9/2 - sqrt(5)/2)^n)/sqrt(5).
G.f.: x/(1 - 9*x + 19*x^2).
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Aug 11 2017

Corrected by Philippe Deléham, Dec 16 2009

A099455 An Alexander sequence for the knot 8_12.

Original entry on oeis.org

1, 7, 36, 168, 755, 3346, 14747, 64848, 284892, 1251103, 5493314, 24118255, 105887532, 464877504, 2040939083, 8960260498, 39337870403, 172703402424, 758212386132, 3328747303735, 14614056052994, 64159460722903, 281676515111412, 1236632261449368, 5429133302704547

Offset: 0

Paul Barry, Oct 16 2004

The denominator is a parameterization of the Alexander polynomial for the knot 8_12. 1/(1-7*x+13*x^2-7*x^3+x^4) is the image of the g.f. of A099453 under the modified Chebyshev transform A(x)->(1/(1+x^2)^2)A(x/(1+x^2)).

Stefano Spezia, Table of n, a(n) for n = 0..1500
Dror Bar-Natan, The Rolfsen Knot Table.
Index entries for linear recurrences with constant coefficients, signature (7,-13,7,-1).

Cf. A099454.

LinearRecurrence[{7,-13,7,-1},{1,7,36,168,755},30] (* Harvey P. Dale, Jan 31 2017 *)

G.f.: (1-x)*(1+x)*(1+x^2)/(1-7*x+13*x^2-7*x^3+x^4). - corrected Nov 24 2012

a(n) = A099454(n) - A099454(n-2).

A192240 Constant term in the reduction of the polynomial (x+3)^n by x^2 -> x+1.

Original entry on oeis.org

1, 3, 10, 37, 149, 636, 2813, 12695, 57922, 265809, 1223521, 5640748, 26026505, 120137307, 554669594, 2561176781, 11826871933, 54615158940, 252210521317, 1164706900879, 5378632571666, 24838652091993, 114705606355625, 529714071477452

Offset: 0

Clark Kimberling, Jun 26 2011

nonn

See A192232.

Robert Israel, Table of n, a(n) for n = 0..1498

Cf. A192232.

seq(eval(rem((x+3)^n,x^2-x-1,x),x=0),n=0..50); # Robert Israel, Mar 14 2023

q[x_] := x + 1;
p[n_, x_] := (x + 3)^n;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),
   x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[
   Last[Most[
     FixedPointList[Expand[#1 /. reductionRules] &, p[n, x]]]], {n, 0,
     30}];
Table[Coefficient[Part[t, n], x, 0], {n, 30}]  (* A192240 *)
Table[Coefficient[Part[t, n], x, 1], {n, 30}]  (* A099453 *)
(* Peter J. C. Moses, Jun 26 2011 *)

Empirical g.f. and recurrence: (1-4*x)/(1-7*x+11*x^2). a(n) = 7*a(n-1) - 11*a(n-2). - Colin Barker, Feb 09 2012
Proof of recurrence: if r(n) == (x+3)^n mod (x^2-x-1), then r(n+j) == (x+1)^(n+j) mod (x^2 - x - 1). Now r(n+2) - 7*r(n+1) + 11*r(n) == ((x+3)^2 - 7*(x+3) + 11)*r(n) == 0 mod (x^2-x-1) since (x+3)^2 - 7*(x+3) + 11 = x^2 - x - 1. - Robert Israel, Mar 14 2023
a(n) = Sum_{i=0..n} (-1)^i*Fibonacci(i+1)*binomial(n,i)*4^(n-i) (conjecture). - Rigoberto Florez, Mar 25 2020

Offset corrected by Robert Israel, Mar 14 2023

A081576 Square array of binomial transforms of Fibonacci numbers, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 5, 8, 3, 0, 1, 7, 20, 21, 5, 0, 1, 9, 38, 75, 55, 8, 0, 1, 11, 62, 189, 275, 144, 13, 0, 1, 13, 92, 387, 905, 1000, 377, 21, 0, 1, 15, 128, 693, 2305, 4256, 3625, 987, 34, 0, 1, 17, 170, 1131, 4955, 13392, 19837, 13125, 2584, 55

Offset: 0

Paul Barry, Mar 22 2003

Array rows are solutions of the recurrence a(n) = (2*k+1)*a(n-1) - A028387(k-1)*a(n-2) where a(0) = 0 and a(1) = 1.

			Square array begins as:
  0, 1,  1,   2,    3,    5,     8, ... A000045;
  0, 1,  3,   8,   21,   55,   144, ... A001906;
  0, 1,  5,  20,   75,  275,  1000, ... A030191;
  0, 1,  7,  38,  189,  905,  4256, ... A099453;
  0, 1,  9,  62,  387, 2305, 13392, ... A081574;
  0, 1, 11,  92,  693, 4955, 34408, ... A081575;
  0, 1, 13, 128, 1131, 9455, 76544, ...
The antidiagonal triangle begins as:
  0;
  0, 1;
  0, 1,  1;
  0, 1,  3,  2;
  0, 1,  5,  8,   3;
  0, 1,  7, 20,  21,   5;
  0, 1,  9, 38,  75,  55,   8;
  0, 1, 11, 62, 189, 275, 144, 13;

G. C. Greubel, Antidiagonal rows n = 0..50, flattened

Array row n: A000045 (n=0), A001906 (n=1), A030191 (n=2), A099453 (n=3), A081574 (n=4), A081575 (n=5).
Array columns k: A005408 (k=3), A077588 (k=4).
Cf. A028387, A081573, A081574, A081575.

A081576:= func< n,k | (&+[Binomial(k,j)*Fibonacci(j)*(n-k)^(k-j): j in [0..k]]) >;
[A081576(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021

T[n_, k_]:= If[n==0, Fibonacci[k], Sum[Binomial[k, j]*Fibonacci[j]*n^(k-j), {j, 0, k}]]; Table[T[n-k, k], {n,0,12}, {k,0,n}] //Flatten (* G. C. Greubel, May 26 2021 *)

def A081576(n,k): return sum( binomial(k,j)*fibonacci(j)*(n-k)^(k-j) for j in (0..k) )
flatten([[A081576(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021

Rows are successive binomial transforms of F(n).
T(n, k) = ( ( (2*n + 1 + sqrt(5))/2 )^k - ( (2*n + 1 - sqrt(5))/2 )^k )/sqrt(5).
From G. C. Greubel, May 26 2021: (Start)
T(n, k) = Sum_{j=0..k} binomial(k,j)*Fibonacci(j)*n^(k-j) with T(0, k) = Fibonacci(k) (square array).
T(n, k) = Sum_{j=0..k} binomial(k,j)*Fibonacci(j)*(n-k)^(k-j) (antidiagonal triangle). (End)

A289783 p-INVERT of the (3^n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 5, 24, 113, 527, 2446, 11325, 52369, 242008, 1117997, 5163891, 23849270, 110142089, 508652653, 2349005592, 10847859961, 50095958215, 231345247934, 1068361195173, 4933730638937, 22784141325656, 105217952251285, 485900111176779, 2243903303473318

Offset: 0

Clark Kimberling, Aug 10 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-11).

Cf. A000244, A289780.

z = 60; s = x/(1 - 3*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000244 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289783 *)

Vec(x*(1 - 2*x) / (1 - 7*x + 11*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

G.f.: (1 - 2 x)/(1 - 7 x + 11 x^2).
a(n) = 7*a(n-1) - 11*a(n-2).
a(n) = (2^(-n-1)*((7-sqrt(5))^(n+1)*(-4+sqrt(5)) + (4+sqrt(5))*(7+sqrt(5))^(n+1))) / (11*sqrt(5)). - Colin Barker, Aug 11 2017
a(n) = A099453(n) - 2*A099453(n-1). - R. J. Mathar, Jul 08 2022
E.g.f.: exp(7*x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Aug 05 2025

A171731 Triangle T : T(n,k)= binomial(n,k)Fibonacci(n-k)= A007318(n,k)A000045(n-k).

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 2, 3, 3, 0, 3, 8, 6, 4, 0, 5, 15, 20, 10, 5, 0, 8, 30, 45, 40, 15, 6, 0, 13, 56, 105, 105, 70, 21, 7, 0, 21, 104, 224, 280, 210, 112, 28, 8, 0, 34, 189, 468, 672, 630, 378, 168, 36, 9, 0, 55, 340, 945, 1560, 1680, 1260, 630, 240, 45, 10, 0

Offset: 0

Philippe Deléham, Dec 16 2009

Diagonal sums : A112576.

Essentially the same as A094440. - Peter Bala, Jan 06 2015

			Triangle begins :
0 ;
1,0 ;
1,2,0 ;
2,3,3,0 ;
3,8,6,4,0 ;
5,15,20,10,5,0 ;
...

Cf. A000045, A007318, A094440.

Flatten[Table[Binomial[n,k]Fibonacci[n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jan 16 2013 *)

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000045(n), A001906(n), A093131(n), A099453(n-1), A081574(n), A081575(n) for x = 0,1,2,3,4,5 respectively. Sum_{k, 0<=k<=n} T(n,k)*2^(n-k) = A014445(n).

cp's OEIS Frontend

A099454 A Chebyshev transform of A099453 associated to the knot 8_12.

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A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)

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A094440 Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.

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A081568 Third binomial transform of Fibonacci(n+1).

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A081574 Fourth binomial transform of Fibonacci numbers F(n).

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A099455 An Alexander sequence for the knot 8_12.

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A171731 Triangle T : T(n,k)= binomial(n,k)Fibonacci(n-k)= A007318(n,k)A000045(n-k).