A290890 -id:A290890 - cp's OEIS frontend

A290901 p-INVERT of the positive integers, where p(S) = 1 - S^3 - S^4.

0, 0, 1, 7, 29, 93, 260, 689, 1845, 5150, 14897, 43663, 127451, 368383, 1056682, 3022366, 8651672, 24818978, 71319058, 205070493, 589550733, 1694075057, 4866102091, 13975547842, 40139685023, 115298782211, 331216158188, 951506566087, 2733431466995

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 (8, -28, 57, -71, 57, -28, 8, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290901 *)

concat(vector(2), Vec(x^2*(1 - x + x^2) / (1 - 8*x + 28*x^2 - 57*x^3 + 71*x^4 - 57*x^5 + 28*x^6 - 8*x^7 + x^8) + O(x^40))) \\ Colin Barker, Aug 18 2017

a(n) = 8*a(n-1) - 28*a(n-2) + 57*a(n-3) - 71*a(n-4) + 57*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

G.f.: x^2*(1 - x + x^2) / (1 - 8*x + 28*x^2 - 57*x^3 + 71*x^4 - 57*x^5 + 28*x^6 - 8*x^7 + x^8). - Colin Barker, Aug 18 2017

A290908 p-INVERT of the positive integers, where p(S) = 1 - 4*S^2.

Original entry on oeis.org

0, 4, 16, 56, 208, 780, 2912, 10864, 40544, 151316, 564720, 2107560, 7865520, 29354524, 109552576, 408855776, 1525870528, 5694626340, 21252634832, 79315912984, 296011017104, 1104728155436, 4122901604640, 15386878263120, 57424611447840, 214311567528244

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, -2, 4, -1)

Cf. A000027, A099486, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 4 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290908 *)
u/4 (* A099486 *)

G.f.: (4 x)/(1 - 4 x + 2 x^2 - 4 x^3 + x^4).
a(n) = 4*a(n-1) - 2*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 4*A099486(n) for n >= 0.

A290919 p-INVERT of the positive integers, where p(S) = (1 - S)^4.

Original entry on oeis.org

4, 18, 72, 271, 976, 3398, 11516, 38179, 124272, 398248, 1259240, 3935420, 12173440, 37314700, 113452128, 342426657, 1026711724, 3059968146, 9069834488, 26748151221, 78518859336, 229505772002, 668173273988, 1938126895864, 5602502738380, 16143099833606

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 (12, -58, 144, -195, 144, -58, 12, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s)^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290919 *)

G.f.: (4 - 30 x + 88 x^2 - 125 x^3 + 88 x^4 - 30 x^5 + 4 x^6)/(1 - 3 x + x^2)^4.
a(n) = 12*a(n-1) - 58*a(n-2) + 144*a(n-3) - 195*a(n-4) + 144*a(n-5) - 58*a(n-6) + 12*a(n-7) - a(n-8).
(a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^4.

A290920 p-INVERT of the positive integers, where p(S) = (1 - S)^5.

Original entry on oeis.org

5, 25, 110, 450, 1746, 6505, 23465, 82435, 283270, 955258, 3169520, 10368490, 33497790, 107028120, 338582738, 1061557195, 3301399385, 10191612315, 31250047480, 95226980516, 288523285450, 869559080385, 2607834545025, 7785230674580, 23142279699355

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 (15, -95, 330, -685, 873, -685, 330, -95, 15, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s)^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290920 *)

Vec((5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5 + O(x^30)) \\ Colin Barker, Aug 24 2017

a(n) = 15*a(n-1) - 95*a(n-2) + 330*a(n-3) - 685*a(n-4) + 873*a(n-5) - 695*a(n-6) + 330*a(n-7) - 95*a(n-8) + 15*a(n-9) - a(n-10).

G.f.: (5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5. - Colin Barker, Aug 24 2017

A290921 p-INVERT of the positive integers, where p(S) = (1 - S)^6.

Original entry on oeis.org

6, 33, 158, 696, 2886, 11425, 43590, 161355, 582340, 2056818, 7130388, 24319054, 81757104, 271353288, 890327048, 2891047695, 9299683770, 29658374355, 93843661530, 294791108106, 919849034686, 2852495485953, 8794877092878, 26971256457596, 82298545175130

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 (18, -141, 630, -1770, 3258, -3989, 3258, -1770, 630, -141, 18, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s)^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290921 *)

Vec((2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6 + O(x^30)) \\ Colin Barker, Aug 24 2017

a(n) = 18*a(n-1) - 141*a(n-2) + 630*a(n-3) - 1770*a(n-4) + 3258*a(n-5) - 3989*a(n-6) + 3258*a(n-7) - 1770*a(n-8) + 630*a(n-9) - 141*a(n-10) + 18*a(n-11) - a(n-12).

G.f.: (2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6. - Colin Barker, Aug 24 2017

A290922 p-INVERT of the positive integers, where p(S) = 1 - S - 2*S^2.

Original entry on oeis.org

1, 5, 20, 75, 279, 1040, 3881, 14485, 54060, 201755, 752959, 2810080, 10487361, 39139365, 146070100, 545141035, 2034494039, 7592835120, 28336846441, 105754550645, 394681356140, 1472970873915, 5497202139519, 20515837684160, 76566148597121, 285748756704325

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 (5, -6, 5, -1)

Cf. A000027, A290890.

I:=[1,5,20,75]; [n le 4 select I[n] else 5*Self(n-1)- 6*Self(n-2)+5*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 19 2017

z = 60; s = x/(1 - x)^2; p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290922 *)
LinearRecurrence[{5, -6, 5, -1}, {1, 5, 20, 75}, 30] (* Vincenzo Librandi, Aug 19 2017 *)

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

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

A290925 p-INVERT of the positive integers, where p(S) = 1 - 3S - 2S^2.

