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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.)

Original entry on oeis.org

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

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Author

Clark Kimberling, Aug 02 2011

Keywords

Comments

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}.

Examples

			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.

Crossrefs

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

Programs

  • Mathematica
    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}]]

Formula

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

A193973 Triangular array: the fission of (p(n,x)) by (q(n,x)), where p(n,x)=x*p(n-1,x)+n+1 with p(0,x)=1, and q(n,x)=x*p(n-1,x)+1 with p(0,x)=1.

Original entry on oeis.org

2, 3, 5, 4, 7, 9, 5, 9, 12, 14, 6, 11, 15, 18, 20, 7, 13, 18, 22, 25, 27, 8, 15, 21, 26, 30, 33, 35, 9, 17, 24, 30, 35, 39, 42, 44, 10, 19, 27, 34, 40, 45, 49, 52, 54, 11, 21, 30, 38, 45, 51, 56, 60, 63, 65, 12, 23, 33, 42, 50, 57, 63, 68, 72, 75, 77, 13, 25, 36, 46
Offset: 0

Views

Author

Clark Kimberling, Aug 10 2011

Keywords

Comments

See A193842 for the definition of fission of two sequences of polynomials or triangular arrays.
This array show the differences of the sequence of triangular numbers (A000217); viz., row n consists of t(n) - t(n-k) for k=1..n-1. - Clark Kimberling, Apr 15 2017

Examples

			First six rows:
  2;
  3,  5;
  4,  7,  9;
  5,  9, 12, 14;
  6, 11, 15, 18, 20;
  7, 13, 18, 22, 25, 27;
  ...

Links

Crossrefs

Cf. A000217, A193664, A193842, A193974.

Programs

  • Maple
    a000217 := proc(n) n*(n+1)/2 end:
seq(print(seq(a000217(n+2) - a000217(n+1-k),k=0..n)),n=0..5); # Georg Fischer, May 03 2022
  • Mathematica
    z = 13;
p[0, x_] := 1; p[n_, x_] := x*p[n - 1, x] + n + 1;
q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193973 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]]  (* A193974 *)

Formula

T(n, k) = A000217(n + 2) - A000217(n + 1 - k), 0 <= k <= n. - Georg Fischer, May 03 2022
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