A128533 -id:A128533 - cp's OEIS frontend

A192872 Constant term in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments.

1, 0, 3, 4, 13, 30, 81, 208, 547, 1428, 3741, 9790, 25633, 67104, 175683, 459940, 1204141, 3152478, 8253297, 21607408, 56568931, 148099380, 387729213, 1015088254, 2657535553, 6957518400

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. 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*p(n-1,x) + e*(x^2)*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, formally, linear recurrence sequences of order 5.  The second Mathematica program below shows initial terms and the recurrence coefficients, which are too long to be included here, which imply these properties:
(1) The numbers a,b,c,f affect initial terms but not the recurrence coefficients, which depend only on u,v,d,e.
(2) If v=0 or e=0, the order of recurrence is <= 3.
(3) If v=0 and e=0, the order of recurrence is 2, and the coefficients are 1+d*u and d*u.
(See A192904 for similar results for other p(n,x).)
...
Examples:
u v a b c d e f seq h.....seq k
1 1 1 2 0 1 1 0 -A121646..A059929
1 1 1 3 0 1 1 0 A128533...A081714
1 1 2 1 0 1 1 0 A081714...A001906
1 1 1 1 1 1 1 0 A000045...A001906
1 1 2 1 1 1 1 0 A129905...A192879
1 1 1 2 1 1 1 0 A061646...A079472
1 1 1 1 0 1 1 1 A192872...A192873
1 1 1 1 1 2 1 1 A192874...A192875
1 1 1 1 1 2 1 1 A192876...A192877
1 1 1 1 1 1 2 1 A192880...A192882
1 1 1 1 1 1 1 1 A166536...A064831
The terms of several of these sequences are products of Fibonacci numbers (A000045), or Fibonacci numbers and Lucas numbers (A000032).

			The coefficients in all the polynomials p(n,x) are Fibonacci numbers (A000045).  The first six 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. 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. A192232, A192744, A192873, A192908 (sums of adjacent terms).

a:=[1,0,3,4];; for n in [5..30] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 06 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1)) )); // G. C. Greubel, Jan 06 2019

(* First program *)
q = x^2; s = x + 1; z = 26;
p[0, x_] := 1; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + 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}] (* A192872 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192873 *)
(* End of 1st program *)
(* ******************************************** *)
(* Second program: much more general *)
(* u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 1; Nine degrees of freedom for user; shown values generate A192872. *)
q = x^2; s = u*x + v; z = 11;
(* will apply reduction (x^2 -> u*x+v) to p(n,x) *)
p[0, x_] := a; p[1, x_] := b*x + c;
(* initial values of polynomial sequence p(n,x) *)
p[n_, x_] := d*x*p[n - 1, x] + e*(x^2)*p[n - 2, x] + f;
(* recurrence for p(n,x) *)
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}];
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}];
Simplify[FindLinearRecurrence[u1]] (* for 0-sequence *)
Simplify[FindLinearRecurrence[u2]] (* for 1-sequence *)
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, 4}]
(* initial values for 0-sequence *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, 4}]
(* initial values for 1-sequence *)
LinearRecurrence[{3,0,-3,1},{1,0,3,4},26] (* Ray Chandler, Aug 02 2015 *)

my(x='x+O('x^30)); Vec((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))) \\ G. C. Greubel, Jan 06 2019

((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 06 2019

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

G.f.: (2*x-1)*(x^2-x+1) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Oct 26 2011

A128535 a(n) = F(n)*L(n-2) where F = Fibonacci and L = Lucas numbers.

Original entry on oeis.org

0, -1, 2, 2, 9, 20, 56, 143, 378, 986, 2585, 6764, 17712, 46367, 121394, 317810, 832041, 2178308, 5702888, 14930351, 39088170, 102334154, 267914297, 701408732, 1836311904, 4807526975, 12586269026, 32951280098, 86267571273, 225851433716, 591286729880

Offset: 0

Axel Harvey, Mar 09 2007

Generally, F(n)*L(n+k) = F(2*n + k) + F(k)*(-1)^(n+1):
if k=0 the sequence is A001906, if k=1 it is A081714.
For n>2, a(n) is twice the area of the triangle with vertices at (F(n-3), F(n-2)), (F(n-1), F(n)), and (L(n), L(n-1)), where F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 22 2014
a(n) is the maximum area of a quadrilateral with lengths of sides in order L(n-2), L(n-2), F(n), F(n) for n>2. - J. M. Bergot, Jan 28 2016

			a(7) = 143 because F(7)*L(5) = 13*11.
G.f. = -x + 2*x^2 + 2*x^3 + 9*x^4 + 20*x^5 + 56*x^6 + 143*x^7 + ...

Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.

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

Cf. A001906, A081714, A128533, A128534.

