A291382 -id:A291382 - cp's OEIS frontend

A291728 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^2.

1, 2, 4, 9, 17, 35, 70, 142, 285, 576, 1160, 2340, 4716, 9510, 19171, 38653, 77926, 157110, 316747, 638599, 1287479, 2595698, 5233196, 10550681, 21271280, 42885152, 86460984, 174314476, 351436368, 708532813, 1428476905, 2879960190, 5806303628, 11706120825

Offset: 0

Clark Kimberling, Sep 08 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).
In the following guide to p-INVERT sequences using s = (1,0,1,0,0,0,0,...) = A154272, in some cases t(1,0,1,0,0,0,0,...) is a shifted (or differently indexed) version of the indicated sequence:
***
p(S)             t(1,0,1,0,0,0,0,...)
1 - S                A000930 (Narayana's cows sequence)
1 - S^2              A002478 (except for 0's)
1 - S^3              A291723
1 - S^5              A291724
(1 - S)^2            A291725
(1 - S)^3            A291726
(1 - S)^4            A291727
1 - S - S^2          A291728
1 - 2S - S^2         A291729
1 - 2S - 2S^2        A291730
(1 - 2S)^2           A291732
(1 - S)(1 - 2S)      A291734
1 - S - S^3          A291735
1 - S^2 - S^3        A291736
1 - S - S^2 - S^3    A291737
1 - S - S^4          A291738
1 - S^3 - S^6        A291739
(1 - S)(1 - S^2)     A291740
(1 - S)(1 + S^2)     A291741

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

Cf. A154272, A290890, A291000, A291382, A291219, A291382.

z = 60; s = x + x^3; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291728 *)

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

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

A291383 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

Original entry on oeis.org

2, 8, 28, 98, 344, 1208, 4240, 14884, 52248, 183408, 643824, 2260040, 7933504, 27849280, 97760384, 343171984, 1204649632, 4228727296, 14844261824, 52108375328, 182918006400, 642104016000, 2254002082560, 7912309005376, 27774878417792, 97499209219328

Offset: 0

Clark Kimberling, Sep 04 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 (2, 4, 4, 2)

Cf. A019590, A291382, A291384.

z = 60; s = x + x^2; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291383 *)
u / 2  (* A291384 *)

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

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

A291387 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 4 S)^2.

Original entry on oeis.org

8, 56, 352, 2096, 12032, 67328, 369664, 2000128, 10696704, 56666112, 297836544, 1555066880, 8073379840, 41709076480, 214558048256, 1099562549248, 5616171483136, 28599668703232, 145249047412736, 735884541427712, 3720035809886208, 18767645931208704

Offset: 0

Clark Kimberling, Sep 04 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.
All terms = 0 mod 8. - Muniru A Asiru, Sep 07 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; 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.: -((8 (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.

A291389 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 5 S)^2.

Original entry on oeis.org

10, 85, 650, 4700, 32750, 222375, 1481250, 9721875, 63062500, 405175000, 2582687500, 16353078125, 102955156250, 644991640625, 4023367968750, 25002220312500, 154848222656250, 956155732421875, 5888138769531250, 36171585068359375, 221714776953125000

Offset: 0

Clark Kimberling, Sep 04 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.
All terms = 0 mod 5. - Muniru A Asiru, Sep 07 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, A291390.

a:=5*[2,17,130,940];; for n in [5..10^2] do a[n]:=10*a[n-1]-15*a[n-2]-50*a[n-3]-25*a[n-4]; od; a;  # Muniru A Asiru, Sep 07 2017

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.: -((5 (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.

A291391 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 6 S)^2.

Original entry on oeis.org

12, 120, 1080, 9180, 75168, 599616, 4691520, 36164880, 275503680, 2078711424, 15559682688, 115688917440, 855249269760, 6291326453760, 46080184338432, 336227628720384, 2445042642140160, 17726787591690240, 128173151784867840, 924487654349822976

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 (12, -24, -72, -36)

Cf. A019590, A291382, A291392.

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 *)

G.f.: -((12 (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.

A291405 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - 2 S^4.

