A192750 -id:A192750 - cp's OEIS frontend

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).

2, 192, 3732480, 105815808000000, 15845956399848960000000000, 64521196676588557133336908800000000000000, 11596208520592232147315615803672416545196288000000000000000000, 254410805372253907145905144265082090216385314644252349615132618240000000000000000000000

Offset: 0

Antti Karttunen, Dec 15 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..10

Cf. A003961, A192750, A192751, A206296, A265752, A265753.

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)));

(definec (A265750 n) (if (zero? n) 2 (* (A003961 (A265750 (- n 1))) (A000079 (+ 2 (* 4 n))))))

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)).

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

Clark Kimberling, Jul 09 2011

nonn

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.

			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,...).

Cf. A192232.

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

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 *)

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

A265752 a(n) = A007814(A265399(n)).

Original entry on oeis.org

0, 1, 0, 2, 1, 1, 1, 3, 0, 2, 2, 2, 3, 2, 1, 4, 5, 1, 8, 3, 1, 3, 13, 3, 2, 4, 0, 3, 21, 2, 34, 5, 2, 6, 2, 2, 55, 9, 3, 4, 89, 2, 144, 4, 1, 14, 233, 4, 2, 3, 5, 5, 377, 1, 3, 4, 8, 22, 610, 3, 987, 35, 1, 6, 4, 3, 1597, 7, 13, 3, 2584, 3, 4181, 56, 2, 10, 3, 4, 6765, 5, 0, 90, 10946, 3, 6, 145, 21, 5, 17711

Offset: 1

Antti Karttunen, Dec 15 2015

nonn

a(n) is the constant term of the reduction by 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 of the encoding).

Completely additive with a(prime(k)) = F(k-2), where F(k) denotes the k-th Fibonacci number, A000045(k) for k >= 0, or A039834(-k) for k <= 0. - Peter Munn, Apr 05 2021, incorporating comment by Antti Karttunen, Dec 15 2015

Antti Karttunen, Table of n, a(n) for n = 1..100

Cf. A007814, A192232, A192750, A206296, A265399, A265750, A265753.

Cf. also A000040, A000045/A039834.

\\ Needs also code from A265398 and A265399.
A265752 = n -> valuation(A265399(n),2);
for(n=1, 100, write("b265752.txt", n, " ", A265752(n)));

(define (A265752 n) (A007814 (A265399 n)))

a(n) = A007814(A265399(n)).
Other identities. For all n >= 1:
a(A000040(n+1)) = A000045(n-1). [Generalized by Peter Munn, Apr 05 2021]
a(A206296(n)) = A192232(n).
a(A265750(n)) = A192750(n).

A192751 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 c_n.

Original entry on oeis.org

0, 1, 7, 18, 39, 75, 136, 237, 403, 674, 1115, 1831, 2992, 4873, 7919, 12850, 20831, 33747, 54648, 88469, 143195, 231746, 375027, 606863, 981984, 1588945, 2571031, 4160082, 6731223, 10891419, 17622760, 28514301, 46137187, 74651618, 120788939

Offset: 0

Clark Kimberling, Jul 09 2011

nonn

Old definition was: coefficient of x in the reduction under 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.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

See A192750 for d_n.

Cf. A000045, A192744, A192232, A022088, A265750, A265752.

(See A192750.)
CoefficientList[Series[x (x^2-4x-1)/((x-1)^2(x^2+x-1)),{x,0,40}],x] (* or *) LinearRecurrence[{3,-2,-1,1},{0,1,7,18},40] (* Harvey P. Dale, Feb 23 2022 *)

G.f.: x*(x^2-4*x-1)/((x-1)^2*(x^2+x-1)). First differences are in A192750. [Colin Barker, Nov 13 2012]
a(n) = 5*Fibonacci(n+3) - (4*n+10). - N. J. A. Sloane, Dec 15 2015
a(n) = A265753(A265750(n)). - Antti Karttunen, Dec 15 2015

Description corrected by Antti Karttunen, Dec 15 2015

Entry revised by N. J. A. Sloane, Dec 15 2015

cp's OEIS Frontend

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).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Scheme

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A265752 a(n) = A007814(A265399(n)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

Scheme

Formula

A192751 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 c_n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions