A192872 -id:A192872 - cp's OEIS frontend

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A192904 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) defined below at Comments.

Original entry on oeis.org

1, 0, 1, 5, 16, 49, 153, 480, 1505, 4717, 14784, 46337, 145233, 455200, 1426721, 4471733, 14015632, 43928817, 137684905, 431542080, 1352570689, 4239325789, 13287204352, 41645725825, 130529073953, 409113752000, 1282274186177

Offset: 0

Clark Kimberling, Jul 12 2011

The titular polynomial is defined by p(n,x) = (x^2)*p(n-1,x) + x*p(n-2,x), with p(0,x) = 1, p(1,x) = x.  The resulting sequence typifies a general class which we shall describe here. Suppose that u,v,a,b,c,d,e,f are numbers used to define these polynomials:
...
q(x) = x^2
s(x) = u*x + v
p(0,x) = a, p(1,x) = b*x + c
p(n,x) = d*(x^2)*p(n-1,x) + e*x*p(n-2,x) + f.
...
We shall assume that u is not 0 and that {d,e} is not {0}.  The reduction of p(n,x) by the repeated substitution q(x) -> s(x), as defined and described at A192232 and A192744, has the form h(n) + k(n)*x.  The numerical sequences h and k are linear recurrence sequences, formally of order 5.  The Mathematica program below, with first line deleted, shows initial terms and recurrence coefficients, which imply these properties:
(1)  the recurrence coefficients depend only on u,v,d,e; the parameters a,b,c,f affect only the initial terms.
(2)  if e=0 or v=0, the order of recurrence is <= 3;
(3)  if e=0 and v=0, the recurrence coefficients are 1+d*u^2 and -d*u^2 (cf. similar results at A192872).
...
Examples:
u v a b c d e f... seq h.....seq k
1 1 1 1 1 1 0 0... A001906..A001519
1 1 1 1 0 0 1 0... A103609..A193609
1 1 1 1 0 1 1 0... A192904..A192905
1 1 1 1 1 1 0 0... A001519..A001906
1 1 1 1 1 1 1 0... A192907..A192907
1 1 1 1 1 1 0 1... A192908..A069403
1 1 1 1 1 1 1 1... A192909..A192910
The terms of these sequences involve Fibonacci numbers, F(n)=A000045(n); e.g.,
A001906: even-indexed F(n)
A001519: odd-indexed F(n)
A103609: (1,1,1,1,2,2,3,3,5,5,8,8,...)

			The first six polynomials and reductions:
1 -> 1
x -> x
x + x^3 -> 1 + 3*x
x^2 + x^3 + x^5 -> 5 + 8*x
x^2 + 2*x^4 + x^5 + x^7 -> 16 + 25*x
x^3 + 2*x^4 + 3*x^6 + x^7 + x^9 -> 49 + 79*x, so that
A192904 = (1,0,1,5,16,49,...) and
A192905 = (0,1,3,8,25,79,...)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,1,1).

Cf. A192232, A192744, A192905, A192872.

a:=[1,0,1,5];; for n in [5..40] do a[n]:=3*a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 10 2019

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4) )); // G. C. Greubel, Jan 10 2019

(* To obtain general results, delete the next line. *)
u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 0;
q = x^2; s = u*x + v; z = 24;
p[0, x_] := a; p[1, x_] := b*x + c;
p[n_, x_] :=  d*(x^2)*p[n - 1, x] + e*x*p[n - 2, x] + f;
Table[Expand[p[n, x]], {n, 0, 8}]
reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u0 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192904 *)
u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192905 *)
Simplify[FindLinearRecurrence[u0]] (* recurrence for 0-sequence *)
Simplify[FindLinearRecurrence[u1]] (* recurrence for 1-sequence *)
LinearRecurrence[{3,0,1,1}, {1,0,1,5}, 40] (* G. C. Greubel, Jan 10 2019 *)

my(x='x+O('x^40)); Vec((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)) \\ G. C. Greubel, Jan 10 2019

((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 10 2019

a(n) = 3*a(n-1) + a(n-3) + a(n-4).

