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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.) 19

%I #24 Feb 19 2015 14:47:04

%S 1,1,3,5,15,33,91,221,583,1465,3795,9653,24831,63441,162763,416525,

%T 1067575,2733673,7003971,17938661,45954543,117709185,301527355,

%U 772364093,1978473511

%N 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.)

%C 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).

%C Since degree(D(p))<degree(p), the result of n applications of D is a constant, which we call the Q-residue of p. If p is a constant to begin with, we define D(p)=p.

%C Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

%C D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

%C D(D(p))=2(x+1)+7(1)+14=2x+23

%C D(D(D(p)))=2(1)+23=25;

%C the Q-residue of p is 25.

%C We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

%C t(0,0)

%C t(1,0)....t(1,1)

%C t(2,0)....t(2,1)....t(2,2)

%C t(3,0)....t(3,1)....t(3,2)....t(3,3)

%C 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.

%C Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

%C Q.....P...................Q-residue of P

%C 1.....1...................A000079, 2^n

%C 1....(x+1)^n..............A007051, (1+3^n)/2

%C 1....(x+2)^n..............A034478, (1+5^n)/2

%C 1....(x+3)^n..............A034494, (1+7^n)/2

%C 1....(2x+1)^n.............A007582

%C 1....(3x+1)^n.............A081186

%C 1....(2x+3)^n.............A081342

%C 1....(3x+2)^n.............A081336

%C 1.....A040310.............A193649

%C 1....(x+1)^n+(x-1)^n)/2...A122983

%C 1....(x+2)(x+1)^(n-1).....A057198

%C 1....(1,2,3,4,...,n)......A002064

%C 1....(1,1,2,3,4,...,n)....A048495

%C 1....(n,n+1,...,2n).......A087323

%C 1....(n+1,n+2,...,2n+1)...A099035

%C 1....p(n,k)=(2^(n-k))*3^k.A085350

%C 1....p(n,k)=(3^(n-k))*2^k.A090040

%C 1....A008288 (Delannoy)...A193653

%C 1....A054142..............A101265

%C 1....cyclotomic...........A193650

%C 1....(x+1)(x+2)...(x+n)...A193651

%C 1....A114525..............A193662

%C More examples:

%C Q...........P.............Q-residue of P

%C (x+1)^n...(x+1)^n.........A000110, Bell numbers

%C (x+1)^n...(x+2)^n.........A126390

%C (x+2)^n...(x+1)^n.........A028361

%C (x+2)^n...(x+2)^n.........A126443

%C (x+1)^n.....1.............A005001

%C (x+2)^n.....1.............A193660

%C A094727.....1.............A193657

%C (k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

%C (k+1).....(x+1)^n.........A112091

%C (x+1)^n...(k+1)...........A029761

%C (k+1)......A049310........A193663

%C (In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

%C A051162...(x+1)^n.........A193658

%C A094727...(x+1)^n.........A193659

%C A049310...(x+1)^n.........A193664

%C A075362....A075362........A193665

%C 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}.

%F Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - _R. J. Mathar_, Feb 19 2015

%e First five rows of Q, coefficients of Fibonacci polynomials (A049310):

%e 1

%e 1...0

%e 1...0...1

%e 1...0...2...0

%e 1...0...3...0...1

%e To obtain a(4)=15, downstep four times:

%e D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]

%e DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)

%e DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)

%e DDDD(x^4+3*x^2+1)=D(1,14)=15.

%t q[n_, k_] := 1;

%t r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];

%t f[n_, x_] := Fibonacci[n + 1, x];

%t p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)

%t v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]

%t Table[v[n], {n, 0, 24}] (* A193649 *)

%t TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]

%t Table[r[k], {k, 0, 8}] (* 2^k *)

%t TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

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

%K nonn

%O 0,3

%A _Clark Kimberling_, Aug 02 2011

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