A094688 -id:A094688 - 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

A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

Original entry on oeis.org

0, 1, 3, 14, 91, 820, 9650, 140601, 2440317, 49109632, 1123595495, 28792920872, 816742025772, 25402428294801, 859492240650847, 31427791175659690, 1234928473553777403, 51893300561135516404, 2322083099525697299278

Offset: 0

Ross La Haye, Dec 14 2004

In what follows a(i,j,k) denotes a three-dimensional array, the terms a(n) are defined as a(n,n,n) in that array. - Joerg Arndt, Jan 03 2021
Previous name was: Three-dimensional array: a(i,j,k) = expansion of x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)), read by a(n,n,n).
a(i,j,k) = the k-th value of the convolution of the Fibonacci numbers (A000045) with the powers of i = Sum_{m=0..k} a(i-1,j,m), both for i = j and i > 0; a(i,j,k) = a(i-1,j,k) + a(j,j,k-1), for i,k > 0; a(i,1,k) = Sum_{m=0..k} a(i-1,0,m), for i > 0. With F = Fibonacci and L = Lucas, then a(1,1,k) = F(k+2) - 1; a(2,1,k) = F(k+3) - 2; a(3,1,k) = L(k+2) - 3; a(4,1,k) = 4*F(k+1) + F(k) - 4; a(1,2,k) = 2^k - F(k+1); a(2,2,k) = 2^(k+1) - F(k+3); a(3,2,k) = 3(2^k - F(k+2)) + F(k); a(4,2,k) = 2^(k+2) - F(k+4) - F(k+2); a(1,3,k) = (3^k + L(k-1))/5, for k > 0; a(2,3,k) = (2 * 3^k - L(k)) /5, for k > 0; a(3,3,k) = (3^(k+1) - L(k+2))/5; a(4,3,k) = (4 * 3^k - L(k+2) - L(k+1))/5, etc..

			a(1,3,3) = 6 because a(1,3,0) = 0, a(1,3,1) = 1, a(1,3,2) = 2 and 4*2 - 2*1 - 3*0 = 6.

G. C. Greubel, Table of n, a(n) for n = 0..385
Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number

a(0, j, k) = A000045(k).
a(1, 2, k+1) - a(1, 2, k) = A099036(k).
a(3, 2, k+1) - a(3, 2, k) = A104004(k).
a(4, 2, k+1) - a(4, 2, k) = A027973(k).
a(1, 3, k+1) - a(1, 3, k) = A099159(k).
a(i, 0, k) = A109754(i, k).
a(i, i+1, 3) = A002522(i+1).
a(i, i+1, 4) = A071568(i+1).
a(2^i-2, 0, k+1) = A118654(i, k), for i > 0.
Sequences of the form a(n, 0, k): A000045(k+1) (n=1), A000032(k) (n=2), A000285(k-1) (n=3), A022095(k-1) (n=4), A022096(k-1) (n=5), A022097(k-1) (n=6), A022098(k-1) (n=7), A022099(k-1) (n=8), A022100(k-1) (n=9), A022101(k-1) (n=10), A022102(k-1) (n=11), A022103(k-1) (n=12), A022104(k-1) (n=13), A022105(k-1) (n=14), A022106(k-1) (n=15), A022107(k-1) (n=16), A022108(k-1) (n=17), A022109(k-1) (n=18), A022110(k-1) (n=19), A088209(k-2) (n=k-2), A007502(k) (n=k-1), A094588(k) (n=k).
Sequences of the form a(1, n, k): A000071(k+2) (n=1), A027934(k-1) (n=2), A098703(k) (n=3).
Sequences of the form a(2, n, k): A001911(k) (n=1), A008466(k+1) (n=2), A106517(k-1) (n=3).
Sequences of the form a(3, n, k): A027961(k) (n=1), A094688(k) (n=3).
Sequences of the form a(4, n, k): A053311(k-1) (n=1), A027974(k-1) (n=2).
Cf. A001622, A001891, A001924, A023537, A104161, A106517.