G.f.: (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4). - Colin Barker, Aug 31 2012

A192878 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

3, 0, 4, 5, 18, 42, 115, 296, 780, 2037, 5338, 13970, 36579, 95760, 250708, 656357, 1718370, 4498746, 11777875, 30834872, 80726748, 211345365, 553309354, 1448582690, 3792438723, 9928733472, 25993761700, 68052551621, 178163893170, 466439127882, 1221153490483

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x + 1, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.

For n>0, also the coefficient of x in the reduction x^2 -> x + 1 of the polynomial A000285(n-1)*x^(n-1). - R. J. Mathar, Jul 12 2011

			The first six polynomials and reductions:
p(0,x) = 3 -> 3
p(1,x) = x -> x
p(2,x) = 4*x^2 -> 4 + 4*x
p(3,x) = 5*x^3 -> 5 + 10*x
p(4,x) = 9*x^4 -> 18 + 27*x
p(5,x) = 14*x^5 -> 42 + 27*x
In general, p(n,x) = A104449(n)*x^n -> A192878(n) + A192879(n)*x.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000285, A192872, A192879.

a:=[3,0,4];; for n in [4..30] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 08 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (3-2*x^2-6*x)/((1+x)*(x^2-3*x+1)) )); // G. C. Greubel, Jan 08 2019

q = x^2; s = x + 1; z = 25;
p[0, x_] := 3; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] :=
FixedPoint[(s PolynomialQuotient @@ #1 +
       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]  (* A192878 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]  (* A192879 *)
FindLinearRecurrence[u1]
FindLinearRecurrence[u2]
LinearRecurrence[{2,2,-1}, {3,0,4}, 30] (* G. C. Greubel, Jan 08 2019 *)

a(n) = round((2^(-n)*(7*(-2)^n-(-4+sqrt(5))*(3+sqrt(5))^n+(3-sqrt(5))^n*(4+sqrt(5))))/5) \\ Colin Barker, Sep 29 2016

Vec((3-2*x^2-6*x)/((1+x)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Sep 29 2016

((3-2*x^2-6*x)/((1+x)*(x^2-3*x+1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 08 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: ( 3 - 2*x^2 - 6*x ) / ( (1+x)*(1 - 3*x + x^2) ). - R. J. Mathar, May 07 2014
a(n) = (2^(-n)*(7*(-2)^n - (-4+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(4+sqrt(5))))/5. - Colin Barker, Sep 29 2016

A192883 Constant term in the reduction by (x^2 -> x + 1) of the polynomial F(n+3)*x^n, where F = A000045 (Fibonacci sequence).

Original entry on oeis.org

2, 0, 5, 8, 26, 63, 170, 440, 1157, 3024, 7922, 20735, 54290, 142128, 372101, 974168, 2550410, 6677055, 17480762, 45765224, 119814917, 313679520, 821223650, 2149991423, 5628750626, 14736260448, 38580030725, 101003831720, 264431464442, 692290561599

Offset: 0

Clark Kimberling, Jul 12 2011

See A192872.

a(n) is also the area of the triangle with vertices at (F(n),F(n+1)), (F(n+1),F(n)), and (F(n+3),F(n+4)) where F(n) = A000045(n). - J. M. Bergot, May 22 2014

			G.f. = 2 + 5*x^2 + 8*x^3 + 26*x^4 + 63*x^5 + 170*x^6 + ... - _Michael Somos_, Mar 18 2022

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001622, A192232, A192744, A192872.

a:=[2,0,5];; for n in [4..30] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 09 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( ( 2-4*x+x^2)/((1+x)*(1-3*x+x^2)) )); // G. C. Greubel, Jan 09 2019

with(combinat):seq(fibonacci(n-1)*fibonacci(n+3), n=0..27): # Gary Detlefs, Oct 19 2011

q = x^2; s = x + 1; z = 28;
p[0, x_] := 2; p[1, x_] := 3 x;
p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] := FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192883 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* minus A121646 *)
LinearRecurrence[{2,2,-1}, {2,0,5}, 30] (* G. C. Greubel, Jan 09 2019 *)
a[ n_] := Fibonacci[n+1]^2 + (-1)^n; (* Michael Somos, Mar 18 2022 *)

a(n) = round((2^(-1-n)*(7*(-1)^n*2^(1+n)+(3-sqrt(5))^(1+n)+(3+sqrt(5))^(1+n)))/5) \\ Colin Barker, Sep 29 2016

