A290890 -id:A290890 - cp's OEIS frontend

A290912 a(n) = (1/6)*A290911(n).

0, 1, 4, 16, 68, 287, 1208, 5088, 21432, 90273, 380236, 1601584, 6745996, 28414655, 119684720, 504121280, 2123397744, 8943915201, 37672461204, 158679314512, 668369521108, 2815224014047, 11857940853032, 49946562182048, 210378775263272, 886131640451169

Offset: 0

Clark Kimberling, Aug 18 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, 0, 4, -1)

Cf. A000027, A290890, A290911.

a:=[0,1,4,16];; for n in [5..30] do a[n]:=4*a[n-1]+4*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018

I:=[0,1,4,16]; [n le 4 select I[n] else 4*Self(n-1)+4*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Sep 13 2018

seq(coeff(series(x/(x^4-4*x^3-4*x+1),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018

z = 60; s = x/(1 - x)^2; p = 1 - 6 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290911 *)
u/6 (* A290912 *)
LinearRecurrence[{4,0,4,-1},{0,1,4,16},30] (* Harvey P. Dale, Sep 18 2022 *)

x='x+O('x^33); concat(0, Vec(x/(1-4*x-4*x^3+x^4))) \\ Altug Alkan, Sep 12 2018

G.f.: x/(1 - 4 x - 4 x^3 + x^4). [Corrected by A.H.M. Smeets, Sep 12 2018]
a(n) = 4*a(n-1) + 4*a(n-3) - a(n-4).
a(n) = (1/6)*A290911(n) for n >= 0.

A290913 p-INVERT of the positive integers, where p(S) = 1 - 7*S^2.

Original entry on oeis.org

0, 7, 28, 119, 532, 2352, 10388, 45913, 202916, 896777, 3963288, 17515680, 77410200, 342112855, 1511961052, 6682082183, 29531331004, 130513137552, 576800248892, 2549157374953, 11265950967908, 49789649104601, 220044376637232, 972481802150208, 4297864230688560

Offset: 0

Clark Kimberling, Aug 18 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, 1, 4, -1)

Cf. A000027, A290890, A290914.

z = 60; s = x/(1 - x)^2; p = 1 - 7 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290913 *)
u/7 (* A290914 *)
LinearRecurrence[{4,1,4,-1},{0,7,28,119},30] (* Harvey P. Dale, Dec 26 2018 *)

G.f.: (7 x)/(1 - 4 x - x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) + a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 7*A290914(n) for n >= 0.

A290914 a(n) = (1/7)*A290913(n).

Original entry on oeis.org

0, 1, 4, 17, 76, 336, 1484, 6559, 28988, 128111, 566184, 2502240, 11058600, 48873265, 215994436, 954583169, 4218761572, 18644733936, 82400035556, 364165339279, 1609421566844, 7112807014943, 31434910948176, 138925971735744, 613980604384080, 2713475226049825

Offset: 0

Clark Kimberling, Aug 18 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, 1, 4, -1)

Cf. A000027, A290890, A290913.

z = 60; s = x/(1 - x)^2; p = 1 - 7 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290913 *)
u/7 (* A290914 *)
LinearRecurrence[{4,1,4,-1},{0,1,4,17},30] (* Harvey P. Dale, May 05 2019 *)

G.f.: x/(1 - 4 x - x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) + a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/7)*A290913(n) for n >= 0.

A290915 p-INVERT of the positive integers, where p(S) = 1 - 8*S^2.

Original entry on oeis.org

0, 8, 32, 144, 672, 3096, 14272, 65824, 303552, 1399848, 6455520, 29770160, 137287520, 633112632, 2919650688, 13464207936, 62091296128, 286339090504, 1320476135328, 6089483698896, 28082152132128, 129503141377112, 597214328432960, 2754102721315680

Offset: 0

Clark Kimberling, Aug 18 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).

See A290890 for a guide to related sequences.

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

Cf. A000027, A290890, A290916.

z = 60; s = x/(1 - x)^2; p = 1 - 8 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290915 *)
u/8 (* A290916 *)

G.f.: (8 x)/(1 - 4 x - 2 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) + 2*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 8*A290916(n) for n >= 0.

A290916 a(n) = (1/8)*A290915(n).

Original entry on oeis.org

0, 1, 4, 18, 84, 387, 1784, 8228, 37944, 174981, 806940, 3721270, 17160940, 79139079, 364956336, 1683025992, 7761412016, 35792386313, 165059516916, 761185462362, 3510269016516, 16187892672139, 74651791054120, 344262840164460, 1587596244438120

Offset: 0

Clark Kimberling, Aug 18 2017

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

Cf. A000027, A290890, A290915.

z = 60; s = x/(1 - x)^2; p = 1 - 8 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290915 *)
u/8 (* A290916 *)

G.f.: x/(1 - 4 x - 2 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) + 2*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/8)*A290915(n) for n >= 0.

A290918 p-INVERT of the positive integers, where p(S) = (1 - S)^3.

