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 1-3 of 3 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

Views

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

A131132 a(n) = a(n-1) + a(n-2) + 1 if n is a multiple of 6, otherwise a(n) = a(n-1) + a(n-2).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 14, 22, 36, 58, 94, 152, 247, 399, 646, 1045, 1691, 2736, 4428, 7164, 11592, 18756, 30348, 49104, 79453, 128557, 208010, 336567, 544577, 881144, 1425722, 2306866, 3732588, 6039454, 9772042, 15811496, 25583539, 41395035, 66978574, 108373609
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2008

Keywords

Comments

Also: convolution of A000045 with the period-6 sequence (0,0,0,0,0,0, 1,...). - R. J. Mathar, May 30 2008
Sequences of the form s(0)=a, s(1)= b, s(n) = s(n-1) + s(n-2) + k if n mod m = p, else s(n) = s(n-1) + s(n-2) have a form s(n) = fibonacci(n-1)*a + fibonacci(n)*b + P(x)*k. a(n) is the P(x) sequence for m=6. s(n) = fib(n)*a + fib(n-1)*b + a(n-6-p)*k. - Gary Detlefs, Dec 05 2010
a(n) is the number of compositions of n where the order of the 2 and the 3 does not matter. - Gregory L. Simay, May 18 2017

Examples

			Since 5 is not a multiple of 6, a(5) = a(4) + a(3) = 5 + 3 = 8. Since 6 is a multiple of 6, a(6) = a(5) + a(4) + 1 = 8 + 5 + 1 = 14. - _Michael B. Porter_, Jul 26 2016

Links

Crossrefs

Cf. A052952, A004695, A080239, A124502, A066983.

Programs

  • Maple
    A131132:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 6 = 1 then t1:=1 else t1:=0; fi: procname(n-1)+procname(n-2)+t1; end; [seq(A131132(n), n=1..100)]; # N. J. A. Sloane, May 25 2008; Typo corrected by R. J. Mathar, May 30 2008
  • Mathematica
    Print[Table[Round[(1 + Sqrt[5])/8 Fibonacci[n + 3]], {n, 0, 50}]] ;
Print[RecurrenceTable[{c[n] == c[n - 1] + c[n - 2] + c[n - 6] - c[n - 7] - c[n - 8], c[0] == 1, c[1] == 1, c[2] == 2, c[3] == 3, c[4] == 5, c[5] == 8, c[6] == 14, c[7] == 22}, c, {n, 0, 50}]] ;  (* John M. Campbell, Jul 07 2016 *)
LinearRecurrence[{1, 1, 0, 0, 0, 1, -1, -1}, {1, 1, 2, 3, 5, 8, 14, 22}, 40] (* Vincenzo Librandi, Jul 07 2016 *)

Formula

O.g.f.: 1/((1-x^6)(1 - x - x^2)). - R. J. Mathar, May 30 2008
a(n) = ((-1)^n-1)/6 + A099837(n+3)/12 + A000045(n+4)/4 + A057079(n)/12. - R. J. Mathar, Dec 07 2010
a(n) = floor(A066983(n+4)/3). - Gary Detlefs, Dec 19 2010
a(n) = round((1 + sqrt(5))/8 A000045(n+3)). - John M. Campbell, Jul 06 2016
a(n) = (number of compositions of n consisting of only 1 or 2 or 6) - (number of compositions with only 7 or ((1 or 2) and 7)) - (number of compositions with only 8 or ((1 or 2) and 8)). The "or" is inclusive. - Gregory L. Simay, May 25 2017

Extensions

More specific name from R. J. Mathar, Dec 09 2009

A192759 Coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

0, 1, 2, 4, 7, 12, 21, 35, 58, 95, 155, 253, 411, 667, 1081, 1751, 2836, 4591, 7431, 12026, 19461, 31492, 50958, 82455, 133418, 215878, 349302, 565186, 914494, 1479686, 2394186, 3873879, 6268072, 10141958, 16410037, 26552002, 42962047
Offset: 0

Views

Author

Clark Kimberling, Jul 09 2011

Keywords

Comments

The titular polynomial is defined recursively by p(n,x)=x*(n-1,x)+floor((n+5)/5) for n>0, where p(0,x)=1. For discussions of polynomial reduction, see A192232 and A192744.

Crossrefs

Cf. A192744, A192232.

Programs

  • Mathematica
    p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + Floor[(n + 5)/5] /; n > 0;
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}]
(* A124502 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
(* A192759 *)

Formula

Conjecture: G.f.: -x / ( (x^2+x-1)*(x^4+x^3+x^2+x+1)*(x-1)^2 ), partial sums of A124502. - R. J. Mathar, May 04 2014
Showing 1-3 of 3 results.

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