Vec((2+x^2-4*x)/((1+x)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Sep 29 2016

{a(n) = fibonacci(n+1)^2 + (-1)^n}; /* Michael Somos, Mar 18 2022 */

((2-4*x+x^2 )/((1+x)*(1-3*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 09 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = Fibonacci(n-1) * Fibonacci(n+3). - Gary Detlefs, Oct 19 2011
a(n) = Fibonacci(n+1)^2 + (-1)^n. - Gary Detlefs, Oct 19 2011
G.f.: ( 2-4*x+x^2 ) / ( (1+x)*(1-3*x+x^2) ). - R. J. Mathar, May 07 2014
a(n) = (2^(-1-n)*(7*(-1)^n*2^(1+n) + (3-sqrt(5))^(1+n) + (3+sqrt(5))^(1+n)))/5. - Colin Barker, Sep 29 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=2} 1/a(n) = 7/18.
Sum_{n>=2} (-1)^n/a(n) = (4/phi - 13/6)/3, where phi is the golden ratio (A001622). (End)
a(n) = a(-2-n) for all n in Z. - Michael Somos, Mar 18 2022

A192914 Constant term in the reduction by (x^2 -> x + 1) of the polynomial C(n)*x^n, where C=A000285.

Original entry on oeis.org

1, 0, 5, 9, 28, 69, 185, 480, 1261, 3297, 8636, 22605, 59185, 154944, 405653, 1062009, 2780380, 7279125, 19057001, 49891872, 130618621, 341963985, 895273340, 2343856029, 6136294753, 16065028224, 42058789925, 110111341545, 288275234716, 754714362597

Offset: 0

Clark Kimberling, Jul 12 2011

See A192872.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A192232, A192744, A192872.

F:=Fibonacci; List([0..30], n -> F(n+1)^2 +F(n)*F(n-3)); # G. C. Greubel, Jan 12 2019

F:=Fibonacci; [F(n+1)^2+F(n)*F(n-3): n in [0..30]]; // Bruno Berselli, Feb 15 2017

q = x^2; s = x + 1; z = 28;
p[0, x_]:= 1; p[1, x_]:= 4 x;
p[n_, x_] := p[n-1, x]*x + p[n-2, x]*x^2;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192914 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* see A192878 *)
LinearRecurrence[{2,2,-1}, {1,0,5}, 30] (* or *) With[{F:= Fibonacci}, Table[F[n+1]^2 +F[n]*F[n-3], {n, 0, 30}]] (* G. C. Greubel, Jan 12 2019 *)

a(n) = round((2^(-1-n)*(3*(-1)^n*2^(2+n)+(3+sqrt(5))^n*(-1+3*sqrt(5))-(3-sqrt(5))^n*(1+3*sqrt(5))))/5) \\ Colin Barker, Sep 29 2016

Vec((1+3*x^2-2*x)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Sep 29 2016

{f=fibonacci}; vector(30, n, n--; f(n+1)^2 +f(n)*f(n-3)) \\ G. C. Greubel, Jan 12 2019

f=fibonacci; [f(n+1)^2 +f(n)*f(n-3) for n in (0..30)] # G. C. Greubel, Jan 12 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1 + 3*x^2 - 2*x)/((1 + x)*(x^2 - 3*x + 1)). - R. J. Mathar, May 08 2014
a(n) = (2^(-1-n)*(3*(-1)^n*2^(2+n) + (3 + sqrt(5))^n*(-1 + 3*sqrt(5)) - (3-sqrt(5))^n*(1 + 3*sqrt(5))))/5. - Colin Barker, Sep 29 2016
a(n) = F(n+1)^2 + F(n)*F(n-3). - Bruno Berselli, Feb 15 2017

A066182 Permutation of the integers with cycle form {1}, {3, 2}, {6, 5, 4}, {10, 9, 8, 7}, ...

