A192969 -id:A192969 - cp's OEIS frontend

A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

0, 1, 3, 9, 20, 40, 74, 131, 225, 379, 630, 1038, 1700, 2773, 4511, 7325, 11880, 19252, 31182, 50487, 81725, 132271, 214058, 346394, 560520, 906985, 1467579, 2374641, 3842300, 6217024, 10059410, 16276523, 26336025, 42612643, 68948766

Offset: 0

Clark Kimberling, Jul 13 2011

The titular polynomials are defined recursively:  p(n,x) = x*p(n-1,x) + 3n - 1, with p(0,x)=1.  For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.
...
The list of examples at A192744 is extended here; the recurrence is given by p(n,x) = x*p(n-1,x) + v(n), with p(0,x)=1, and the reduction of p(n,x) by x^2 -> x+1 is represented by u1 + u2*x:
...
If v(n)=        n, then u1=A001595, u2=A104161.
If v(n)=      n-1, then u1=A001610, u2=A066982.
If v(n)=     3n-1, then u1=A171516, u2=A192951.
If v(n)=     3n-2, then u1=A192746, u2=A192952.
If v(n)=     2n-1, then u1=A111314, u2=A192953.
If v(n)=      n^2, then u1=A192954, u2=A192955.
If v(n)=   -1+n^2, then u1=A192956, u2=A192957.
If v(n)=    1+n^2, then u1=A192953, u2=A192389.
If v(n)=   -2+n^2, then u1=A192958, u2=A192959.
If v(n)=    2+n^2, then u1=A192960, u2=A192961.
If v(n)=    n+n^2, then u1=A192962, u2=A192963.
If v(n)=   -n+n^2, then u1=A192964, u2=A192965.
If v(n)= n(n+1)/2, then u1=A030119, u2=A192966.
If v(n)= n(n-1)/2, then u1=A192967, u2=A192968.
If v(n)= n(n+3)/2, then u1=A192969, u2=A192970.
If v(n)=     2n^2, then u1=A192971, u2=A192972.
If v(n)=   1+2n^2, then u1=A192973, u2=A192974.
If v(n)=  -1+2n^2, then u1=A192975, u2=A192976.
If v(n)=  1+n+n^2, then u1=A027181, u2=A192978.
If v(n)=  1-n+n^2, then u1=A192979, u2=A192980.
If v(n)=  (n+1)^2, then u1=A001891, u2=A053808.
If v(n)=  (n-1)^2, then u1=A192981, u2=A192982.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (3, -2, -1, 1).

Cf. A000045, A192232, A192744.

F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # G. C. Greubel, Jul 12 2019

I:=[0, 1, 3, 9]; [n le 4 select I[n] else 3*Self(n-1)-2*Self(n-2)-1*Self(n-3)+Self(n-4): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

F:=Fibonacci; [F(n+4)+2*F(n+2)-(3*n+5): n in [0..40]]; // G. C. Greubel, Jul 12 2019

(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + 3n - 1;
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}] (* A171516 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192951 *)
(* Additional programs *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,9},40] (* Vincenzo Librandi, Nov 16 2011 *)
With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-(3*n+5), {n,0,40}]] (* G. C. Greubel, Jul 12 2019 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,-1,-2,3]^n*[0;1;3;9])[1,1] \\ Charles R Greathouse IV, Mar 22 2016

vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2)-(3*n+5)) \\ G. C. Greubel, Jul 12 2019

f=fibonacci; [f(n+4)+2*f(n+2)-(3*n+5) for n in (0..40)] # G. C. Greubel, Jul 12 2019

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).
From Bruno Berselli, Nov 16 2011: (Start)
G.f.: x*(1+2*x^2)/((1-x)^2*(1 - x - x^2)).
a(n) = ((25+13*t)*(1+t)^n + (25-13*t)*(1-t)^n)/(10*2^n) - 3*n - 5 = A000285(n+2) - 3*n - 5  where t=sqrt(5). (End)
a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (3*n+5). - G. C. Greubel, Jul 12 2019

