A289780 -id:A289780 - cp's OEIS frontend

A291012 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)(1 - 2S).

2, 7, 22, 68, 208, 632, 1912, 5768, 17368, 52232, 156952, 471368, 1415128, 4247432, 12746392, 38247368, 114758488, 344308232, 1032990232, 3099101768, 9297567448, 27893226632, 83680728472, 251044282568, 753137042008, 2259419514632, 6778275321112

Offset: 0

Clark Kimberling, Aug 23 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 A291000 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,-6).

Cf. A000012, A033453, A289780, A291000.

[2] cat [8*3^(n-1) - 2^(n-1): n in [1..40]]; // G. C. Greubel, Jun 04 2023

z = 60; s = x/(1-x); p = (1-s)^2(1-2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* this sequence *)
LinearRecurrence[{5,-6}, {2,7,22}, 40] (* G. C. Greubel, Jun 04 2023 *)

Vec((2 -3*x -x^2)/((1-2*x)*(1-3*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017

[8*3^(n-1) - 2^(n-1) - int(n==0)/6 for n in range(41)] # G. C. Greubel, Jun 04 2023

G.f.: (2 - 3 x - x^2)/(1 - 5*x + 6*x^2).
a(n) = 5*a(n-1) - 6*a(n-2) for n >= 4.
a(n) = (16*3^n - 3*2^n) / 6 for n > 0. - Colin Barker, Aug 23 2017
E.g.f.: (1/6)*(-1 - 3*exp(2*x) + 16*exp(3*x)). - G. C. Greubel, Jun 04 2023

A291024 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 2 S^2)^2.

Original entry on oeis.org

0, 4, 8, 24, 64, 172, 456, 1200, 3136, 8148, 21064, 54216, 139008, 355196, 904840, 2298720, 5825408, 14729636, 37168008, 93612408, 235369664, 590852172, 1481051720, 3707411472, 9268764096, 23145174388, 57732471752, 143857070376, 358113876352, 890666303260

Offset: 0

Clark Kimberling, Aug 24 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 A291000 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..999
Index entries for linear recurrences with constant coefficients, signature (4,-2,-4,-1).

Cf. A000012, A033453, A289780, A291000, A291142.

z = 60; s = x/(1 - x); p = 1 - 3 s^2 + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291024 *)
u/4 (* A291142 *)

concat(0, Vec(4*x*(1 - 2*x) / (1 - 2*x - x^2)^2 + O(x^30))) \\ Colin Barker, Aug 24 2017

G.f.: -((4 (-x + 2 x^2))/(-1 + 2 x + x^2)^2).
a(n) = 4*a(n-1) - 2 a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
a(n) = 4*A291142(n) for n >= 0.
a(n) = ((1+sqrt(2))^n*(3*sqrt(2) + 2*(-1+sqrt(2))*n) - (1-sqrt(2))^n*(3*sqrt(2) + 2*(1+sqrt(2))*n)) / 4. - Colin Barker, Aug 24 2017
E.g.f.: exp(x)*x*cosh(sqrt(2)*x) + 3*exp(x)*sinh(sqrt(2)*x)/sqrt(2). - Stefano Spezia, Jun 07 2025

A289782 p-INVERT of the Lucas numbers (A000032), where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 9, 35, 146, 593, 2428, 9911, 40495, 165399, 675637, 2759792, 11273144, 46048100, 188095781, 768327108, 3138436438, 12819777601, 52365789305, 213901984464, 873739509697, 3569021260182, 14578615958179, 59550231769665, 243248749683441, 993614171826023

Offset: 0

Clark Kimberling, Aug 10 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 A289780 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, -7, 1)

Cf. A000032, A289780.

z = 60; s = (2 - x) x/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000032 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289782 *)

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

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

A289783 p-INVERT of the (3^n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 5, 24, 113, 527, 2446, 11325, 52369, 242008, 1117997, 5163891, 23849270, 110142089, 508652653, 2349005592, 10847859961, 50095958215, 231345247934, 1068361195173, 4933730638937, 22784141325656, 105217952251285, 485900111176779, 2243903303473318

Offset: 0

Clark Kimberling, Aug 10 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).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-11).

Cf. A000244, A289780.

z = 60; s = x/(1 - 3*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000244 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289783 *)

Vec(x*(1 - 2*x) / (1 - 7*x + 11*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

G.f.: (1 - 2 x)/(1 - 7 x + 11 x^2).
a(n) = 7*a(n-1) - 11*a(n-2).
a(n) = (2^(-n-1)*((7-sqrt(5))^(n+1)*(-4+sqrt(5)) + (4+sqrt(5))*(7+sqrt(5))^(n+1))) / (11*sqrt(5)). - Colin Barker, Aug 11 2017
a(n) = A099453(n) - 2*A099453(n-1). - R. J. Mathar, Jul 08 2022
E.g.f.: exp(7*x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Aug 05 2025

A289784 p-INVERT of the (4^n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 6, 35, 201, 1144, 6477, 36557, 205950, 1158967, 6517653, 36638504, 205911129, 1157068585, 6501305814, 36527449211, 205222232433, 1152978556888, 6477584595765, 36391668781013, 204450911709582, 1148616498546991, 6452981164440861, 36253117007574920

Offset: 0

Clark Kimberling, Aug 10 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).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9, -19)

Cf. A000302, A289780.

z = 60; s = x/(1 - 4*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000302 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289784 *)

Vec(x*(1 - 3*x) / (1 - 9*x + 19*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

G.f.: (1 - 3 x)/(1 - 9 x + 19 x^2).
a(n) = 9*a(n-1) - 19*a(n-2).
a(n) = (2^(-2-n)*((9-sqrt(5))^(n+1)*(-11+3*sqrt(5)) + (9+sqrt(5))^(n+1)*(11+3*sqrt(5)))) / (19*sqrt(5)). - Colin Barker, Aug 11 2017
a(n) = A081574(n+1)-3*A081574(n). - R. J. Mathar, Jul 08 2022

A289796 a(n) = (1/3)*A289795(n).