Original entry on oeis.org

1, 3, 2, 6, 4, 5, 10, 7, 8, 9, 15, 11, 12, 13, 14, 21, 16, 17, 18, 19, 20, 28, 22, 23, 24, 25, 26, 27, 36, 29, 30, 31, 32, 33, 34, 35, 45, 37, 38, 39, 40, 41, 42, 43, 44, 55, 46, 47, 48, 49, 50, 51, 52, 53, 54, 66, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 78, 67, 68, 69, 70, 71

Offset: 1

Wouter Meeussen, Dec 15 2001

Arrange natural numbers 1,2,3,4,5,... as a triangle like A000027, then rotate each row of triangle one step right. - Antti Karttunen, May 07 2002

As a rectangular array, a(n) is the natural interspersion of the sequence of triangular numbers; see A192872. [Clark Kimberling, Aug 12 2011]

			Northwest corner, when sequence is formatted as the natural interspersion of the sequence (1,3,6,10,15,...) of triangular numbers:
  1...3...6...10...15
  2...4...7...11...16
  5...8...12..17...23
  9...13..18..24...31     [ _Clark Kimberling_, Aug 12 2011 ]

Inverse permutation: A066181.

Cf. A000027, A192872.

FromCycles[Table[n(n-1)/2+Range[n, 1, -1], {n, 13}]]

from math import isqrt, comb
def A066182(n): return -1+n+comb(isqrt(n<<3)+3>>1,2)-comb(isqrt(n-1<<3)+3>>1,2) # Chai Wah Wu, Jun 09 2025

a(n) = -1+n+binomial(A002024(n)+1,2)-binomial(A002024(n-1)+1,2) where A002024(n) is round(sqrt(2*n)). - Brian Tenneson, Feb 03 2017

A192879 Coefficient of x in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

0, 1, 4, 10, 27, 70, 184, 481, 1260, 3298, 8635, 22606, 59184, 154945, 405652, 1062010, 2780379, 7279126, 19057000, 49891873, 130618620, 341963986, 895273339, 2343856030, 6136294752, 16065028225, 42058789924, 110111341546, 288275234715, 754714362598

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x+1, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.

A192879 is also generated as the coefficient sequence of x in the reduction x^2->x+1 of the polynomial v(n,x) defined by v(0,x) = 2, v(1,x) = x+1, and v(n,x) = x*v(n-1,x) + 2*(x^2)*v(n-1,x) + 1, for n>0, v(n,x) = F(n)*x^(n-1) + L(n)*x^n, where F(n) = A000045(n) (Fibonacci numbers) and L(n) = A000032(n) (Lucas numbers).

			The first six polynomials and reductions:
  p(0,x) = 3 -> 3
  p(1,x) = x -> x
  p(2,x) = 4*x^2 -> 4+4*x
  p(3,x) = 5*x^3 -> 5+10*x
  p(4,x) = 9*x^4 -> 18+27*x
  p(5,x) = 14*x^5 -> 42+27*x
In general, p(n,x) = (A104449(n))*x^n -> A192878(n) + A192879(n)*x.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A192872, A192878.

a:=[0,1,4];; for n in [4..40] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Jan 07 2019

I:=[0,1,4]; [n le 3 select I[n] else 2*Self(n-1) +2*Self(n-2) -Self(n-3): n in [1..40]]; // G. C. Greubel, Jan 07 2019

with(combinat); seq( fibonacci(2*n) + fibonacci(n)*fibonacci(n-1), n=0..40); # G. C. Greubel, Feb 13 2020

(See A192878.)
LinearRecurrence[{2,2,-1}, {0,1,4}, 30] (* G. C. Greubel, Jan 07 2019 *)
a[n_] := a[n] = 2*a[n-1]+2*a[n - 2]-a[n-3]; a[0] = 0; a[1]=1; a[2]=4; Table[a[n], {n,0,40}] (* Rigoberto Florez, Feb 06 2020 *)
Table[Fibonacci[n]*Fibonacci[n-1]+Fibonacci[2n], {n,0,40}] (* Rigoberto Florez, Feb 06 2020 *)

a(n) = round((2^(-1-n)*((-1)^n*2^(1+n)+(3+sqrt(5))^n*(-1+3*sqrt(5))-(3-sqrt(5))^n*(1+3*sqrt(5))))/5) \\ Colin Barker, Sep 29 2016

concat(0, Vec(x*(1+2*x)/((1+x)*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Sep 29 2016

