A292400 -id:A292400 - cp's OEIS frontend

A292399 p-INVERT of (1,2,3,5,8,...) (distinct Fibonacci numbers), where p(S) = (1 - S)^2.

2, 7, 22, 69, 212, 644, 1936, 5772, 17088, 50288, 147232, 429136, 1245888, 3604544, 10396160, 29900992, 85784064, 245548800, 701402624, 1999734016, 5691409408, 16172221440, 45885403136, 130011401216, 367902195712, 1039836672000, 2935713865728, 8279592292352

Offset: 0

Clark Kimberling, Sep 30 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,0,-8,-4)

Cf. A000045, A155112, A292400.

z = 60; s = x (x + 1)/(1 - x - x^2); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292399 *)

G.f.: -(((1 + x) (-2 + 3 x + 3 x^2))/(-1 + 2 x + 2 x^2)^2).
a(n) = 4*a(n-1) - 8*a(n-3) - 4*a(n-4) for n >= 5.
a(n) = Sum_{k=0..n+1} (k+1) * A155112(n+1,k). - Alois P. Heinz, Sep 29 2022

A111587 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.

Original entry on oeis.org

1, 4, 9, 20, 49, 120, 289, 696, 1681, 4060, 9801, 23660, 57121, 137904, 332929, 803760, 1940449, 4684660, 11309769, 27304196, 65918161, 159140520, 384199201, 927538920, 2239277041, 5406093004, 13051463049, 31509019100, 76069501249

Offset: 0

Creighton Dement, Aug 08 2005

Let (b(n)) be the p-INVERT of (1,2,2,2,2,2,...) using p(S) = 1 - S^2; then
b(0) = 0 and b(n) = a(n-1) for n >= 1; see A292400. - Clark Kimberling, Sep 30 2017
Floretion Algebra Multiplication Program, FAMP Code: 2kbasekseq[J+G] with J = + j' + k' + 'ii' and G = + .5'ii' + .5'jj' + .5'kk' + .5e

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

Cf. A019898, A046729, A090390, A111588.

I:=[1,4,9,20]; [n le 4 select I[n] else 2*Self(n-1) +2*Self(n-3)+Self(n-4): n in [1..35]]; // Vincenzo Librandi, Oct 01 2017

LinearRecurrence[{2,0,2,1},{1,4,9,20},30] (* Harvey P. Dale, Jul 26 2011 *)
CoefficientList[Series[(x + 1)^2 / ((x^2 + 1) (1 - 2 x - x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Oct 01 2017 *)

a(2n) = A090390(n+1), a(2n+1) = A046729(n+1);
G.f.: (x+1)^2/((x^2+1)*(1-2*x-x^2)). [sign flipped by R. J. Mathar, Nov 10 2009]
a(n) = A057077(n+1)/2 - A001333(n+2)/2. - R. J. Mathar, Nov 10 2009

A292401 p-INVERT of (1,0,2,0,2,0,2,0,2,0,...), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 3, 8, 17, 34, 71, 144, 289, 578, 1147, 2264, 4449, 8706, 16975, 32992, 63937, 123586, 238323, 458600, 880753, 1688482, 3231639, 6175728, 11785313, 22460802, 42754283, 81290424, 154396097, 292953858, 555334047, 1051781312, 1990373249, 3763583618

Offset: 0

Clark Kimberling, Sep 30 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

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

Cf. A176742, A292400.

z = 60; s = x (x^2 + 1)/(1 - x^2); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* abs. values of A176742 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292401 *)

G.f.: -(((1 + x^2) (-2 + x + 2 x^2 + x^3))/(-1 + x + x^2 + x^3)^2).

a(n) = 2*a(n-1) + a(n-2) - 3*a(n-4) - 2*a(n-5) - s(n-6) for n >= 7.

cp's OEIS Frontend

A292399 p-INVERT of (1,2,3,5,8,...) (distinct Fibonacci numbers), where p(S) = (1 - S)^2.

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A111587 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.

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A292401 p-INVERT of (1,0,2,0,2,0,2,0,2,0,...), where p(S) = (1 - S)^2.

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cp's OEIS Frontend

A292399 p-INVERT of (1,2,3,5,8,...) (distinct Fibonacci numbers), where p(S) = (1 - S)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A111587 a(n) = 2*a(n-1) + 2*a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A292401 p-INVERT of (1,0,2,0,2,0,2,0,2,0,...), where p(S) = (1 - S)^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A111587 a(n) = 2a(n-1) + 2a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.