A192970 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

Original entry on oeis.org

0, 1, 3, 9, 21, 44, 85, 156, 276, 476, 806, 1347, 2230, 3667, 6001, 9787, 15923, 25862, 41955, 68006, 110170, 178406, 288828, 467509, 756636, 1224469, 1981455, 3206301, 5188161, 8394896, 13583521, 21978912, 35562960, 57542432, 93105986

Offset: 0

Clark Kimberling, Jul 13 2011

The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) + n(n+3)/2, with p(0,x)=1. For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,1,2,-1).

Cf. A000032, A000045, A192232, A192744, A192951, A192970.

F:=Fibonacci;; List([0..40], n-> 2*F(n+4)+F(n+2)-(n^2+7*n+14)/2); # G. C. Greubel, Jul 24 2019

[Fibonacci(n+4)+Lucas(n+3)-(n^2+7*n+14)/2: n in [0..40]]; // Vincenzo Librandi, Jul 13 2019

(* First progream *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + n*(n+3)/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}] (* A192969 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192970 *)
(* Additional programs *)
CoefficientList[Series[x*(1-x+2*x^2-x^3)/((1-x-x^2)*(1-x)^3), {x,0,40}], x] (* Vincenzo Librandi, Jul 13 2019 *)
Table[LucasL[n+3]+Fibonacci[n+4]-(n^2+7*n+14)/2, {n,0,40}] (* G. C. Greubel, Jul 24 2019 *)

vector(40, n, n--; f=fibonacci; 2*f(n+4)+f(n+2)-(n^2+7*n+14)/2) \\ G. C. Greubel, Jul 24 2019

f=fibonacci; [2*f(n+4)+f(n+2)-(n^2+7*n+14)/2 for n in (0..40)] # G. C. Greubel, Jul 24 2019

a(n) = 4*a(n-1) - 5*a(n-2) + a(n-3) + 2*a(n-4) - a(n-5).
G.f.: x*(1-x+2*x^2-x^3)/((1-x-x^2)*(1-x)^3). - R. J. Mathar, May 11 2014
a(n) = Fibonacci(n+4) + Lucas(n+3) - (n^2 + 7*n + 14)/2. - Ehren Metcalfe, Jul 13 2019

A210728 a(n) = a(n-1) + a(n-2) + n + 2 with n>1, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 7, 14, 27, 48, 83, 140, 233, 384, 629, 1026, 1669, 2710, 4395, 7122, 11535, 18676, 30231, 48928, 79181, 128132, 207337, 335494, 542857, 878378, 1421263, 2299670, 3720963, 6020664, 9741659, 15762356, 25504049, 41266440, 66770525, 108037002, 174807565

Offset: 0

Alex Ratushnyak, May 10 2012

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

Cf. A065220: a(n)=a(n-1)+a(n-2)+n-5, a(0)=1,a(1)=2 (except first 2 terms).
Cf. A168043: a(n)=a(n-1)+a(n-2)+n-3, a(0)=1,a(1)=2 (except first 2 terms).
Cf. A131269: a(n)=a(n-1)+a(n-2)+n-2, a(0)=1,a(1)=2.
Cf. A000126: a(n)=a(n-1)+a(n-2)+n-1, a(0)=1,a(1)=2.
Cf. A104161: a(n)=a(n-1)+a(n-2)+n,   a(0)=1,a(1)=2 (except the first term).
Cf. A192969: a(n)=a(n-1)+a(n-2)+n+1, a(0)=1,a(1)=2.
Cf. A210729: a(n)=a(n-1)+a(n-2)+n+3, a(0)=1,a(1)=2.