Original entry on oeis.org

0, 2, 4, 8, 24, 52, 120, 290, 672, 1592, 3760, 8824, 20800, 48976, 115296, 271588, 639488, 1505816, 3546032, 8350064, 19662944, 46302800, 109033952, 256754760, 604609280, 1423740736, 3352643712, 7894846528, 18590881280, 43778039424, 103089066752

Offset: 0

Clark Kimberling, Sep 07 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, 2, 4, 4, 8, 12, 8, 2)

Cf. A019590, A291382, A291406.

z = 60; s = x + x^2; p = 1 - 2 s^2 - 2 s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291405 *)
u / 2 (* A291406 *)

G.f.: -((2 x (1 + x)^2 (1 + x^2 + 2 x^3 + x^4))/(-1 + 2 x^2 + 4 x^3 + 4 x^4 + 8 x^5 + 12 x^6 + 8 x^7 + 2 x^8)).

a(n) = 2*a(n-2) + 4*a(n-3) + 4*a(n-4) + 5*a(n-5) + 12*a(n-6) + 8*a(n-7) + 2*a(n-8) for n >= 9.

A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

Original entry on oeis.org

1, 3, 6, 11, 21, 39, 70, 126, 224, 394, 690, 1201, 2079, 3585, 6158, 10541, 17991, 30623, 51996, 88092, 148944, 251364, 423492, 712369, 1196557, 2007135, 3362598, 5626847, 9405465, 15705447, 26200066, 43667802, 72719312, 121000846, 201185334, 334265089

Offset: 0

Clark Kimberling, Sep 07 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 (1,2,1,-2,-3,-1)

Cf. A019590, A275439, A291382.

a:=[1,3,6,11,21,39];;
for n in [7..10^2] do a[n]:=a[n-1]+2*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]- a[n-6]; od; a; # Muniru A Asiru, Sep 10 2017

z = 60; s = x + x^2; p = (1 - s)(1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291408 *)

G.f.: -(((1 + x) (-1 - x + 2 x^3 + x^4))/((-1 + x + x^2)^2 (1 + x + x^2))).
a(n) = a(n-1) + 2*a(n-2) + a(n-3) - 2*a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.
a(n) = (1/2) * A275439(n+4). - Alois P. Heinz, May 20 2025

A291417 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + 2 S^2.

Original entry on oeis.org

4, 18, 76, 322, 1360, 5744, 24256, 102428, 432528, 1826456, 7712656, 32568568, 137528704, 580748416, 2452351488, 10355650832, 43729255232, 184657419808, 779760883392, 3292730050592, 13904353779456, 58714516845824, 247936332973056, 1046971490364864

Offset: 0

Clark Kimberling, Sep 07 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, 2, -4, -2)

Cf. A019590, A291382, A291462.

z = 60; s = x + x^2; p = 1 - 4 s + 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291417 *)
u / 2  (*A291462)

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

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

A291379 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^4.

Original entry on oeis.org

0, 0, 0, 1, 4, 6, 4, 2, 8, 28, 56, 71, 68, 94, 228, 497, 808, 1044, 1352, 2316, 4608, 8264, 12592, 17717, 26968, 47044, 83912, 138417, 211052, 319850, 517356, 881918, 1483336, 2377252, 3700648, 5853067, 9605596, 15991378, 26154700, 41734433, 66214096

Offset: 0

Clark Kimberling, Sep 04 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, 0, 0, 1, 4, 6, 4, 1)

Cf. A019590, A291382.

z = 60; s = x + x^2; p = (1 - s)^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291379 *)

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

a(n) = a(n-4) + 4*a(n-5) + 6*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.

A291384 a(n) = (1/2)*A291383(n).

Original entry on oeis.org

1, 4, 14, 49, 172, 604, 2120, 7442, 26124, 91704, 321912, 1130020, 3966752, 13924640, 48880192, 171585992, 602324816, 2114363648, 7422130912, 26054187664, 91459003200, 321052008000, 1127001041280, 3956154502688, 13887439208896, 48749604609664

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 (2, 4, 4, 2)

Cf. A019590, A291382, A291383.

z = 60; s = x + x^2; p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291383 *)
u / 2  (* A291384 *)

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

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

cp's OEIS Frontend

A291728 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^2.

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A291383 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

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A291387 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 4 S)^2.

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A291389 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 5 S)^2.

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A291391 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 6 S)^2.

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A291405 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - 2 S^4.

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A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

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A291417 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + 2 S^2.

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