[Fibonacci(n)*Lucas(n-2): n in [0..30]]; // Vincenzo Librandi, Feb 20 2013

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<0,-1,2>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

Table[Fibonacci[i]LucasL[i-2], {i,0,30}] (* Harvey P. Dale, Feb 16 2011 *)
LinearRecurrence[{2, 2, -1}, {0, -1, 2}, 40] (* Vincenzo Librandi, Feb 20 2013 *)
a[ n_] := Fibonacci[2 n - 2] + (-1)^n; (* Michael Somos, May 26 2014 *)

a(n)=([0,1,0; 0,0,1; -1,2,2]^n*[0;-1;2])[1,1] \\ Charles R Greathouse IV, Feb 01 2016

a(n) = round(((-1)^n+(2^(-1-n)*(-(3-sqrt(5))^n*(3+sqrt(5))-(-3+sqrt(5))*(3+sqrt(5))^n))/sqrt(5))) \\ Colin Barker, Sep 28 2016

a(n) = F(2*(n-1)) - (-1)^(n+1), assuming F(0)=0 and L(0)=2.
From R. J. Mathar, Apr 16 2009: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(-1+4*x)/((1+x)*(x^2-3*x+1)). (End)
a(n+1) = - A116697(2*n). - Reinhard Zumkeller, Feb 25 2011
a(-n) = - A128533(n). - Michael Somos, May 26 2014
0 = a(n)*(+4*a(n) + a(n+1) - 17*a(n+2)) + a(n+1)*(-14*a(n+1) + a(n+2)) + a(n+2)*(+4*a(n+2)) for all n in Z. - Michael Somos, May 26 2014
a(n) = ((-1)^n+(2^(-1-n)*(-(3-sqrt(5))^n*(3+sqrt(5))-(-3+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5)). - Colin Barker, Sep 28 2016

More terms from Harvey P. Dale, Feb 16 2011

A128534 a(n) = Fibonacci(n)*Lucas(n-1).

Original entry on oeis.org

0, 2, 1, 6, 12, 35, 88, 234, 609, 1598, 4180, 10947, 28656, 75026, 196417, 514230, 1346268, 3524579, 9227464, 24157818, 63245985, 165580142, 433494436, 1134903171, 2971215072, 7778742050, 20365011073, 53316291174, 139583862444, 365435296163, 956722026040, 2504730781962

Offset: 0

Axel Harvey, Mar 08 2007

Generally, F(n)*L(n+k) = F(2*n + k) + F(k)*(-1)^(n+1). If k=0 the sequence is A001906; if k=1 it is A081714.
a(n) is the maximum area of a quadrilateral with lengths of sides in order F(n), F(n), L(n-1), L(n-1) for n>1. - J. M. Bergot, Jan 28 2016
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Lucas numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			a(5) = 35 because F(5)*L(4) = 5*7.

Robert Israel, Table of n, a(n) for n = 0..2370
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045, A001906, A081714, A128533, A128535.

[Fibonacci(n)*Lucas(n-1): n in [0..30]]; // G. C. Greubel, Dec 21 2017

seq(combinat:-fibonacci(2*n-1)+(-1)^(n+1),n=0..50); # Robert Israel, Jan 28 2016

Table[Fibonacci[n] LucasL[n - 1], {n, 0, 31}] (* Michael De Vlieger, Jan 29 2016 *)

concat( 0, Vec(-x*(-2+3*x)/((1+x)*(x^2-3*x+1)) + O(x^40))) \\ Michel Marcus, Jan 28 2016

a(n) = F(2*n - 1) + (-1)^(n+1), assuming F(0)=0 and L(0)=2.
From R. J. Mathar, Apr 16 2009: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: x*(2-3*x)/((1+x)*(x^2-3*x+1)). (End)
a(n) = (2^(-1-n)*(-5*(-1)^n*2^(1+n) - (-5+sqrt(5))*(3+sqrt(5))^n + (3-sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Apr 05 2016
a(n+1) = A081714(n) + 2*(-1)^n. - A.H.M. Smeets, Feb 26 2022

More terms from Michel Marcus, Jan 28 2016

A186679 First differences of A116697.

Original entry on oeis.org

0, -3, 4, -4, 7, -14, 22, -33, 54, -90, 145, -232, 376, -611, 988, -1596, 2583, -4182, 6766, -10945, 17710, -28658, 46369, -75024, 121392, -196419, 317812, -514228, 832039, -1346270, 2178310, -3524577, 5702886, -9227466, 14930353, -24157816, 39088168, -63245987, 102334156, -165580140

Offset: 0

Reinhard Zumkeller, Feb 25 2011

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

Cf. A000045, A075193, A081714, A116697, A128533.

A186679:= func< n | (-1)^n*Fibonacci(n+2) - (-1)^Floor(n/2) >;
[A186679(n): n in [0..40]]; // G. C. Greubel, Aug 24 2025

Table[(-1)^n*Fibonacci[n+2] -(-1)^Floor[n/2], {n,0,40}] (* G. C. Greubel, Aug 24 2025 *)

def A186679(n): return (-1)**n*fibonacci(n+2) -(-1)**(n//2)
print([A186679(n) for n in range(41)]) # G. C. Greubel, Aug 24 2025

a(n) = A116697(n+1) - A116697(n).
a(2*n) = A128533(n).
a(2*n+1) = A081714(n+1).
a(n+2) = A075193(n+2) - a(n).
G.f.: x*(-3+x)/((1+x-x^2)*(1+x^2)). - Colin Barker, Sep 08 2012
From G. C. Greubel, Aug 24 2025: (Start)
a(n) = (-1)^n*Fibonacci(n+2) - (-1)^floor(n/2).
E.g.f.: exp(-x/2)*(cosh(p*x) - (3/sqrt(5))*sinh(p*x)) - cos(x) - sin(x), where 2*p = sqrt(5). (End)

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A192872 Constant term in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments.

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A128535 a(n) = F(n)*L(n-2) where F = Fibonacci and L = Lucas numbers.

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A128534 a(n) = Fibonacci(n)*Lucas(n-1).

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A186679 First differences of A116697.

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