RecurrenceTable[{a[0] == 1, a[1] == 2, a[n] == a[n - 1] + a[n - 2] + n + 2}, a, {n, 36}] (* Bruno Berselli, Jun 27 2012 *)
nxt[{n_,a_,b_}]:={n+1,b,a+b+n+3}; NestList[nxt,{1,1,2},40][[;;,2]] (* Harvey P. Dale, Aug 26 2024 *)

G.f.: (1-x+3*x^2-2*x^3)/((1-x)^2*(1-x-x^2)). - Bruno Berselli, Jun 27 2012
a(n) = ((5+sqrt(5))*(1+sqrt(5))^(n+1)-(5-sqrt(5))*(1-sqrt(5))^(n+1))/(2^(n+1)*sqrt(5))-n-5. - Bruno Berselli, Jun 27 2012
a(n) = -n-5+A022112(n+1). R. J. Mathar, Jul 03 2012

A210729 a(n) = a(n-1) + a(n-2) + n + 3 with n>1, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 8, 16, 31, 55, 95, 160, 266, 438, 717, 1169, 1901, 3086, 5004, 8108, 13131, 21259, 34411, 55692, 90126, 145842, 235993, 381861, 617881, 999770, 1617680, 2617480, 4235191, 6852703, 11087927, 17940664, 29028626, 46969326, 75997989, 122967353

Offset: 0

Alex Ratushnyak, May 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A065220: a(n)=a(n-1)+a(n-2)+n-5, a(0)=1,a(1)=2 (except first 2 terms).
Cf. A168043: a(n)=a(n-1)+a(n-2)+n-3, a(0)=1,a(1)=2 (except first 2 terms).
Cf. A131269: a(n)=a(n-1)+a(n-2)+n-2, a(0)=1,a(1)=2.
Cf. A000126: a(n)=a(n-1)+a(n-2)+n-1, a(0)=1,a(1)=2.
Cf. A104161: a(n)=a(n-1)+a(n-2)+n,   a(0)=1,a(1)=2 (except the first term).
Cf. A192969: a(n)=a(n-1)+a(n-2)+n+1, a(0)=1,a(1)=2.
Cf. A210728: a(n)=a(n-1)+a(n-2)+n+2, a(0)=1,a(1)=2.
Cf. A000032, A000045.

F:=Fibonacci;; List([0..40], n-> 2*F(n+3)+3*F(n+1)-n-6); # G. C. Greubel, Jul 09 2019

[3*Fibonacci(n+1)+2*Fibonacci(n+3)-n-6: n in [0..40]]; // Vincenzo Librandi, Jul 18 2013

Table[3*Fibonacci[n+1]+2*Fibonacci[n+3]-n-6,{n,0,40}] (* Vaclav Kotesovec, May 13 2012 *)

vector(40, n, n--; f=fibonacci; 2*f(n+3)+3*f(n+1)-n-6) \\ G. C. Greubel, Jul 09 2019

prpr, prev = 1,2
for n in range(2, 99):
    current = prev+prpr+n+3
    print(prpr, end=',')
    prpr = prev
    prev = current

f=fibonacci; [2*f(n+3)+3*f(n+1)-n-6 for n in (0..40)] # G. C. Greubel, Jul 09 2019

G.f.: (1-x+4*x^2-3*x^3)/((1-x-x^2)*(1-x)^2).
a(n) = 3*Fibonacci(n+1)+2*Fibonacci(n+3)-n-6. - Vaclav Kotesovec, May 13 2012
a(n) = 2*Lucas(n+2) + Fibonacci(n+1) - (n+6). - G. C. Greubel, Jul 09 2019

cp's OEIS Frontend

A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Magma

Mathematica

PARI

PARI

Sage

Formula

A192970 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

A210728 a(n) = a(n-1) + a(n-2) + n + 2 with n>1, a(0)=1, a(1)=2.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A210729 a(n) = a(n-1) + a(n-2) + n + 3 with n>1, a(0)=1, a(1)=2.

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Python

Sage

Formula