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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.)

%I A193649 #24 Feb 19 2015 14:47:04
%S A193649 1,1,3,5,15,33,91,221,583,1465,3795,9653,24831,63441,162763,416525,
%T A193649 1067575,2733673,7003971,17938661,45954543,117709185,301527355,
%U A193649 772364093,1978473511
%N 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.)
%C A193649 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 A193649 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 A193649 Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.
%C A193649   D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14
%C A193649   D(D(p))=2(x+1)+7(1)+14=2x+23
%C A193649   D(D(D(p)))=2(1)+23=25;
%C A193649   the Q-residue of p is 25.
%C A193649 We may regard the sequence Q of polynomials as the triangular array formed by coefficients:
%C A193649 t(0,0)
%C A193649 t(1,0)....t(1,1)
%C A193649 t(2,0)....t(2,1)....t(2,2)
%C A193649 t(3,0)....t(3,1)....t(3,2)....t(3,3)
%C A193649   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 A193649 Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:
%C A193649 Q.....P...................Q-residue of P
%C A193649 1.....1...................A000079, 2^n
%C A193649 1....(x+1)^n..............A007051, (1+3^n)/2
%C A193649 1....(x+2)^n..............A034478, (1+5^n)/2
%C A193649 1....(x+3)^n..............A034494, (1+7^n)/2
%C A193649 1....(2x+1)^n.............A007582
%C A193649 1....(3x+1)^n.............A081186
%C A193649 1....(2x+3)^n.............A081342
%C A193649 1....(3x+2)^n.............A081336
%C A193649 1.....A040310.............A193649
%C A193649 1....(x+1)^n+(x-1)^n)/2...A122983
%C A193649 1....(x+2)(x+1)^(n-1).....A057198
%C A193649 1....(1,2,3,4,...,n)......A002064
%C A193649 1....(1,1,2,3,4,...,n)....A048495
%C A193649 1....(n,n+1,...,2n).......A087323
%C A193649 1....(n+1,n+2,...,2n+1)...A099035
%C A193649 1....p(n,k)=(2^(n-k))*3^k.A085350
%C A193649 1....p(n,k)=(3^(n-k))*2^k.A090040
%C A193649 1....A008288 (Delannoy)...A193653
%C A193649 1....A054142..............A101265
%C A193649 1....cyclotomic...........A193650
%C A193649 1....(x+1)(x+2)...(x+n)...A193651
%C A193649 1....A114525..............A193662
%C A193649 More examples:
%C A193649 Q...........P.............Q-residue of P
%C A193649 (x+1)^n...(x+1)^n.........A000110, Bell numbers
%C A193649 (x+1)^n...(x+2)^n.........A126390
%C A193649 (x+2)^n...(x+1)^n.........A028361
%C A193649 (x+2)^n...(x+2)^n.........A126443
%C A193649 (x+1)^n.....1.............A005001
%C A193649 (x+2)^n.....1.............A193660
%C A193649 A094727.....1.............A193657
%C A193649 (k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)
%C A193649 (k+1).....(x+1)^n.........A112091
%C A193649 (x+1)^n...(k+1)...........A029761
%C A193649 (k+1)......A049310........A193663
%C A193649 (In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)
%C A193649 A051162...(x+1)^n.........A193658
%C A193649 A094727...(x+1)^n.........A193659
%C A193649 A049310...(x+1)^n.........A193664
%C A193649 A075362....A075362........A193665
%C A193649 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 A193649 Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - _R. J. Mathar_, Feb 19 2015
%e A193649 First five rows of Q, coefficients of Fibonacci polynomials (A049310):
%e A193649 1
%e A193649 1...0
%e A193649 1...0...1
%e A193649 1...0...2...0
%e A193649 1...0...3...0...1
%e A193649 To obtain a(4)=15, downstep four times:
%e A193649 D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
%e A193649 DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
%e A193649 DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
%e A193649 DDDD(x^4+3*x^2+1)=D(1,14)=15.
%t A193649 q[n_, k_] := 1;
%t A193649 r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
%t A193649 f[n_, x_] := Fibonacci[n + 1, x];
%t A193649 p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
%t A193649 v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
%t A193649 Table[v[n], {n, 0, 24}]    (* A193649 *)
%t A193649 TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
%t A193649 Table[r[k], {k, 0, 8}]  (* 2^k *)
%t A193649 TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]
%Y A193649 Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).
%K A193649 nonn
%O A193649 0,3
%A A193649 _Clark Kimberling_, Aug 02 2011