A291382 -id:A291382 - cp's OEIS frontend

A291385 a(n) = (1/4)*A073388(n+1).

1, 4, 14, 47, 152, 480, 1488, 4548, 13744, 41152, 122272, 360944, 1059584, 3095552, 9005568, 26101824, 75404544, 217191424, 623928832, 1788071680, 5113137152, 14592352256, 41569120256, 118219097088, 335685021696, 951817715712, 2695241605120, 7622609858560

Offset: 0

Clark Kimberling, Sep 04 2017

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. A019590, A291382, A073388, A155112.

z = 60; s = x + x^2; p = (1 - 2 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A073388 *)
u / 2  (* A291385 *)
LinearRecurrence[{4,0,-8,-4},{1,4,14,47},30] (* Harvey P. Dale, Aug 24 2022 *)

G.f.: -(((1 + x) (-1 + x + 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 * A155112(n+1,k). - Alois P. Heinz, Sep 29 2022

A291388 a(n) = (1/8)*A291387(n).

Original entry on oeis.org

1, 7, 44, 262, 1504, 8416, 46208, 250016, 1337088, 7083264, 37229568, 194383360, 1009172480, 5213634560, 26819756032, 137445318656, 702021435392, 3574958587904, 18156130926592, 91985567678464, 465004476235776, 2345955741401088, 11813573860786176

Offset: 0

Clark Kimberling, Sep 04 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,-32,-16)

Cf. A019590, A291382, A291388.

a:=8*[1,7,44,262];; for n in [5..10^2] do a[n]:=8*a[n-1]+-8*a[n-2]-32*a[n-3]-16*a[n-4]; od;
A291388:=(1/8)*a;  # Muniru A Asiru, Sep 07 2017

z = 60; s = x + x^2; p = (1 - 4 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291387 *)
u / 8  (* A291388 *)

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

a(n) = 8*a(n-1) - 8*a(n-2) - 32*a(n-3) - 16*a(n-4) for n >= 5.

A291390 a(n) = (1/5)*A291389(n).

Original entry on oeis.org

2, 17, 130, 940, 6550, 44475, 296250, 1944375, 12612500, 81035000, 516537500, 3270615625, 20591031250, 128998328125, 804673593750, 5000444062500, 30969644531250, 191231146484375, 1177627753906250, 7234317013671875, 44342955390625000, 271252632343750000

Offset: 0

Clark Kimberling, Sep 06 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10, -15, -50, -25)

Cf. A019590, A291382, A291389.

z = 60; s = x + x^2; p = (1 - 5 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291389 *)
u / 5  (* A291390 *)

G.f.: -(((1 + x) (-2 + 5 x + 5 x^2))/(-1 + 5 x + 5 x^2)^2).

a(n) = 10*a(n-1) - 15*a(n-2) - 50*a(n-3) - 25*a(n-4) for n >= 5.

A291392 a(n) = (1/12)*A291391(n).

Original entry on oeis.org

1, 10, 90, 765, 6264, 49968, 390960, 3013740, 22958640, 173225952, 1296640224, 9640743120, 71270772480, 524277204480, 3840015361536, 28018969060032, 203753553511680, 1477232299307520, 10681095982072320, 77040637862485248, 554445497303525376

Offset: 0

Clark Kimberling, Sep 06 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (12, -24, -72, -36)

Cf. A019590, A291382, A291391.

z = 60; s = x + x^2; p = (1 - 6 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291391 *)
u / 12  (* A291392 *)
LinearRecurrence[{12,-24,-72,-36},{1,10,90,765},30] (* Harvey P. Dale, Nov 24 2023 *)

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

a(n) = 12*a(n-1) - 24*a(n-2) - 72*a(n-3) - 36*a(n-4) for n >= 5.

A291393 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).

Original entry on oeis.org

3, 10, 29, 83, 232, 643, 1771, 4862, 13321, 36455, 99696, 272535, 744839, 2035358, 5561381, 15195075, 41515496, 113425323, 309888403, 846638398, 2313071313, 6319448079, 17265085152, 47169141487, 128868574671, 352075628734, 961888724621, 2627929220939

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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 (3, 1, -4, -2)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s)(1 - 2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291393 *)

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

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

A291394 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 3 S).

