A090040 -id:A090040 - 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

A083065 4th row of number array A083064.

Original entry on oeis.org

1, 4, 19, 94, 469, 2344, 11719, 58594, 292969, 1464844, 7324219, 36621094, 183105469, 915527344, 4577636719, 22888183594, 114440917969, 572204589844, 2861022949219, 14305114746094, 71525573730469, 357627868652344

Offset: 0

Paul Barry, Apr 21 2003

Inverse binomial transform of A090040. [Paul Curtz, Jan 11 2009]
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=7, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). [Milan Janjic, Feb 21 2010]
For an integer x, consider the sequence P(x) of polynomials p_{1}, p_{2}, p_{3}, . . . defined by p_{1} = x-1, p_{n+1} = x*p_{1} - 1. P(5) = This sequence. P(1), P(2), P(3), P(4) are A000004, A123412, A007051, A007583 resp. [K.V.Iyer, Jun 22 2010]
It appears that if s(n) is a first order rational sequence of the form s(0)=2, s(n)= (3*s(n-1)+2)/(2*s(n-1)+3), n>0, then s(n)=2*a(n)/(2*a(n)-1), n>0.
An Engel expansion of 5/3 to the base b := 5/4 as defined in A181565, with the associated series expansion 5/3 = b + b^2/4 + b^3/(4*19) + b^4/(4*19*94) + .... Cf. A007051. - Peter Bala, Oct 29 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-5).

Cf. A007583, A083066, A007051.

[(3*5^n+1)/4: n in [0..30]]; // Vincenzo Librandi, Nov 04 2011

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*5-1 od: seq(a[n], n=1..22); # Zerinvary Lajos, Feb 22 2008

CoefficientList[Series[(1-2x)/((1-5x)(1-x)),{x,0,30}],x] (* or *) LinearRecurrence[{6,-5},{1,4},30] (* Harvey P. Dale, Jul 27 2022 *)

a(n) = (3*5^n+1)/4.
G.f.: (1-2*x)/((1-5*x)(1-x)).
E.g.f.: (3*exp(5*x) + exp(x))/4.
a(n) = 5*a(n-1)-1 with n>0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 6*a(n-1)-5*a(n-2). - Vincenzo Librandi, Nov 04 2011
a(n) = 5^n - Sum_{i=0..n-1} 5^i. - Bruno Berselli, Jun 20 2013

A154410 a(n) = 5(36^n + 2^n)/2.

Original entry on oeis.org

10, 50, 280, 1640, 9760, 58400, 350080, 2099840, 12597760, 75584000, 453498880, 2720983040, 16325877760, 97955225600, 587731271680, 3526387466240, 21158324469760, 126949946163200, 761699675668480, 4570198051389440, 27421188303093760

Offset: 0

Paul Curtz, Jan 09 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-12).

[5*(3*6^n+2^n)/2: n in [0..30]]; // Vincenzo Librandi, Aug 07 2011

LinearRecurrence[{8,-12},{10,50},30] (* Harvey P. Dale, Apr 27 2018 *)

a(n) = 8*a(n-1) - 12*a(n-2).
a(n) = 6*a(n-1) - 10*2^(n-1).
a(n) = A154407(n+1) - A154407(n).
a(n) = 10*A090040(n).
G.f.: 10*(1-3*x)/((1-2*x)*(1-6*x)). - Jaume Oliver Lafont, Aug 30 2009
E.g.f.: (5/2)*( exp(2*x) + 3*exp(6*x) ). - G. C. Greubel, Sep 16 2016

Entries corrected and extended by Paolo P. Lava, Jan 20 2009

Comments turned into formulas by R. J. Mathar, Sep 07 2009

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A154529 A090040 mod 9.

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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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A083065 4th row of number array A083064.

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A154410 a(n) = 5*(3*6^n + 2^n)/2.

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A154410 a(n) = 5(36^n + 2^n)/2.