A101220:= func< n | (&+[n^k*Fibonacci(n-k): k in [0..n]]) >;
[A101220(n): n in [0..30]]; // G. C. Greubel, Jun 01 2025

Join[{0}, Table[Sum[Fibonacci[n-k]*n^k, {k, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 03 2021 *)

a(n)=sum(k=0,n,fibonacci(n-k)*n^k) \\ Joerg Arndt, Jan 03 2021

def A101220(n): return sum(n^k*fibonacci(n-k) for k in range(n+1))
print([A101220(n) for n in range(31)]) # G. C. Greubel, Jun 01 2025

a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1; a(i, j, k) = ((j+1)*a(i, j, k-1)) - ((j-1)*a(i, j, k-2)) - (j*a(i, j, k-3)), for k > 2.
a(i, j, k) = Fibonacci(k) + i*a(j, j, k-1), for i, k > 0.
a(i, j, k) = (Phi^k - (-Phi)^-k + i(((j^k - Phi^k) / (j - Phi)) - ((j^k - (-Phi)^-k) / (j - (-Phi)^-1)))) / sqrt(5), where Phi denotes the golden mean/ratio (A001622).
i^k = a(i-1, i, k) + a(i-2, i, k+1).
A104161(k) = Sum_{m=0..k} a(k-m, 0, m).
a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1, a(i, j, 3) = i*(j+1) + 2; a(i, j, k) = (j+2)*a(i, j, k-1) - 2*j*a(i, j, k-2) - a(i, j, k-3) + j*a(i, j, k-4), for k > 3. a(i, j, 0) = 0, a(i, j, 1) = 1; a(i, j, k) = a(i, j, k-1) + a(i, j, k-2) + i * j^(k-2), for k > 1.
G.f.: x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)).
a(n, n, n) = Sum_{k=0..n} Fibonacci(n-k) * n^k. - Ross La Haye, Jan 14 2006
Sum_{m=0..k} binomial(k,m)*(i-1)^m = a(i-1,i,k) + a(i-2,i,k+1), for i > 1. - Ross La Haye, May 29 2006
From Ross La Haye, Jun 03 2006: (Start)
a(3, 3, k+1) - a(3, 3, k) = A106517(k).
a(1, 1, k) = A001924(k) - A001924(k-1), for k > 0.
a(2, 1, k) = A001891(k) - A001891(k-1), for k > 0.
a(3, 1, k) = A023537(k) - A023537(k-1), for k > 0.
Sum_{j=0..i+1} a(i-j+1, 0, j) - Sum_{j=0..i} a(i-j, 0, j) = A001595(i). (End)
a(i,j,k) = a(j,j,k) + (i-j)*a(j,j,k-1), for k > 0.
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jan 03 2021

New name from Joerg Arndt, Jan 03 2021

A098703 a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.

Original entry on oeis.org

0, 1, 2, 6, 17, 50, 148, 441, 1318, 3946, 11825, 35454, 106328, 318929, 956698, 2869950, 8609617, 25828474, 77484812, 232453449, 697358750, 2092073666, 6276216817, 18828643686, 56485920112, 169457742625, 508373199218, 1525119551286

Offset: 0

Ross La Haye, Oct 27 2004

Sums of antidiagonals of A090888.
Partial sums of A099159.
Form an array with m(0,n) = A000045(n), the Fibonacci numbers, and m(i,j) = Sum_{kJ. M. Bergot, May 27 2013

			a(2) = 2 because 3^2 = 9, Luc(1) = 1 and (9 + 1) / 5 = 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Eric Weisstein, Golden Ratio
Eric Weisstein, Lucas Number
Eric Weisstein, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A000045, A000244, A000285, A001622, A001654, A022383, A052964, A084178, A090888, A094688, A099159.

