cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

1, 1, 3, 8, 29, 133, 762, 5215, 41257, 369032, 3676209, 40333241, 483094250, 6271446691, 87705811341, 1314473334832, 21017294666173, 357096406209005, 6424799978507178, 122024623087820183, 2439706330834135361, 51219771117454755544
Offset: 0

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Author

Clark Kimberling, Jul 09 2011

Keywords

Comments

The titular polynomial is defined recursively by p(n,x)=x*p(n-1,x)+n! for n>0, where p(0,x)=1; see the Example. For an introduction to polynomial reduction, see A192232. The discussion at A192232 Comments continues here:
...
Let R(p,q,s) denote the "reduction of polynomial p by q->s" as defined at A192232. Suppose that q(x)=x^k for some k>0 and that s(x)=s(k,0)*x^(k-1)+s(k,1)*x^(k-2)+...+s(k,k-2)x+s(k,k-1).
...
First, we shall take p(x)=x^n, where n>=0; the results will be used to formulate R(p,q,s) for general p. Represent R(x^n,q,s) by
...
R(x^n)=s(n,0)*x^(k-1)+s(n,1)*x^(k-2)+...+s(n,k-2)*x+s(n,k-1).
...
Then each of the sequences u(n)=s(n,h), for h=0,1,...,k-1, satisfies this linear recurrence relation:
...
u(n)=s(k,0)*u(n-1)+s(k,1)*u(n-2)+...+s(k,k-2)*u(n-k-1)+s(k,k-1)*u(n-k), with initial values tabulated here:
...
n: ..s(n,0)...s(n,1)..s(n,2).......s(n,k-2)..s(n,k-1)
0: ....0........0.......0..............0.......1
1: ....0........0.......0..............1.......0
...
k-2: ..0........1.......0..............0.......0
k-1: ..0........0.......0..............0.......0
k: ..s(k,0)...s(k,1)..s(k,2).......s(k,k-2)..s(k,k-1)
...
That completes the formulation for p(x)=x^n. Turning to the general case, suppose that
...
p(n,x)=p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n-1)*x+p(n,n)
...
is a polynomial of degree n>=0. Then the reduction denoted by (R(p(n,x) by x^k -> s(x)) is the polynomial of degree k-1 given by the matrix product P*S*X, where P=(p(n,0)...p(n,1).........p(n-k)...p(n,n-k+1); X has all 0's except for main diagonal (x^(k-1), x^(k-2)...x,1); and S has
...
row 1: ... s(n,0) ... s(n,1) ...... s(n,k-2) . s(n,k-1)
row 2: ... s(n-1,0) . s(n-1,1) .... s(n-1,k-2) s(n-1,k-1)
...
row n-k+1: s(k,0).... s(k,1) ...... s(k,k-2) ..s(k,k-1)
row n-k+2: p(n,n-k+1) p(n,n-k+2) .. p(n,n-1) ..p(n,n)
*****
As a class of examples, suppose that (v(n)), for n>=0, is a sequence, that p(0,x)=1, and p(n,x)=v(n)+p(n-1,x) for n>0. If q(x)=x^2 and s(x)=x+1, and we write the reduction R(p(n,x)) as u1(n)*x+u2(n), then the sequences u1 and u2 are convolutions with the Fibonacci sequence, viz., let F=(0,1,1,2,3,5,8,...)=A000045 and let G=(1,0,1,1,2,3,5,8...); then u1=G**v and u2=F**v, where ** denotes convolution. Examples (with a few exceptions for initial terms):
...
If v(n)=n! then u1=A192744, u2=A192745.
If v(n)=n+1 then u1=A000071, u2=A001924.
If v(n)=2n then u1=A014739, u2=A027181.
If v(n)=2n+1 then u1=A001911, u2=A001891.
If v(n)=3n+1 then u1=A027961, u2=A023537.
If v(n)=3n+2 then u1=A192746, u2=A192747.
If v(n)=3n then u1=A154691, u2=A192748.
If v(n)=4n+1 then u1=A053311, u2=A192749.
If v(n)=4n+2 then u1=A192750, u2=A192751.
If v(n)=4n+3 then u1=A192752, u2=A192753.
If v(n)=4n then u1=A147728, u2=A023654.
If v(n)=5n+1 then u1=A192754, u2=A192755.
If v(n)=5n then u1=A166863, u2=A192756.
If v(n)=floor((n+1)tau) then u1=A192457, u2=A023611.
If v(n)=floor((n+2)/2) then u1=A052952, u2=A129696.
If v(n)=floor((n+3)/3) then u1=A004695, u2=A178982.
If v(n)=floor((n+4)/4) then u1=A080239, u2=A192758.
If v(n)=floor((n+5)/5) then u1=A124502, u2=A192759.
If v(n)=n+2 then u1=A001594, u2=A192760.
If v(n)=n+3 then u1=A022318, u2=A192761.
If v(n)=n+4 then u1=A022319, u2=A192762.
If v(n)=2^n then u1=A027934, u2=A008766.
If v(n)=3^n then u1=A106517, u2=A094688.

Examples

			The first five polynomials and their reductions:
1 -> 1
1+x -> 1+x
2+x+x^2 -> 3+2x
6+2x+x^2+x^3 -> 8+5x
24+6x+2x^2+x^3+x^4 -> 29+13x, so that
A192744=(1,1,3,8,29,...) and A192745=(0,1,2,5,13,...).
		

Crossrefs

Cf. A192232.