Original entry on oeis.org

3, 17, 92, 495, 2661, 14304, 76891, 413329, 2221860, 11943663, 64203453, 345127232, 1855239875, 9972887313, 53609499612, 288179176047, 1549114207525, 8327301302176, 44763611772699, 240627889663761, 1293501104827044, 6953246818258415, 37377348295412093

Offset: 0

Clark Kimberling, Aug 19 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 (7, -10, 7, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290925 *)

G.f.: (3 - 4 x + 3 x^2)/(1 - 7 x + 10 x^2 - 7 x^3 + x^4).

a(n) = 7*a(n-1) - 10*a(n-2) + 7*a(n-3) - a(n-4).

A290926 p-INVERT of the positive integers, where p(S) = (1 - S^2)^2.

Original entry on oeis.org

0, 2, 8, 23, 64, 182, 520, 1475, 4152, 11624, 32408, 90028, 249272, 688140, 1894600, 5203665, 14260968, 39004962, 106486512, 290226621, 789776888, 2146082610, 5823823120, 15784464728, 42731452816, 115556460982, 312175750152, 842537682283, 2271900155120

Offset: 0

Clark Kimberling, Aug 19 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 (8, -26, 48, -59, 48, -26, 8, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s^2)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290926 *)

G.f.: (2 x - 8 x^2 + 11 x^3 - 8 x^4 + 2 x^5)/(1 - 4 x + 5 x^2 - 4 x^3 + x^4)^2.

a(n) = 8*a(n-1) - 26*a(n-2) + 48*a(n-3) - 59*a(n-4) + 48*a(n-5) - 26*a(n-6) + 8*a(n-7) - a(n-8).

A290927 p-INVERT of the positive integers, where p(S) = (1 - S^2)^3.

Original entry on oeis.org

0, 3, 12, 36, 108, 331, 1008, 3027, 8992, 26502, 77592, 225806, 653544, 1882224, 5396776, 15411399, 43847688, 124331457, 351448620, 990586686, 2784612380, 7808372811, 21845061504, 60983031772, 169897677504, 472435652577, 1311365875700, 3633925019190

Offset: 0

Clark Kimberling, Aug 19 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 (12, -63, 196, -414, 636, -731, 636, -414, 196, -63, 12, -1)

Cf. A000027, A033453, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s^2)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290927 *)

concat(0, Vec(x*(3 - 24*x + 81*x^2 - 156*x^3 + 193*x^4 - 156*x^5 + 81*x^6 - 24*x^7 + 3*x^8) / ((1 - 3*x + x^2)^3*(1 - x + x^2)^3) + O(x^30))) \\ Colin Barker, Aug 19 2017

a(n) = 12*a(n-1) - 63*a(n-2) + 196*a(n-3) - 414*a(n-4) + 636*a(n-5) - 731*a(n-6) + 636*a(n-7) - 414*a(n-8) + 196*a(n-9) - 63*a(n-10) + 12*a(n-11) - a(n-12).

G.f.: x*(3 - 24*x + 81*x^2 - 156*x^3 + 193*x^4 - 156*x^5 + 81*x^6 - 24*x^7 + 3*x^8) / ((1 - 3*x + x^2)^3*(1 - x + x^2)^3). - Colin Barker, Aug 19 2017

A290928 p-INVERT of the positive integers, where p(S) = (1 - S^3)^2.

Original entry on oeis.org

0, 0, 2, 12, 42, 115, 288, 738, 2020, 5751, 16362, 45697, 125538, 342318, 933050, 2547630, 6960042, 18990309, 51699042, 140439411, 380871538, 1031705466, 2792009100, 7548723827, 20389716864, 55020917232, 148334534420, 399562167420, 1075432476492

Offset: 0

Clark Kimberling, Aug 19 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 (12,-66,222,-507,822,-965,822,-507,222,-66,12,-1)

Cf. A000027, A033453, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290928 *)
LinearRecurrence[{12, -66, 222, -507, 822, -965, 822, -507, 222, -66, 12, -1}, {0, 0, 2, 12, 42, 115, 288, 738, 2020, 5751, 16362, 45697}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

concat(vector(2), Vec(x^2*(2 - 12*x + 30*x^2 - 41*x^3 + 30*x^4 - 12*x^5 + 2*x^6) / ((1 - 3*x + x^2)^2*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)^2) + O(x^30))) \\ Colin Barker, Aug 19 2017

a(n) = 12*a(n-1) - 66*a(n-2) + 222*a(n-3) - 507*a(n-4) + 822*a(n-5) - 965*a(n-6) + 822*a(n-7) - 507*a(n-8) + 222*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12).

G.f.: x^2*(2 - 12*x + 30*x^2 - 41*x^3 + 30*x^4 - 12*x^5 + 2*x^6) / ((1 - 3*x + x^2)^2*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)^2). - Colin Barker, Aug 19 2017

cp's OEIS Frontend

A290901 p-INVERT of the positive integers, where p(S) = 1 - S^3 - S^4.

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A290908 p-INVERT of the positive integers, where p(S) = 1 - 4*S^2.

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A290919 p-INVERT of the positive integers, where p(S) = (1 - S)^4.

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A290920 p-INVERT of the positive integers, where p(S) = (1 - S)^5.

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A290921 p-INVERT of the positive integers, where p(S) = (1 - S)^6.

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A290922 p-INVERT of the positive integers, where p(S) = 1 - S - 2*S^2.

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A290925 p-INVERT of the positive integers, where p(S) = 1 - 3*S - 2*S^2.

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A290926 p-INVERT of the positive integers, where p(S) = (1 - S^2)^2.

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A290925 p-INVERT of the positive integers, where p(S) = 1 - 3S - 2S^2.