A166863 -id:A166863 - cp's OEIS frontend

A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

1, 1, 3, 8, 29, 133, 762, 5215, 41257, 369032, 3676209, 40333241, 483094250, 6271446691, 87705811341, 1314473334832, 21017294666173, 357096406209005, 6424799978507178, 122024623087820183, 2439706330834135361, 51219771117454755544

Offset: 0

Clark Kimberling, Jul 09 2011

nonn

The titular polynomial is defined recursively by p(n,x)=x*p(n-1,x)+n! for n>0, where p(0,x)=1; see the Example.  For an introduction to polynomial reduction, see A192232.  The discussion at A192232 Comments continues here:
...
Let R(p,q,s) denote the "reduction of polynomial p by q->s" as defined at A192232.  Suppose that q(x)=x^k for some k>0 and that s(x)=s(k,0)*x^(k-1)+s(k,1)*x^(k-2)+...+s(k,k-2)x+s(k,k-1).
...
First, we shall take p(x)=x^n, where n>=0; the results will be used to formulate R(p,q,s) for general p.  Represent R(x^n,q,s) by
...
R(x^n)=s(n,0)*x^(k-1)+s(n,1)*x^(k-2)+...+s(n,k-2)*x+s(n,k-1).
...
Then each of the sequences u(n)=s(n,h), for h=0,1,...,k-1, satisfies this linear recurrence relation:
...
u(n)=s(k,0)*u(n-1)+s(k,1)*u(n-2)+...+s(k,k-2)*u(n-k-1)+s(k,k-1)*u(n-k), with initial values tabulated here:
...
n: ..s(n,0)...s(n,1)..s(n,2).......s(n,k-2)..s(n,k-1)
0: ....0........0.......0..............0.......1
1: ....0........0.......0..............1.......0
...
k-2: ..0........1.......0..............0.......0
k-1: ..0........0.......0..............0.......0
k: ..s(k,0)...s(k,1)..s(k,2).......s(k,k-2)..s(k,k-1)
...
That completes the formulation for p(x)=x^n.  Turning to the general case, suppose that
...
p(n,x)=p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n-1)*x+p(n,n)
...
is a polynomial of degree n>=0.  Then the reduction denoted by (R(p(n,x) by x^k -> s(x)) is the polynomial of degree k-1 given by the matrix product P*S*X, where P=(p(n,0)...p(n,1).........p(n-k)...p(n,n-k+1); X has all 0's except for main diagonal (x^(k-1), x^(k-2)...x,1); and S has
...
row 1: ... s(n,0) ... s(n,1) ...... s(n,k-2) . s(n,k-1)
row 2: ... s(n-1,0) . s(n-1,1) .... s(n-1,k-2) s(n-1,k-1)
...
row n-k+1: s(k,0).... s(k,1) ...... s(k,k-2) ..s(k,k-1)
row n-k+2: p(n,n-k+1) p(n,n-k+2) .. p(n,n-1) ..p(n,n)
*****
As a class of examples, suppose that (v(n)), for n>=0, is a sequence, that p(0,x)=1, and p(n,x)=v(n)+p(n-1,x) for n>0.  If q(x)=x^2 and s(x)=x+1, and we write the reduction R(p(n,x)) as u1(n)*x+u2(n), then the sequences u1 and u2 are convolutions with the Fibonacci sequence, viz., let F=(0,1,1,2,3,5,8,...)=A000045 and let G=(1,0,1,1,2,3,5,8...); then u1=G**v and u2=F**v, where ** denotes convolution.  Examples (with a few exceptions for initial terms):
...
If v(n)=n! then u1=A192744, u2=A192745.
If v(n)=n+1 then u1=A000071, u2=A001924.
If v(n)=2n then u1=A014739, u2=A027181.
If v(n)=2n+1 then u1=A001911, u2=A001891.
If v(n)=3n+1 then u1=A027961, u2=A023537.
If v(n)=3n+2 then u1=A192746, u2=A192747.
If v(n)=3n then u1=A154691, u2=A192748.
If v(n)=4n+1 then u1=A053311, u2=A192749.
If v(n)=4n+2 then u1=A192750, u2=A192751.
If v(n)=4n+3 then u1=A192752, u2=A192753.
If v(n)=4n then u1=A147728, u2=A023654.
If v(n)=5n+1 then u1=A192754, u2=A192755.
If v(n)=5n then u1=A166863, u2=A192756.
If v(n)=floor((n+1)tau) then u1=A192457, u2=A023611.
If v(n)=floor((n+2)/2) then u1=A052952, u2=A129696.
If v(n)=floor((n+3)/3) then u1=A004695, u2=A178982.
If v(n)=floor((n+4)/4) then u1=A080239, u2=A192758.
If v(n)=floor((n+5)/5) then u1=A124502, u2=A192759.
If v(n)=n+2 then u1=A001594, u2=A192760.
If v(n)=n+3 then u1=A022318, u2=A192761.
If v(n)=n+4 then u1=A022319, u2=A192762.
If v(n)=2^n then u1=A027934, u2=A008766.
If v(n)=3^n then u1=A106517, u2=A094688.