Original entry on oeis.org

4, 17, 66, 254, 968, 3679, 13962, 52957, 200812, 761396, 2886768, 10944725, 41494856, 157319353, 596443614, 2261290498, 8573204920, 32503490435, 123230092830, 467200760741, 1771292578424, 6715480046152, 25460317920096, 96527393973769, 365963135802988

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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, -6, -3)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291394 *)

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

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

A291395 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 2 S)(1 - 3 S).

Original entry on oeis.org

5, 24, 103, 425, 1704, 6715, 26153, 101052, 388303, 1486337, 5673840, 21616915, 82244873, 312603348, 1187325847, 4507385921, 17104894344, 64893555547, 246150297257, 933554883084, 3540272085535, 13424640644225, 50903370755040, 193007618806051, 731797403031305

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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 (5, -1, -12, -6)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - 2s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291395 *)

G.f.: -(((1 + x) (-5 + 6 x + 6 x^2))/((-1 + 2 x + 2 x^2) (-1 + 3 x + 3 x^2))).

a(n) = 5*a(n-1) - a(n-2) - 12*a(n-3) - 6*a(n-4) for n >= 5.

A291396 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).

Original entry on oeis.org

6, 31, 140, 596, 2440, 9751, 38344, 149147, 575794, 2211278, 8460912, 32289105, 122994890, 467887343, 1778208080, 6753481344, 25636583768, 97283620659, 369070501684, 1399909005427, 5309251592646, 20133801242298, 76346423589984, 289487843638333

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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, -5, -16, 7, 18, 6)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s)(1 - 2s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291396 *)

G.f.: -(((1 + x) (6 - 11 x - 5 x^2 + 12 x^3 + 6 x^4))/((-1 + x + x^2) (-1 + 2 x + 2 x^2) (-1 + 3 x + 3 x^2))).

a(n) = 6*a(n-1) - 5*a(n-2) - 16*a(n-3) + 7*a(n-4) + 18*a(n-5) + 6*a(n-6) for n >= 7.

A291397 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

Original entry on oeis.org

1, 2, 4, 10, 22, 46, 97, 209, 452, 972, 2084, 4472, 9608, 20645, 44345, 95238, 204552, 439366, 943734, 2027046, 4353861, 9351633, 20086392, 43143592, 92668072, 199041584, 427521184, 918272425, 1972356577, 4236422746, 9099408124, 19544609858, 41979848918

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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 (1, 1, 1, 3, 3, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291397 *)

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

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

A291398 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^3.

Original entry on oeis.org

0, 1, 3, 5, 9, 19, 39, 76, 150, 301, 600, 1191, 2370, 4721, 9396, 18696, 37212, 74069, 147417, 293398, 583956, 1162257, 2313237, 4604037, 9163443, 18238042, 36299229, 72246487, 143792475, 286190708, 569606421, 1133689810, 2256387135, 4490895817, 8938246848

Offset: 0

Clark Kimberling, Sep 06 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 A291382 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 (0, 1, 3, 4, 3, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291398 *)
LinearRecurrence[{0,1,3,4,3,1},{0,1,3,5,9,19},40] (* Harvey P. Dale, Dec 13 2017 *)

G.f.: -((x (1 + x)^2 (1 + x + x^2))/(-1 + x^2 + 3 x^3 + 4 x^4 + 3 x^5 + x^6)).

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

cp's OEIS Frontend

A291385 a(n) = (1/4)*A073388(n+1).

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A291388 a(n) = (1/8)*A291387(n).

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A291390 a(n) = (1/5)*A291389(n).

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A291392 a(n) = (1/12)*A291391(n).

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A291393 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).

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A291394 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 3 S).

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A291395 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 2 S)(1 - 3 S).

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Mathematica

Formula

A291396 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).

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Author

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Crossrefs

Programs

Mathematica

Formula

A291397 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

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