A290890 -id:A290890 - cp's OEIS frontend

A290929 p-INVERT of the positive integers, where p(S) = (1 - S)(1 - S^2).

1, 4, 13, 39, 114, 330, 948, 2703, 7655, 21554, 60389, 168468, 468199, 1296826, 3581185, 9862749, 27096216, 74277342, 203200986, 554869701, 1512575195, 4116813032, 11188568267, 30367047720, 82316338381, 222875101936, 602784607477, 1628612506131

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, -18, 23, -18, 7, -1)

Cf. A000027, A033453, A290890.

I:=[1,4,13,39,114,330]; [n le 6 select I[n] else 7*Self(n-1)-18*Self(n-2)+23*Self(n-3)-18*Self(n-4)+7*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Aug 20 2017

z = 60; s = x/(1 - x)^2; p = (1 - s)(1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290929 *)
LinearRecurrence[{7, -18, 23, -18, 7, -1}, {1, 4, 13, 39, 114, 330}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

Vec((1 - 3*x + 3*x^2 - 3*x^3 + x^4) / ((1 - 3*x + x^2)^2*(1 - x + x^2)) + O(x^30)) \\ Colin Barker, Aug 19 2017

a(n) = 7*a(n-1) - 18*a(n-2) + 23*a(n-3) - 18*a(n-4) + 7*a(n-5) - a(n-6).

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

A290930 p-INVERT of the positive integers, where p(S) = (1 - S^2)(1 - 2*S^2).

Original entry on oeis.org

0, 3, 12, 37, 116, 372, 1188, 3763, 11860, 37261, 116760, 365056, 1139224, 3549635, 11045804, 34335421, 106633804, 330916268, 1026277180, 3181108619, 9855901108, 30524529485, 94506627952, 292521594048, 905220237168, 2800700318291, 8663793207244

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, -25, 44, -54, 44, -25, 8, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = (1 - s^2)(1 - 2s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290930 *)

G.f.: (3 x - 12 x^2 + 16 x^3 - 12 x^4 + 3 x^5)/(1 - 8 x + 25 x^2 - 44 x^3 + 54 x^4 - 44 x^5 + 25 x^6 - 8 x^7 + x^8).

a(n) = 8*a(n-1) - 25*a(n-2) + 44*a(n-3) - 54*a(n-4) + 44*a(n-5) - 25*a(n-6) + 8*a(n-7) - a(n-8).

A291025 p-INVERT of the positive integers, where p(S) = 1 - 3*S + S^2.

Original entry on oeis.org

3, 14, 62, 273, 1200, 5271, 23146, 101626, 446181, 1958880, 8600043, 37756502, 165760934, 727733433, 3194937360, 14026596927, 61580365906, 270353629378, 1186921889997, 5210892012480, 22877154557139, 100436585338334, 440942410322894, 1935850452749409

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, -13, 7, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291025 *)
LinearRecurrence[{7,-13,7,-1},{3,14,62,273},30] (* Harvey P. Dale, Jun 22 2022 *)

G.f.: (3 - 7 x + 3 x^2)/(1 - 7 x + 13 x^2 - 7 x^3 + x^4).

a(n) = 7*a(n-1) - 13*a(n-2) + 7*a(n-3) - a(n-4).

A291026 p-INVERT of the positive integers, where p(S) = 1 - 4*S + S^2.

Original entry on oeis.org

4, 23, 128, 711, 3948, 21920, 121700, 675673, 3751296, 20826953, 115629868, 641969344, 3564171060, 19788040311, 109861881472, 609945846247, 3386378699324, 18800948912352, 104381615697460, 579519775642745, 3217455182279552, 17863096800262569

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,-15,8,-1).

Cf. A000027, A033453, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291026 *)
LinearRecurrence[{8,-15,8,-1},{4,23,128,711},30] (* Harvey P. Dale, May 18 2024 *)

G.f.: (4 - 9 x + 4 x^2)/(1 - 8 x + 15 x^2 - 8 x^3 + x^4).

a(n) = 8*a(n-1) - 15*a(n-2) + 8*a(n-3) - a(n-4).

A291027 p-INVERT of the positive integers, where p(S) = 1 - 5*S + S^2.

Original entry on oeis.org

5, 34, 226, 1501, 9968, 66195, 439582, 2919134, 19385099, 128730656, 854861845, 5676882210, 37698479330, 250344342349, 1662462010576, 11039913707011, 73312769785118, 486848208799710, 3233013554202907, 21469477452590144, 142572387761274149, 946780646936461346

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 (9, -17, 9, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 5 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291027 *)

G.f.: (5 - 11 x + 5 x^2)/(1 - 9 x + 17 x^2 - 9 x^3 + x^4).

a(n) = 9*a(n-1) - 17*a(n-2) + 9*a(n-3) - a(n-4).

A291028 p-INVERT of the positive integers, where p(S) = 1 - 6*S + S^2.

