A290890 -id:A290890 - cp's OEIS frontend

A290917 p-INVERT of the positive integers, where p(S) = (1 - S)^2.

2, 7, 22, 67, 200, 588, 1708, 4913, 14018, 39725, 111922, 313752, 875702, 2434747, 6746350, 18636343, 51340988, 141089508, 386857888, 1058572325, 2891193242, 7882921697, 21458980582, 58330331952, 158339542250, 429274563823, 1162435429318, 3144299295403

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 (6, -11, 6, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290917 *)
LinearRecurrence[{6,-11,6,-1},{2,7,22,67},30] (* Harvey P. Dale, Jul 22 2024 *)

Vec((2 - x)*(1 - 2*x) / (1 - 3*x + x^2)^2 + O(x^30)) \\ Colin Barker, Aug 24 2017

G.f.: (2 - 5 x + 2 x^2)/(1 - 3 x + x^2)^2.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4).
(a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^2.
a(n) = (((3-sqrt(5))/2)^n * (-3+sqrt(5)) * (-5+7*sqrt(5)-5*n) + 2^(-n) * (3+sqrt(5))^(n+1) * (5+7*sqrt(5)+5*n)) / 50. - Colin Barker, Aug 24 2017

A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 7, 12, 23, 41, 77, 140, 258, 470, 861, 1570, 2867, 5225, 9526, 17352, 31607, 57547, 104766, 190684, 347029, 631476, 1148985, 2090427, 3803044, 6918379, 12585209, 22892932, 41641932, 75744383, 137772396, 250592150, 455792833, 829016539

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.

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

Cf. A000045, A289975, A289780.

z = 60; s = x^4/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,0,1,2,3,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289977 *)
LinearRecurrence[{2,1,-2,0,-1,-1,0,1},{0,0,0,1,1,2,3,7},40] (* Harvey P. Dale, Jul 14 2018 *)

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

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

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

A290900 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^3.

Original entry on oeis.org

0, 1, 5, 17, 51, 149, 439, 1308, 3916, 11728, 35093, 104943, 313773, 938199, 2805439, 8389163, 25086356, 75016104, 224321012, 670787533, 2005857561, 5998122649, 17936209267, 53634716681, 160384099011, 479597177352, 1434141243492, 4288517958652

Offset: 0

Clark Kimberling, Aug 17 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
Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, Flattened Catalan Words, arXiv:2405.05357 [math.CO], 2024. See p. 22.
Index entries for linear recurrences with constant coefficients, signature (6,-14,19,-14,6,-1).

Cf. A000027, A033453, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290900 *)

concat(0, Vec(x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6) + O(x^40))) \\ Colin Barker, Aug 18 2017

a(n) = 6*a(n-1) - 14*a(n-2) + 19*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6).

G.f.: x*(1 - x + x^2) / (1 - 6*x + 14*x^2 - 19*x^3 + 14*x^4 - 6*x^5 + x^6). - Colin Barker, Aug 18 2017

A290904 p-INVERT of the positive integers, where p(S) = 1 - 2*S^2.

Original entry on oeis.org

0, 2, 8, 24, 72, 222, 688, 2128, 6576, 20322, 62808, 194120, 599960, 1854270, 5730912, 17712288, 54742624, 169190722, 522910632, 1616137848, 4994929128, 15437616926, 47712391952, 147462678768, 455756685840, 1408587979170, 4353463496440, 13455066133672

Offset: 0

Clark Kimberling, Aug 17 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, -4, 4, -1)

Cf. A000027, A290890, A290905.

z = 60; s = x/(1 - x)^2; p = 1 - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290903 *)
u/2 (* A290905 *)

G.f.: (2 x)/(1 - 4 x + 4 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 4*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 2*A290905(n) for n >= 0.

A290905 a(n) = (1/2)*A290904(n).

Original entry on oeis.org

0, 1, 4, 12, 36, 111, 344, 1064, 3288, 10161, 31404, 97060, 299980, 927135, 2865456, 8856144, 27371312, 84595361, 261455316, 808068924, 2497464564, 7718808463, 23856195976, 73731339384, 227878342920, 704293989585, 2176731748220, 6727533066836, 20792502889884

Offset: 0

Clark Kimberling, Aug 17 2017

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

Cf. A000027, A290890, A290904.

z = 60; s = x/(1 - x)^2; p = 1 - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290904 *)
u/2 (* A290905 *)

G.f.: x/(1 - 4 x + 4 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 4*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/2)*A290904(n) for n >= 0.

