A291219 -id:A291219 - cp's OEIS frontend

A291242 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - S^2 + S^3.

2, 5, 13, 35, 91, 241, 631, 1662, 4362, 11470, 30127, 79179, 208023, 546633, 1436257, 3773939, 9916134, 26055432, 68461966, 179888381, 472667065, 1241962303, 3263330095, 8574599917, 22530279167, 59199680826, 155550750026, 408719050346, 1073934109927

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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,-5,-4,2,1)

Cf. A000035, A291219.

I:=[2,5,13,35,91,241]; [n le 6 select I[n] else 2*Self(n-1)+4*Self(n-2)-5*Self(n-3)-4*Self(n-4)+2*Self(n-5)+Self(n-6): n in [1..30]]; // Vincenzo Librandi, Aug 29 2017

z = 60; s = x/(1 - x^2); p = 1 - 2 s - s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291242 *)
LinearRecurrence[{2, 4, -5, -4, 2, 1}, {2, 5, 13, 35, 91, 241}, 30] (* Vincenzo Librandi, Aug 29 2017 *)

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

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

A291243 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + S^2.

Original entry on oeis.org

3, 8, 24, 71, 210, 621, 1836, 5428, 16047, 47440, 140247, 414612, 1225716, 3623579, 10712370, 31668929, 93622704, 276776352, 818232603, 2418937120, 7151092203, 21140739568, 62498266944, 184763326671, 546214936050, 1614772594421, 4773744472356, 14112597876668

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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,-3,-1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - 3 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291243 *)

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

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

A291244 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 4 S + S^2.

Original entry on oeis.org

4, 15, 60, 239, 952, 3792, 15104, 60161, 239628, 954465, 3801740, 15142752, 60315260, 240242367, 956911980, 3811486495, 15181573232, 60469889136, 240858271816, 959365196977, 3821257929948, 15220493940369, 60624914631700, 241475755550400, 961824703141876

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - 4 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291244 *)

G.f.: (4 - x - 4*x^2)/(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) for n >= 5.

A291245 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 5 S + S^2.

Original entry on oeis.org

5, 24, 120, 599, 2990, 14925, 74500, 371876, 1856265, 9265776, 46251265, 230868900, 1152410620, 5752399899, 28713814350, 143328549649, 715442152480, 3571217840400, 17826174791885, 88981552487776, 444162405876285, 2217091490069376, 11066885918992400

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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,-5,-1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - 5 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291245 *)

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

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

A291246 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 6 S + S^2.

Original entry on oeis.org

6, 35, 210, 1259, 7548, 45252, 271296, 1626481, 9751122, 58460185, 350482050, 2101219272, 12597285450, 75523579487, 452780964690, 2714524435655, 16274188816248, 97567447965516, 584938949030724, 3506841484816717, 21024308981321682, 126045494230596949

Offset: 0

Clark Kimberling, Aug 28 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 A291219 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 (6,1,-6,-1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - 6 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291246 *)

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

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

A291247 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.

Original entry on oeis.org

1, 2, 5, 10, 24, 49, 112, 238, 526, 1142, 2491, 5442, 11842, 25873, 56344, 122975, 268042, 584633, 1274820, 2779895, 6062306, 13219186, 28827703, 62861754, 137082358, 298927682, 651861824, 1421488867, 3099781932, 6759580078, 14740333285, 32143687954

Offset: 0

Clark Kimberling, Aug 29 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 A291219 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, 5, -2, -9, 2, 5, -1, -1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 + s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291247 *)

G.f.: (1 + x - 2 x^2 - 3 x^3 + 2 x^4 + x^5 - x^6)/(1 - x - 5 x^2 + 2 x^3 + 9 x^4 - 2 x^5 - 5 x^6 + x^7 + x^8).

a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 9*a(n-4) + 2*a(n-5) + 5*a(n-6) - a(n-7) - a(n-8) for n >= 9.

A291248 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.

Original entry on oeis.org

1, 2, 5, 12, 27, 65, 146, 346, 788, 1845, 4239, 9865, 22758, 52818, 122072, 282954, 654528, 1516221, 3508817, 8125763, 18808494, 43550500, 100815652, 233418699, 540371471, 1251079052, 2896357943, 6705591388, 15524220275, 35941069252, 83208225215

Offset: 0

Clark Kimberling, Aug 29 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 A291219 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, 6, -3, -12, 3, 12, -3, -6, 1, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 - s^4 + s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291248 *)

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

a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 12*a(n-4) + 3*a(n-5) + 12*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.

A291249 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.

Original entry on oeis.org

1, 2, 5, 10, 23, 47, 102, 214, 452, 955, 2003, 4223, 8854, 18610, 39032, 81896, 171752, 360103, 754985, 1582497, 3316978, 6951684, 14568692, 30530311, 63977107, 134063288, 280920507, 588643384, 1233430247, 2584481968, 5415381139, 11347029572, 23775710791

Offset: 0

Clark Kimberling, Aug 29 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 A291219 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, 6, -3, -14, 3, 14, -3, -6, 1, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 + s^4 + s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291249 *)
LinearRecurrence[{1,6,-3,-14,3,14,-3,-6,1,1},{1,2,5,10,23,47,102,214,452,955},40] (* Harvey P. Dale, Jul 21 2018 *)

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

a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 14*a(n-4) + 3*a(n-5) + 14*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.

A291250 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2 S^2 + 2 S^3.

Original entry on oeis.org

1, 3, 4, 13, 17, 52, 69, 203, 272, 781, 1053, 2976, 4029, 11267, 15296, 42469, 57765, 159596, 217361, 598499, 815860, 2241165, 3057025, 8383872, 11440897, 31340691, 42781588, 117100285, 159881873, 437378260, 597260133, 1633244795, 2230504928, 6097779229

Offset: 0

Clark Kimberling, Aug 29 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 A291219 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, 5, -4, -5, 1, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - s - 2 s^2 + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291250 *)
LinearRecurrence[{1,5,-4,-5,1,1},{1,3,4,13,17,52},40] (* Harvey P. Dale, May 13 2019 *)

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

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

A291251 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + 2 S^3.

Original entry on oeis.org

0, 3, -2, 15, -18, 76, -126, 405, -802, 2241, -4884, 12696, -29100, 72903, -171490, 421683, -1005030, 2448356, -5873706, 14243001, -34280258, 82936965, -199930344, 483172656, -1165648152, 2815517835, -6794932418, 16408304343, -39606671610, 95629756540

Offset: 0

Clark Kimberling, Aug 29 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).
Putting s = (0, -1, 0, -1, 0, -1, ...) gives (|a(n)|).  For given s, it would be of interest to know conditions on p that imply that t(s) has terms that are all positive (or all nonnegative, or strictly increasing, or alternating, as in the present case.)
See A291219 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, 6, -2, -6, 0, 1)

Cf. A000035, A291219.

z = 60; s = x/(1 - x^2); p = 1 - 3 s^2 + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291251 *)

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

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

cp's OEIS Frontend

A291242 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 2 S - S^2 + S^3.

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A291243 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + S^2.

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A291244 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 4 S + S^2.

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A291245 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 5 S + S^2.

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A291246 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 6 S + S^2.

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A291247 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.

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A291248 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.

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A291249 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.

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