Original entry on oeis.org

6, 47, 362, 2787, 21456, 165180, 1271644, 9789793, 75367038, 580215573, 4466808294, 34387867640, 264736107506, 2038079457267, 15690220398162, 120791667500967, 929918545909756, 7159007901103540, 55113853093361544, 424295774604244773, 3266454697733704038

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 (10,-19,10,-1).

Cf. A000027, A290890.

I:=[6,47,362,2787]; [n le 4 select I[n] else 10*Self(n-1)-19*Self(n-2)+10*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Aug 20 2017

z = 60; s = x/(1 - x)^2; p = 1 - 6 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291028 *)
LinearRecurrence[{10, -19, 10, -1}, {6, 47, 362, 2787}, 40] (* Vincenzo Librandi, Aug 20 2017 *)

G.f.: (6 - 13 x + 6 x^2)/(1 - 10 x + 19 x^2 - 10 x^3 + x^4).

a(n) = 10*a(n-1) - 19*a(n-2) + 10*a(n-3) - a(n-4).

A291029 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3.

Original entry on oeis.org

1, 4, 15, 55, 198, 706, 2510, 8923, 31737, 112918, 401799, 1429744, 5087461, 18102522, 64413263, 229198253, 815544198, 2901909494, 10325718678, 36741486569, 130735386073, 465189151460, 1655259161187, 5889825416864, 20957469541173, 74571909803996

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, -18, 25, -18, 7, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291029 *)

G.f.: (1 - 3 x + 5 x^2 - 3 x^3 + x^4)/(1 - 7 x + 18 x^2 - 25 x^3 + 18 x^4 - 7 x^5 + x^6).

a(n) = 7*a(n-1) - 18*a(n-2) + 25*a(n-3) - 18*a(n-4) + 7*a(n-5) - a(n-6).

A291031 p-INVERT of the positive integers, where p(S) = 1 - 3S + 2S^3.

Original entry on oeis.org

3, 15, 70, 321, 1461, 6624, 29967, 135399, 611318, 2758881, 12447753, 56154744, 253306119, 1142572767, 5153589754, 23244956169, 104843981505, 472885383744, 2132882300571, 9620044596687, 43389716584682, 195702453488433, 882684641446989, 3981207177094608

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 (9, -27, 36, -27, 9, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 3 s + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291031 *)

G.f.: (3 - 12 x + 16 x^2 - 12 x^3 + 3 x^4)/(1 - 9 x + 27 x^2 - 36 x^3 + 27 x^4 - 9 x^5 + x^6).

a(n) = 9*a(n-1) - 27*a(n-2) + 36*a(n-3) - 27*a(n-4) + 90*a(n-5) - a(n-6).

A291032 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3 + S^4.

Original entry on oeis.org

1, 4, 15, 54, 188, 645, 2208, 7570, 25982, 89190, 306095, 1050268, 3603276, 12361763, 42409154, 145491117, 499126660, 1712311759, 5874263702, 20152234481, 69134134820, 237171010852, 813636681973, 2791253840066, 9575645985794, 32850107071454, 112695214040224

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 (9, -33, 68, -87, 68, -33, 9, -1)

Cf. A000027, A290890.

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] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291032 *)

G.f.: (1 - 5 x + 12 x^2 - 17 x^3 + 12 x^4 - 5 x^5 + x^6)/(1 - 9 x + 33 x^2 - 68 x^3 + 87 x^4 - 68 x^5 + 33 x^6 - 9 x^7 + x^8).

a(n) = 9*a(n-1) - 33*a(n-2) + 68*a(n-3) - 87*a(n-4) + 68*a(n-5) - 33*a(n-6) + 9*a(n-7) - a(n-8).

A291034 p-INVERT of the positive integers, where p(S) = 1 - 7*S.

Original entry on oeis.org

7, 63, 560, 4977, 44233, 393120, 3493847, 31051503, 275969680, 2452675617, 21798110873, 193730322240, 1721774789287, 15302242781343, 135998410242800, 1208683449403857, 10742152634391913, 95470690260123360, 848494059706718327, 7540975847100341583

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 (9, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 7 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291034 *)

G.f.: 7/(1 - 9 x + x^2).
a(n) = 9*a(n-1) - a(n-2).
a(n) = 7*A018913(n) for n >= 1.

cp's OEIS Frontend

A290929 p-INVERT of the positive integers, where p(S) = (1 - S)(1 - S^2).

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A290930 p-INVERT of the positive integers, where p(S) = (1 - S^2)(1 - 2*S^2).

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A291025 p-INVERT of the positive integers, where p(S) = 1 - 3*S + S^2.

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A291026 p-INVERT of the positive integers, where p(S) = 1 - 4*S + S^2.

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A291027 p-INVERT of the positive integers, where p(S) = 1 - 5*S + S^2.

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A291028 p-INVERT of the positive integers, where p(S) = 1 - 6*S + S^2.

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A291029 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3.

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A291031 p-INVERT of the positive integers, where p(S) = 1 - 3*S + 2*S^3.

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A291031 p-INVERT of the positive integers, where p(S) = 1 - 3S + 2S^3.