			The first five polynomials and their reductions:
1 -> 1
1+x -> 1+x
2+x+x^2 -> 3+2x
6+2x+x^2+x^3 -> 8+5x
24+6x+2x^2+x^3+x^4 -> 29+13x, so that
A192744=(1,1,3,8,29,...) and A192745=(0,1,2,5,13,...).

Cf. A192232.

A192744p := proc(n,x)
    option remember;
    if n = 0 then
        1;
    else
        x*procname(n-1,x)+n! ;
        expand(%) ;
    end if;
end proc:
A192744 := proc(n)
    local p;
    p := A192744p(n,x) ;
    while degree(p,x) > 1 do
        p := algsubs(x^2=x+1,p) ;
        p := expand(p) ;
    end do:
    coeftayl(p,x=0,0) ;
end proc: # R. J. Mathar, Dec 16 2015

q = x^2; s = x + 1; z = 40;
p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n!;
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}]
  (* A192744 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192745 *)

G.f.: (1-x)/(1-x-x^2)/Q(0), where Q(k)= 1 - x*(k+1)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
Conjecture: a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2) +3*a(n-3) +(-n+2)*a(n-4)=0. - R. J. Mathar, May 04 2014
Conjecture: (-n+2)*a(n) +(n^2-n-1)*a(n-1) +(-n^2+3*n-3)*a(n-2) -(n-1)^2*a(n-3)
=0. - R. J. Mathar, Dec 16 2015

A154691 Expansion of (1+x+x^2) / ((1-x)*(1-x-x^2)).

Original entry on oeis.org

1, 3, 7, 13, 23, 39, 65, 107, 175, 285, 463, 751, 1217, 1971, 3191, 5165, 8359, 13527, 21889, 35419, 57311, 92733, 150047, 242783, 392833, 635619, 1028455, 1664077, 2692535, 4356615, 7049153, 11405771, 18454927, 29860701, 48315631, 78176335

Offset: 0

R. J. Mathar, Jan 14 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tomislav Došlić and Biserka Kolarec, On Log-Definite Tempered Combinatorial Sequences, Mathematics (2025) Vol. 13, Iss. 7, 1179.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A001595, A006355, A055389, A066629, A078642, A166863.

a154691 n = a154691_list !! n
a154691_list = 1 : zipWith (+)
                   a154691_list (drop 2 $ map (* 2) a000045_list)
-- Reinhard Zumkeller, Nov 17 2013

A154691:= func< n | 2*Fibonacci(n+3) - 3 >;
[A154691(n): n in [0..40]]; // G. C. Greubel, Jan 18 2025

A154691 := proc(n) coeftayl( (1+x+x^2)/(1-x-x^2)/(1-x),x=0,n) ; end proc:

Fibonacci[Range[3,60]]*2 -3 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *)
CoefficientList[Series[(1 + x + x^2)/((1 - x - x^2)(1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 18 2012 *)

Vec((1+x+x^2) / ((1-x-x^2)*(1-x)) + O(x^60)) \\ Colin Barker, Feb 01 2017

def A154691(n): return 2*fibonacci(n+3) - 3
print([A154691(n) for n in range(41)]) # G. C. Greubel, Jan 18 2025

a(n+1) - a(n) = A006355(n+3) = A055389(n+3).
a(n) = A066629(n-1) + A066629(n).
a(n) = A006355(n+4) - 3 = A078642(n+1) - 3.
a(n+1) = a(n) + 2*A000045(n+2). - Reinhard Zumkeller, Nov 17 2013
From Colin Barker, Feb 01 2017: (Start)
a(n) = -3 + (2^(1-n)*((1-r)^n*(-2+r) + (1+r)^n*(2+r))) / r where r=sqrt(5).
a(n) = 2*a(n-1) - a(n-3) for n>2. (End)
a(n) = 2*Fibonacci(n+3) - 3. - Greg Dresden, Oct 10 2020
E.g.f.: 4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 3*exp(x). - Stefano Spezia, Apr 09 2025

A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.