Original entry on oeis.org

1, 8, 54, 361, 2420, 16227, 108802, 729512, 4891347, 32796280, 219897701, 1474404984, 9885824398, 66284043461, 444431768220, 2979896612959, 19980083465882, 133965632756376, 898234023419479, 6022621953315440, 40381430948778393, 270755823312682408

Offset: 0

Clark Kimberling, Aug 12 2017

See A289780 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 (7, -3, 7, -1)

Cf. A289795, A289780.

z = 60; s = 3*x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008585 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289795 *)
u/3  (* A289796 *)

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

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

A289797 p-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 5, 21, 84, 330, 1291, 5052, 19784, 77500, 303608, 1189372, 4659245, 18252027, 71500068, 280092848, 1097230105, 4298267549, 16837948391, 65960645632, 258392925744, 1012223324455, 3965263584006, 15533444957104, 60850409347588, 238374187312038

Offset: 0

Clark Kimberling, Aug 12 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 A289780 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 (7, -17, 23, -16, 6, -1)

Cf. A000217, A289780.

z = 60; s = x/(1 - x)^3; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000217 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289797 *)
LinearRecurrence[{7,-17,23,-16,6,-1},{1,5,21,84,330,1291},30] (* Harvey P. Dale, Jul 10 2020 *)

G.f.: (1 - 2 x + 3 x^2 - x^3)/(1 - 7 x + 17 x^2 - 23 x^3 + 16 x^4 - 6 x^5 + x^6).

a(n) = 7*a(n-1) - 17*a(n-2) + 23*a(n-3) - 16*a(n-4) + 6*a(n-5) - a(n-6).

A289798 p-INVERT of (-1 + 2^n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 5, 22, 93, 387, 1602, 6623, 27377, 113174, 467877, 1934315, 7996978, 33061703, 136686153, 565097958, 2336269341, 9658775347, 39932014114, 165089847535, 682526498529, 2821750872886, 11665888441301, 48229967585083, 199395852702354, 824356889826903

Offset: 0

Clark Kimberling, Aug 12 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 A289780 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 (7, -15, 14, -4)

Cf. A000225, A033453, A289780.

z = 60; s = x/((1 - 2*x)*(1 - x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000225 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289798 *)

G.f.: (1 - 2 x + 2 x^2)/(1 - 7 x + 15 x^2 - 14 x^3 + 4 x^4).

a(n) = 7*a(n-1) - 15*a(n-2) + 14*a(n-3) - 4*a(n-4).

A289799 p-INVERT of (n^3), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 10, 62, 377, 2232, 13015, 75898, 444014, 2601503, 15244128, 89303905, 523084546, 3063814838, 17945741321, 105115487400, 615706236199, 3606449444722, 21124456768934, 123734572586495, 724763983514112, 4245239506761217, 24866107799273146, 145650985218990062

Offset: 0

Clark Kimberling, Aug 12 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 A289780 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, -27, 55, -36, 55, -27, 9, -1)

Cf. A000578, A289780.

z = 60; s = x*(1 + 4*x + x^2)/(1 - x)^4; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000578 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289799 *)
LinearRecurrence[{9,-27,55,-36,55,-27,9,-1},{1,10,62,377,2232,13015,75898,444014},30] (* Harvey P. Dale, Jan 07 2024 *)

G.f.: (1 + x - x^2 + 34 x^3 - x^4 + x^5 + x^6)/(1 - 9 x + 27 x^2 - 55 x^3 + 36 x^4 - 55 x^5 + 27 x^6 - 9 x^7 + x^8).

a(n) = 9*a(n-1) - 27*a(n-2) + 55*a(n-3) - 36*a(n-4) + 55*a(n-5) - 27*a(n-6) + 9*a(n-7) - a(n-8).

A289800 p-INVERT of the central binomial coefficients (A000984), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 17, 75, 336, 1517, 6879, 31276, 142439, 649431, 2963266, 13528285, 61785007, 282257992, 1289734455, 5894167695, 26939918564, 123142940445, 562928407213, 2573477722376, 11765383864555, 53790586563231, 245933621620228, 1124446028551665, 5141224466008849

Offset: 0

Clark Kimberling, Aug 12 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 A289780 for a guide to related sequences.

Cf. A000984, A289780.

z = 60; s = x/Sqrt[1 - 4 x]; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000984 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289800 *)

cp's OEIS Frontend

A291012 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)*(1 - 2*S).

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A291024 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 2 S^2)^2.

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A289782 p-INVERT of the Lucas numbers (A000032), where p(S) = 1 - S - S^2.

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A289783 p-INVERT of the (3^n), where p(S) = 1 - S - S^2.

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A289784 p-INVERT of the (4^n), where p(S) = 1 - S - S^2.

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A289796 a(n) = (1/3)*A289795(n).

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A289797 p-INVERT of the triangular numbers (A000217), where p(S) = 1 - S - S^2.

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A289798 p-INVERT of (-1 + 2^n), where p(S) = 1 - S - S^2.

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A291012 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)(1 - 2S).