(x*(1+2*x)/((1+x)*(1-3*x+x^2))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 07 2019

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), with a(0) = 0, a(1) = 1, a(2) = 4.
G.f.: x * (1+2*x) / ((1+x) * (1-3*x+x^2)). - Colin Barker, Jun 18 2012
a(n) = (2^(-1-n) * ((-1)^n*2^(1+n) + (3+sqrt(5))^n * (-1+3*sqrt(5)) - (3-sqrt(5))^n * (1+3*sqrt(5))))/5. - Colin Barker, Sep 29 2016
a(n) = F(n-1)*F(n) + F(2n), where F(n) is a Fibonacci number. - Rigoberto Florez, Feb 06 2020
E.g.f.: (exp(-x) + exp(3*x/2) * (3*sqrt(5)*sinh(sqrt(5)*x/2) - cosh(sqrt(5)*x/2)))/5. - Stefano Spezia, Feb 06 2020
a(n)*F(n) = the number of ways to tile a 3-arm starfish (with n-1 cells on each arm and one cell in the center) using squares and dominos. - Greg Dresden and Hasita Kanamarlapudi, Oct 02 2023

A192873 Coefficient of x in the reduction by (x^2->x+1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

0, 1, 2, 7, 18, 49, 128, 337, 882, 2311, 6050, 15841, 41472, 108577, 284258, 744199, 1948338, 5100817, 13354112, 34961521, 91530450, 239629831, 627359042, 1642447297, 4299982848, 11257501249, 29472520898, 77160061447, 202007663442, 528862928881, 1384581123200

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x, and p(n,x) = x*p(n-1,x) + (x^2)*p(n-1,x) + 1. See A192872.

First differences give A236428. - Richard R. Forberg, Feb 23 2014

			The coefficients of all the polynomials p(n,x) are Fibonacci numbers (A000045).  The first 6 and their reductions:
p(0,x) = 1 -> 1
p(1,x) = x -> x
p(2,x) = 1 +2*x^2 -> 3 +2*x
p(3,x) = 1 +x +3*x^3 -> 4 +7*x
p(4,x) = 1 +x +2*x^2 +5*x^4 -> 13 +18*x
p(5,x) = 1 +x +2*x^2 +3*x^3 +8*x^5 -> 30 +49*x
G.f. = x + 2*x^2 + 7*x^3 + 18*x^4 + 49*x^5 + 128*x^6 + 337*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A192872, A192232, A192744.

a:=[0,1,2,7];; for n in [5..40] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 07 2019

I:=[0,1,2,7]; [n le 4 select I[n] else 3*Self(n-1) - 3*Self(n-3) +Self(n-4): n in [1..40]]; // G. C. Greubel, Jan 07 2019

seq(coeff(series(x*(x^2-x+1)/((1-x)*(1+x)*(x^2-3*x+1)),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Jan 08 2019

(See A192872.)
a[ n_] := SeriesCoefficient[ x * (1 - x + x^2) / ((1 - x^2) * (1 - 3*x + x^2)), {x, 0, Abs @ n}]; (* Michael Somos, Apr 08 2014 *)
LinearRecurrence[{3,0,-3,1}, {0,1,2,7}, 40] (* G. C. Greubel, Jan 07 2019 *)

concat(0, Vec(-x*(x^2-x+1)/((x-1)*(x+1)*(x^2-3*x+1)) + O(x^40))) \\ Colin Barker, Apr 01 2014

(x*(x^2-x+1)/((1-x^2)*(x^2-3*x+1))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 07 2019

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).
G.f.: x*(x^2-x+1) / ((1-x)*(1+x)*(x^2-3*x+1)). - Colin Barker, Apr 01 2014
a(n) = (1/10) * (4L(2*n) - 3*(-1)^n - 5), with L(n) the Lucas numbers (A000032). - Ralf Stephan, Apr 06 2014
a(-n) = a(n) for all n in Z. - Michael Somos, Apr 08 2014

More terms from Colin Barker, Apr 01 2014

A192874 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

1, 0, 4, 6, 26, 72, 246, 774, 2532, 8150, 26434, 85448, 276654, 895054, 2896788, 9373678, 30334682, 98163784, 317666758, 1027987894, 3326644036, 10765237670, 34837054674, 112735054856, 364818336766, 1180576879422

Offset: 0

Clark Kimberling, Jul 11 2011

nonn

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,6,-5,-6,4).