Original entry on oeis.org

1983, 1984, 1986, 1989, 1994, 2002, 2015, 2036, 2070, 2125, 2214, 2358, 2591, 2968, 3578, 4565, 6162, 8746, 12927, 19692, 30638, 48349, 77006, 123374, 198399, 319792, 516210, 834021, 1348250, 2180290, 3526559, 5704868, 9229446, 14932333, 24159798

Offset: 1

Geoff Ahiakwo, Oct 22 2009

Starting at some a(1)=s and creating further terms with the recurrence a(n)=a(n-1)+A000045(n) defines a family of sequences with recurrences a(n)= 2*a(n-1) -a(n-3).
The generating functions are x*( s+(1-s)*x+(1-s)*x^2 )/((1-x) * (1-x-x^2)).
The terms are a(n) = A000045(n+2)+s-2 = s + A001911(n-1) = (2*s+1+k)/2 where k=A166863(n-1), n>=1.
Examples: Up to offsets, s=1 yields A000071, s=2 yields A000045 shifted left thrice, s=3 yields A001611 shifted left thrice, s=4 yields A018910.
I appreciate the editing by R. J. Mathar. However I would like further analysis of the following formula. The sequence which I call GAP can have any integer as its first term, not just 1983. Thus a(1) can be 0, 1, 2, 3,... Then a(2) is always a(1)+ 1, while a(3) is a(1) + k(n)/2; where k(n) = k(n-2)+ k(n-1)+4 (This is a separate sequence submitted for consideration). [Geoff Ahiakwo, Nov 19 2009]

			For s=1983, n=3, we have k= A166863(2)= 5, a(3) = (2s+1+k)/2 = (2*1983+1+5)/2 = 1986.
For n=3, a(3)= a(1)+ k(3)/2 = a(1)+ [K(3-2)+ k(3-1)]/2 + 2 = a(1)+ 1 + 2 thus if a(1)is 0, a(3)= 3; if a(1)= 5, a(3)= 8; if a(1)=1983, a(3)= 1986, etc. [_Geoff Ahiakwo_, Nov 19 2009]

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

Cf. A001911, A000071

LinearRecurrence[{2, 0, -1}, {1983, 1984, 1986}, 100] (* G. C. Greubel, May 27 2016 *)

a(n) = 2*a(n-1) - a(n-3).
G.f.: x*(-1983 + 1982*x + 1982*x^2)/((1-x)*(x^2+x-1)).
Let a(n)= a(1)+ k(n)/2, then G.f.: k(n)= k(n-2)+ k(n-1) + 4. - Geoff Ahiakwo, Nov 19 2009

Definition and comments edited by R. J. Mathar, Oct 26 2009

A192756 Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

0, 1, 6, 17, 38, 75, 138, 243, 416, 699, 1160, 1909, 3124, 5093, 8282, 13445, 21802, 35327, 57214, 92631, 149940, 242671, 392716, 635497, 1028328, 1663945, 2692398, 4356473, 7049006, 11405619, 18454770, 29860539, 48315464, 78176163

Offset: 0

Clark Kimberling, Jul 09 2011

nonn

The titular polynomial is defined recursively by p(n,x)=x*(n-1,x)+5n for n>0, where p(0,x)=1. For discussions of polynomial reduction, see A192232 and A192744.

Cf. A192744, A192232.

q = x^2; s = x + 1; z = 40;
p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + 5 n;
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}]
(* A166863 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
(* A192756 *)

Conjecture: G.f.: -x*(1+3*x+x^2) / ( (x^2+x-1)*(x-1)^2 ). a(n) = A001924(n)+3*A001924(n-1)+A001924(n-2). Partial sums of A166863. - R. J. Mathar, May 04 2014

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A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

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A154691 Expansion of (1+x+x^2) / ((1-x)*(1-x-x^2)).

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A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.

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A192756 Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

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