I:=[0,1,2]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-3*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 18 2018

f[n_] := (3^n + Fibonacci[n] + Fibonacci[n - 2])/5; Table[ f[n], {n, 0, 27}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{4, -2, -3}, {0, 1, 2}, 30] (* Jean-François Alcover, Feb 17 2018 *)

def A098703(n): return (3**n + lucas_number2(n-1,1,-1))//5
print([A098703(n) for n in range(21)]) # G. C. Greubel, Jun 02 2025

a(n) = (((1 + sqrt(5))^n - (1 - sqrt(5))^n) / (2^n*sqrt(5))) + ((3^n - (((1 + sqrt(5)) / 2)^(n+1) + ((1 - sqrt(5)) / 2)^(n+1))) / 5).
a(n) = (3^n + (((1 + sqrt(5)) / 2)^(n-1) + ((1 - sqrt(5)) / 2)^(n-1))) / 5.
a(n) = (3^n + A000032(n-1))/5 = A000045(n) + (3^n - A000032(n+1))/5.
a(n) = (3^n + A000045(n) + A000045(n-2))/5.
a(n) = (3^n + 4*A000045(n) - A000045(n+2))/5.
a(n) = Sum_{k=0...n-1} (A000045(k)*3^(n-k-1) - A000045(k-2)*2^(n-k-1)).
a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
a(n) = A000045(n) + A094688(n-1).
a(n) = 3^1 * a(n-1) - A000045(n-3), for n > 2.
a(n) = 3^2 * a(n-2) - A000285(n-4), for n > 3.
a(n) = 3^3 * a(n-3) - A022383(n-5), for n > 4.
Limit_{n -> oo} a(n) / a(n-1) = 3.
From Ross La Haye, Dec 21 2004: (Start)
a(n) = A101220(1,3,n).
Binomial transform of unsigned A084178.
Binomial transform of signed A084178 gives the Fibonacci oblongs (A001654). (End)
G.f.: x*(1-2*x)/((1-3*x)*(1-x-x^2)). - Ross La Haye, Aug 09 2005
a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3^(n-2) for n > 1. - Ross La Haye, Aug 20 2005
Binomial transform of A052964 beginning {0,1,0,3,1,10,...}. - Ross La Haye, May 31 2006

More terms from Robert G. Wilson v, Nov 04 2004

More terms from Ross La Haye, Dec 21 2004

A106517 Convolution of Fibonacci(n-1) and 3^n.

Original entry on oeis.org

1, 3, 10, 31, 95, 288, 869, 2615, 7858, 23595, 70819, 212512, 637625, 1913019, 5739290, 17218247, 51655351, 154967040, 464902717, 1394710735, 4184136386, 12552415923, 37657258715, 112971793856, 338915410225, 1016746277043

Offset: 0

Paul Barry, May 05 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A000045, A000244, A094688, A101220.

Diagonal sums of number triangle A106516.

I:=[1,3,10]; [n le 3 select I[n] else 4*Self(n-1) -2*Self(n-2) -3*Self(n-3): n in [1..41]]; // G. C. Greubel, Aug 05 2021

LinearRecurrence[{4,-2,-3},{1,3,10},30] (* Harvey P. Dale, Oct 08 2014 *)

a(n) = sum(k=0, n, fibonacci(n-k-1) * 3^k); \\ Michel Marcus, Aug 06 2021

[(2*3^(n+1) - lucas_number2(n+1, 1, -1))/5 for n in (0..40)] # G. C. Greubel, Aug 05 2021

G.f.: (1-x)/((1-x-x^2)*(1-3*x)).
a(n) = Sum_{k=0..n} Fibonacci(n-k-1) * 3^k.
a(n) = A101220(2, 3, n+1). - Ross La Haye, Jul 25 2005
a(n) = A101220(3, 3, n+1) - A101220(3, 3, n). - Ross La Haye, May 31 2006
a(n) = (1/5)*(6*3^n - Lucas(n+1)). - Ralf Stephan, Nov 16 2010
Sum_{k=0..n} a(k) = A094688(n+1). - G. C. Greubel, Aug 05 2021

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.

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A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

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A098703 a(n) = (3^n + phi^(n-1) + (-phi)^(1-n)) / 5, where phi denotes the golden ratio A001622.

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A106517 Convolution of Fibonacci(n-1) and 3^n.

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A099167 G.f.: (1+x^2)/((1-3x)(1-x-x^2)).

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