OFFSET
0,3
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))<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.
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....(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....cyclotomic...........A193650
1....(x+1)(x+2)...(x+n)...A193651
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
(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)
(k+1).....(x+1)^n.........A112091
(x+1)^n...(k+1)...........A029761
(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)
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}.
FORMULA
Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015
EXAMPLE
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.
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}]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Aug 02 2011
STATUS
approved