Programs

  • Maple
    A192744p := proc(n,x)
        option remember;
        if n = 0 then
            1;
        else
            x*procname(n-1,x)+n! ;
            expand(%) ;
        end if;
    end proc:
    A192744 := proc(n)
        local p;
        p := A192744p(n,x) ;
        while degree(p,x) > 1 do
            p := algsubs(x^2=x+1,p) ;
            p := expand(p) ;
        end do:
        coeftayl(p,x=0,0) ;
    end proc: # R. J. Mathar, Dec 16 2015
  • Mathematica
    q = x^2; s = x + 1; z = 40;
    p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n!;
    Table[Expand[p[n, x]], {n, 0, 7}]
    reduce[{p1_, q_, s_, x_}] :=
    FixedPoint[(s PolynomialQuotient @@ #1 +
           PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
    t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
    u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
      (* A192744 *)
    u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
      (* A192745 *)

Formula

G.f.: (1-x)/(1-x-x^2)/Q(0), where Q(k)= 1 - x*(k+1)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
Conjecture: a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2) +3*a(n-3) +(-n+2)*a(n-4)=0. - R. J. Mathar, May 04 2014
Conjecture: (-n+2)*a(n) +(n^2-n-1)*a(n-1) +(-n^2+3*n-3)*a(n-2) -(n-1)^2*a(n-3)
=0. - R. J. Mathar, Dec 16 2015

A265753 a(n) = A007949(A265399(n)).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 2, 0, 2, 1, 3, 1, 5, 2, 2, 0, 8, 2, 13, 1, 3, 3, 21, 1, 2, 5, 3, 2, 34, 2, 55, 0, 4, 8, 3, 2, 89, 13, 6, 1, 144, 3, 233, 3, 3, 21, 377, 1, 4, 2, 9, 5, 610, 3, 4, 2, 14, 34, 987, 2, 1597, 55, 4, 0, 6, 4, 2584, 8, 22, 3, 4181, 2, 6765, 89, 3, 13, 5, 6, 10946, 1, 4, 144, 17711, 3, 9, 233, 35, 3, 28657, 3, 7, 21
Offset: 1

Views

Author

Antti Karttunen, Dec 15 2015

Keywords

Comments

a(n) = Coefficient of x in the reduction under x^2->x+1 of the polynomial encoded in the prime factorization of n. (Assuming here only polynomials with nonnegative integer coefficients, see e.g. A206296 for the details).
Completely additive with a(prime(k)) = F(k-1), where F(k) denotes the k-th Fibonacci number, A000045(k). - Peter Munn, Mar 29 2021, incorporating comment by Antti Karttunen, Dec 15 2015

Crossrefs

Programs

Formula

a(n) = A007949(A265399(n)).
Other identities. For all n >= 1:
a(A000040(n)) = A000045(n-1). [Generalized by Peter Munn, Mar 29 2021]
a(A206296(n)) = A112576(n).
a(A265750(n)) = A192751(n).

A265750 Prime factorization representation of polynomials defined recursively by p(0,x)=1 and for n>0: p(n,x) = x*p(n-1,x) + 4n+2. (See A192750).

Original entry on oeis.org

2, 192, 3732480, 105815808000000, 15845956399848960000000000, 64521196676588557133336908800000000000000, 11596208520592232147315615803672416545196288000000000000000000, 254410805372253907145905144265082090216385314644252349615132618240000000000000000000000
Offset: 0

Views

Author

Antti Karttunen, Dec 15 2015

Keywords

Crossrefs

Programs

  • PARI
    A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ Using code of Michel Marcus
    A265750(n) = if(0==n, 2, A003961(A265750(n-1)) * 2^((4*n)+2));
    for(n=0, 10, write("b265750.txt", n, " ", A265750(n)));
    
  • Scheme
    (definec (A265750 n) (if (zero? n) 2 (* (A003961 (A265750 (- n 1))) (A000079 (+ 2 (* 4 n))))))

Formula

a(0) = 2; for n >= 1, a(n) = A003961(a(n-1)) * 2^((4*n)+2).
Other identities. For all n >= 1:
A192750(n) = A265752(a(n)).
A192751(n) = A265753(a(n)).

A192750 Define a pair of sequences c_n, d_n by c_0=0, d_0=1 and thereafter c_n = c_{n-1}+d_{n-1}, d_n = c_{n-1}+4*n+2; sequence here is d_n.

Original entry on oeis.org

1, 6, 11, 21, 36, 61, 101, 166, 271, 441, 716, 1161, 1881, 3046, 4931, 7981, 12916, 20901, 33821, 54726, 88551, 143281, 231836, 375121, 606961, 982086, 1589051, 2571141, 4160196, 6731341, 10891541, 17622886, 28514431, 46137321, 74651756
Offset: 0

Views

Author

Clark Kimberling, Jul 09 2011

Keywords

Comments

Old definition was: constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined recursively by p(n,x) = x*p(n-1,x) + 4n+2 for n>0, with p(0,x)=1.
For discussions of polynomial reduction, see A192232 and A192744.

Crossrefs

Programs

  • Mathematica
    q = x^2; s = x + 1; z = 40;
    p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + 4 n + 2;
    Table[Expand[p[n, x]], {n, 0, 7}]
    reduce[{p1_, q_, s_, x_}] :=
    FixedPoint[(s PolynomialQuotient @@ #1 +
           PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
    t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
    u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
      (* A192750 *)
    u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
      (* A192751 *)
    LinearRecurrence[{2,0,-1},{1,6,11},40] (* Harvey P. Dale, Dec 03 2023 *)

Formula

G.f.: ( 1+4*x-x^2 ) / ( (x-1)*(x^2+x-1) ). The first differences are in A022088. - R. J. Mathar, May 04 2014
a(n) = 5*Fibonacci(n+2)-4. - Gerry Martens, Jul 04 2015
a(n) = A265752(A265750(n)). - Antti Karttunen, Dec 15 2015

Extensions

Entry revised by N. J. A. Sloane, Dec 15 2015
Showing 1-4 of 4 results.