A290906 p-INVERT of the positive integers, where p(S) = 1 - 3*S^2.

Original entry on oeis.org

0, 3, 12, 39, 132, 456, 1572, 5409, 18612, 64053, 220440, 758640, 2610840, 8985147, 30922188, 106418031, 366235308, 1260390744, 4337606988, 14927778921, 51373622388, 176801189997, 608457401520, 2093992746720, 7206429919920, 24800769855603, 85351303248012

Offset: 0

Clark Kimberling, Aug 17 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.

Index entries for linear recurrences with constant coefficients, signature (4, -3, 4, -1)

Cf. A000027, A290890, A290907.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290906 *)
u/3 (* A290907 *)

G.f.: (3 x)/(1 - 4 x + 3 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 3*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 3*A290907(n) for n >= 0.

A290907 a(n) = (1/3)*A290906(n).

Original entry on oeis.org

0, 1, 4, 13, 44, 152, 524, 1803, 6204, 21351, 73480, 252880, 870280, 2995049, 10307396, 35472677, 122078436, 420130248, 1445868996, 4975926307, 17124540796, 58933729999, 202819133840, 697997582240, 2402143306640, 8266923285201, 28450434416004, 97911543452733

Offset: 0

Clark Kimberling, Aug 17 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -3, 4, -1)

Cf. A000027, A290890, A290906.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290906 *)
u/3 (* A290907 *)

G.f.: x/(1 - 4 x + 3 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 3*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/3)*A290906(n) for n >= 0.

A290909 p-INVERT of the positive integers, where p(S) = 1 - 5*S^2.

Original entry on oeis.org

0, 5, 20, 75, 300, 1200, 4780, 19045, 75900, 302475, 1205400, 4803680, 19143320, 76288725, 304020900, 1211564475, 4828248580, 19241224720, 76678887300, 305575754325, 1217760780300, 4852941691355, 19339630115120, 77071046136000, 307138560414000

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, A290910.

z = 60; s = x/(1 - x)^2; p = 1 - 5 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290909 *)
u/5 (* A290910 *)

G.f.: (5 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) = 5*A290910(n) for n >= 0.

A290910 a(n) = (1/5)*A290909(n), n>= 0.

Original entry on oeis.org

0, 1, 4, 15, 60, 240, 956, 3809, 15180, 60495, 241080, 960736, 3828664, 15257745, 60804180, 242312895, 965649716, 3848244944, 15335777460, 61115150865, 243552156060, 970588338271, 3867926023024, 15414209227200, 61427712082800, 244797754857825

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, A290909.

z = 60; s = x/(1 - x)^2; p = 1 - 5 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290909 *)
u/5 (* A290910 *)
LinearRecurrence[{4,-1,4,-1},{0,1,4,15},30] (* Harvey P. Dale, Feb 19 2018 *)

concat([0], Vec(1/(1 - 4*x + x^2 - 4*x^3 + x^4) + O(x^30))) \\ Andrew Howroyd, Feb 26 2018

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) = (1/5)*A290909(n) for n >= 0.

A290911 p-INVERT of the positive integers, where p(S) = 1 - 6*S^2.

Original entry on oeis.org

0, 6, 24, 96, 408, 1722, 7248, 30528, 128592, 541638, 2281416, 9609504, 40475976, 170487930, 718108320, 3024727680, 12740386464, 53663491206, 226034767224, 952075887072, 4010217126648, 16891344084282, 71147645118192, 299679373092288, 1262272651579632

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, 0, 4, -1)

Cf. A000027, A290890, A290912.

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 *)

G.f.: (6 x)/(1 - 4 x - 4 x^3 + x^4).
a(n) = 4*a(n-1) + 4*a(n-3) - a(n-4).
a(n) = 6*A290912(n) for n >= 0.

cp's OEIS Frontend

A290917 p-INVERT of the positive integers, where p(S) = (1 - S)^2.

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A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.

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A290900 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^3.

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A290904 p-INVERT of the positive integers, where p(S) = 1 - 2*S^2.

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A290905 a(n) = (1/2)*A290904(n).

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A290906 p-INVERT of the positive integers, where p(S) = 1 - 3*S^2.

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A290907 a(n) = (1/3)*A290906(n).

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A290909 p-INVERT of the positive integers, where p(S) = 1 - 5*S^2.

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