Original entry on oeis.org

3, 12, 43, 147, 486, 1566, 4944, 15351, 47009, 142278, 426315, 1266300, 3732705, 10928910, 31806583, 92069229, 265215756, 760621914, 2172669846, 6183333681, 17538237677, 49590486888, 139817553417, 393157465848, 1102792703055, 3086146454592, 8617872504643

Offset: 0

Clark Kimberling, Aug 18 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9, -30, 45, -30, 9, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290918 *)

G.f.: (3 - 15 x + 25 x^2 - 15 x^3 + 3 x^4)/(1 - 3 x + x^2)^3.
a(n) = 9*a(n-1) - 30*a(n-2) + 45*a(n-3) - 30*a(n-4) + 9*a(n-5) - a(n-6).
(a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^3.

A290923 p-INVERT of the positive integers, where p(S) = 1 - 2S - 2S^2.

Original entry on oeis.org

2, 10, 46, 208, 938, 4230, 19078, 86048, 388106, 1750490, 7895302, 35610480, 160615298, 724429270, 3267420814, 14737172032, 66469626002, 299800475370, 1352201455582, 6098885514512, 27508034668634, 124070532153830, 559600027205398, 2523985228499040

Offset: 0

Clark Kimberling, Aug 19 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -8, 6, -1)

Cf. A000027, A033453, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290923 *)
u/2   (* A290924 *)

G.f.: (2 (1 - x + x^2))/(1 - 6 x + 8 x^2 - 6 x^3 + x^4).

a(n) = 6*a(n-1) - 8*a(n-2) + 6*a(n-3) - a(n-4).

A290924 a(n) = (1/2)*A290923(n).

Original entry on oeis.org

1, 5, 23, 104, 469, 2115, 9539, 43024, 194053, 875245, 3947651, 17805240, 80307649, 362214635, 1633710407, 7368586016, 33234813001, 149900237685, 676100727791, 3049442757256, 13754017334317, 62035266076915, 279800013602699, 1261992614249520, 5692013155802701

Offset: 0

Clark Kimberling, Aug 19 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -8, 6, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290923 *)
u/2   (* A290924 *)
LinearRecurrence[{6,-8,6,-1},{1,5,23,104},30] (* Harvey P. Dale, Nov 10 2022 *)

G.f.: (1 - x + x^2)/(1 - 6 x + 8 x^2 - 6 x^3 + x^4).

a(n) = 6*a(n-1) - 8*a(n-2) + 6*a(n-3) - a(n-4).

A290992 p-INVERT of (0,0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by two zeros, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 4, 7, 14, 27, 48, 82, 140, 242, 420, 726, 1250, 2153, 3720, 6446, 11184, 19408, 33676, 58431, 101378, 175861, 304988, 528800, 916714, 1589091, 2754612, 4775074, 8277754, 14350253, 24878304, 43131381, 74777890, 129645147, 224770632

Offset: 0

Clark Kimberling, Aug 21 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).

See A290890 for a guide to related sequences.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,0,-2,1,0,1).

Cf. A000027, A033453, A289780, A290890, A290990, A290991.

R:=PowerSeriesRing(Integers(), 60); [0,0,0] cat Coefficients(R!( x^3*(1-2*x+x^2+x^4)/(1-4*x+6*x^2-4*x^3+2*x^5-x^6-x^8) )); // G. C. Greubel, Apr 12 2023

z = 60; s = x^4/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,0,1,2,3,4,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290992 *)

concat(vector(3), Vec(x^3*(1 - 2*x + x^2 + x^4) / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^5 - x^6 - x^8) + O(x^50))) \\ Colin Barker, Aug 24 2017

def f(x): return x^3*(1-2*x+x^2+x^4)/(1-4*x+6*x^2-4*x^3+2*x^5-x^6-x^8)
def A290992_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( f(x) ).list()
A290992_list(60) # G. C. Greubel, Apr 12 2023

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 2*a(n-5) + a(n-6) + a(n-8).

G.f.: x^3*(1 - 2*x + x^2 + x^4) / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^5 - x^6 - x^8). - Colin Barker, Aug 24 2017

A291033 p-INVERT of the positive integers, where p(S) = 1 - 6*S.

Original entry on oeis.org

6, 48, 378, 2976, 23430, 184464, 1452282, 11433792, 90018054, 708710640, 5579667066, 43928625888, 345849340038, 2722866094416, 21437079415290, 168773769227904, 1328753074407942, 10461250826035632, 82361253533877114, 648428777444981280, 5105068966025973126

Offset: 0

Clark Kimberling, Aug 19 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).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 6 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291033 *)

G.f.: 6/(1 - 8 x + x^2).
a(n) = 8*a(n-1) - a(n-2).
a(n) = 6*A001090(n) for n >= 1.

cp's OEIS Frontend

A290912 a(n) = (1/6)*A290911(n).

Views

Author

Keywords

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Formula

A290913 p-INVERT of the positive integers, where p(S) = 1 - 7*S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A290914 a(n) = (1/7)*A290913(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A290915 p-INVERT of the positive integers, where p(S) = 1 - 8*S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A290916 a(n) = (1/8)*A290915(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A290918 p-INVERT of the positive integers, where p(S) = (1 - S)^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A290923 p-INVERT of the positive integers, where p(S) = 1 - 2*S - 2*S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A290924 a(n) = (1/2)*A290923(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

A290923 p-INVERT of the positive integers, where p(S) = 1 - 2S - 2S^2.