Cf. A192872, A192875.

a:=[1,0,4,6,26];; for n in [6..30] do a[n]:=2*a[n-1]+6*a[n-2] - 5*a[n-3]-6*a[n-4]+4*a[n-5]; od; a; # G. C. Greubel, Jan 08 2019

I:=[1,0,4,6,26]; [n le 5 select I[n] else 2*Self(n-1)+6*Self(n-2) -5*Self(n-3)-6*Self(n-4)+4*Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 08 2019

q = x^2; s = x + 1; z = 26;
p[0, x_] := 1; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + 2*p[n - 2, x]*x^2 + 1;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] := FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192874 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192875 *)
LinearRecurrence[{2,6,-5,-6,4}, {1,0,4,6,26}, 30] (* G. C. Greubel, Jan 08 2019 *)

my(x='x+O('x^30)); Vec((x^2-x+1)*(4*x^2+x-1)/((x-1)*(x^2-x-1)*( 4*x^2+2*x-1))) \\ G. C. Greubel, Jan 08 2019

((x^2-x+1)*(4*x^2+x-1)/((x-1)*(x^2-x-1)*( 4*x^2+2*x-1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 08 2019

a(n) = 2*a(n-1) + 6*a(n-2) - 5*a(n-3) - 6*a(n-4) + 4*a(n-5).

G.f.: (x^2-x+1)*(4*x^2+x-1) / ( (x-1)*(x^2-x-1)*(4*x^2+2*x-1) ). - R. J. Mathar, May 06 2014

A192875 Coefficient of x in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

0, 1, 3, 11, 37, 119, 391, 1257, 4087, 13195, 42757, 138271, 447615, 1448249, 4687071, 15166963, 49082501, 158832391, 513995543, 1663319433, 5382623015, 17418520571, 56367538373, 182409150671, 590288468367, 1910213517529

Offset: 0

Clark Kimberling, Jul 11 2011

nonn

The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,6,-5,-6,4).

Cf. A192872, A192874.

a:=[0,1,3,11,37];; for n in [6..30] do a[n]:=2*a[n-1]+6*a[n-2] - 5*a[n-3]-6*a[n-4]+4*a[n-5]; od; a; # G. C. Greubel, Jan 08 2019

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!( x*(1+2*x)*(1-x+x^2)/((1-x)*(1+ x-x^2)*(1-2*x-4*x^2)) )); // G. C. Greubel, Jan 08 2019

seq(coeff(series(x*(1+2*x)*(1-x+x^2)/((1-x)*(1+x-x^2)*(1-2*x-4*x^2)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Jan 08 2019

(See A192874.)
LinearRecurrence[{2,6,-5,-6,4}, {0,1,3,11,37}, 30] (* G. C. Greubel, Jan 08 2019 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+2*x)*(1-x+x^2)/((1-x)*(1+ x-x^2)*(1-2*x-4*x^2)))) \\ G. C. Greubel, Jan 08 2019

(x*(1+2*x)*(1-x+x^2)/((1-x)*(1+ x-x^2)*(1-2*x-4*x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 08 2019

a(n) = 2*a(n-1) + 6*a(n-2) - 5*a(n-3) - 6*a(n-4) + 4*a(n-5).

G.f.: x*(1+2*x)*(1-x+x^2) / ( (1-x)*(1+x-x^2)*(1-2*x-4*x^2)). - R. J. Mathar, May 06 2014

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A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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A192904 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) defined below at Comments.

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A192878 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) given in Comments.

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A192883 Constant term in the reduction by (x^2 -> x + 1) of the polynomial F(n+3)*x^n, where F = A000045 (Fibonacci sequence).

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A192914 Constant term in the reduction by (x^2 -> x + 1) of the polynomial C(n)*x^n, where C=A000285.

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A066182 Permutation of the integers with cycle form {1}, {3, 2}, {6, 5, 4}, {10, 9, 8, 7}, ...

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