A289780 -id:A289780 - cp's OEIS frontend

A289801 p-INVERT of the tetrahedral numbers (A000292), where p(S) = 1 - S - S^2.

1, 6, 29, 133, 597, 2661, 11856, 52878, 235986, 1053345, 4701627, 20985035, 93662073, 418038721, 1865820223, 8327671681, 37168717729, 165894342774, 740432630793, 3304756826019, 14750048986898, 65833571645931, 293833543748968, 1311460845206801

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, -31, 62, -74, 57, -28, 8, -1)

Cf. A000292, A289780.

z = 60; s = x/(1 - x)^4; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000292 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289801 *)

G.f.: (1 - 3 x + 6 x^2 - 4 x^3 + x^4)/(1 - 9 x + 31 x^2 - 62 x^3 + 74 x^4 - 57 x^5 + 28 x^6 - 8 x^7 + x^8).

a(n) = 9*a(n-1) - 31*a(n-2) + 62*a(n-3) - 74*a(n-4) + 57*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

A289802 p-INVERT of the quarter-squares (A002620), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 15, 53, 185, 643, 2234, 7764, 26988, 93819, 326149, 1133811, 3941521, 13702079, 47633109, 165588965, 575643853, 2001134880, 6956629199, 24183622175, 84070541130, 292257951771, 1015988587832, 3531923782817, 12278174929397, 42683134990390

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 (5, -5, -4, 12, -5, -4, 4, -1)

Cf. A002620, A289780.

z = 60; s = x/((1 - x)^2*(1 - x^2)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A002620 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289802 *)

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

a(n) = 5*a(n-1) - 5*a(n-2) - 4*a(n-3) + 12*a(n-4) - 5*a(n-5) - 4*a(n-6) + 4*a(n-7) - a(n-8).

A289804 p-INVERT of the even bisection (A001519) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 3, 9, 29, 96, 321, 1077, 3617, 12149, 40802, 137009, 459991, 1544169, 5183201, 17396800, 58387097, 195950657, 657602545, 2206838633, 7405775266, 24852220929, 83398067755, 279861976377, 939138581941, 3151475258656, 10575403936625, 35487807890381

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, 9, 1)

Cf. A001519, A289780, A289803.

z = 60; s = x (1 - 2*x)/(1 - 3*x + x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A001519 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289804 *)

G.f.: (-1 + 4 x - 3 x^2 - 2 x^3)/(-1 + 7 x - 15 x^2 + 9 x^3 + x^4).

a(n) = 7*a(n-1) - 15*a(n-2) + 9*a(n-3) + a(n-4).

A289805 p-INVERT of A103889, where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 9, 36, 153, 624, 2584, 10632, 43865, 180774, 745347, 3072528, 12666854, 52218790, 215273737, 887468000, 3658604277, 15082652352, 62178493132, 256331858332, 1056732372729, 4356396740786, 17959318086575, 74037587378784, 305221185520298, 1258279413185322

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 (4,2,-8,8,-2,1)

Cf. A103889, A289780.

z = 60; s = x*(x^2 - x + 2)/((x - 1)^2*(1 + x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A103889 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289805 *)
LinearRecurrence[{4,2,-8,8,-2,1},{2,9,36,153,624,2584},30] (* Harvey P. Dale, Jun 30 2021 *)

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

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

A289806 p-INVERT of (1,1,2,2,3,3,...) (A008619), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 3, 9, 26, 74, 211, 600, 1708, 4860, 13832, 39364, 112029, 318827, 907366, 2582312, 7349121, 20915193, 59523497, 169400608, 482104856, 1372044007, 3904762096, 11112739032, 31626246588, 90006565434, 256153755080, 728999555983, 2074692805003, 5904462080604

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 (3, 1, -5, 2, 2, -1)

Cf. A008619, A289780.

z = 60; s = x/((1 - x) (1 - x^2)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008619 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289806 *)

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

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

A289807 p-INVERT of (1,2,2,3,3,4,4,...) (A080513), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 13, 42, 133, 424, 1348, 4291, 13653, 43449, 138261, 439979, 1400101, 4455420, 14178073, 45117606, 143573662, 456881476, 1453892534, 4626590576, 14722780217, 46850970327, 149089600359, 474434334814, 1509749422360, 4804338875098, 15288412556740

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 (3, 2, -5, 1, 2, -1)

Cf. A080513, A289780.

z = 60; s = x (1 + x - x^2)/((1 - x)^2*(1 + x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A080513 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289807 *)
LinearRecurrence[{3,2,-5,1,2,-1},{1,4,13,42,133,424},30] (* Harvey P. Dale, Aug 20 2024 *)

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

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

A289809 p-INVERT of (1,2,1,3,1,4,1,5,...) (A133622), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 12, 38, 114, 354, 1076, 3311, 10120, 31043, 95044, 291284, 892242, 2733804, 8375092, 25659298, 78610859, 240840496, 737856017, 2260561368, 6925635380, 21217961710, 65005083598, 199154984626, 610147638720, 1869298875531, 5726938575936, 17545523113507

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 + ^2c(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 (1, 7, 1, -9, -3, 5, 1, -1)

Cf. A133622, A289780.

z = 60; s = x (1 + 2 x - x^2 - x^3)/(1 - x^2)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A133622 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289809 *)

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

A289810 p-INVERT of A081696, where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 3, 10, 37, 141, 548, 2155, 8543, 34062, 136393, 547957, 2207144, 8908901, 36021499, 145853606, 591277797, 2399421839, 9745388640, 39611178893, 161109065899, 655647568024, 2669558849029, 10874316446699, 44313536385428, 180644362403905, 736631134007651

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. A081696, A289780.

z = 60; s = x/(x + Sqrt[1 - 4*x]); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A081696 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289810 *)

A289843 p-INVERT of (1,0,2,0,3,0,4,0,5,...) (A027656), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 2, 5, 13, 29, 73, 168, 410, 962, 2317, 5483, 13131, 31193, 74509, 177311, 423025, 1007505, 2402354, 5723761, 13644587, 32514730, 77501115, 184698088, 440216833, 1049148789, 2500520812, 5959478837, 14203542282, 33851496564, 80679640434, 192285583548

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 (1, 5, -2, -6, 1, 4, 0, -1)

Cf. A027656, A289780.

z = 60; s = x/(1 - x^2)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A027656 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289843 *)

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

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

A289844 p-INVERT of A175676 (starting at n=3), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 2, 3, 7, 16, 31, 64, 134, 274, 567, 1168, 2405, 4967, 10232, 21094, 43505, 89672, 184892, 381203, 785886, 1620327, 3340606, 6887304, 14199737, 29275538, 60357622, 124439898, 256558196, 528948160, 1090536002, 2248364880, 4635470266, 9556979689, 19703689739

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) + c(1)*x + c(2)*x^2 + ... 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.

Index entries for linear recurrences with constant coefficients, signature (1, 1, 4, -2, 0, -6, 1, 0, 4, 0, 0, -1)

Cf. A175686, A289780.

z = 60; s = x/((x - 1)^2*(1 + x + x^2)^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A175676, shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289844 *)

Vec((1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12) + O(x^60)) \\ Colin Barker, Aug 13 2017

a(n) = a(n-1) + a(n-2) + 4*a(n-3) - 2*a(n-4) - 6*a(n-7) + a(n-8) + 4*a(n-10) - a(n-13).

G.f.: (1 + x - 2*x^3 + x^6) / (1 - x - x^2 - 4*x^3 + 2*x^4 + 6*x^6 - x^7 - 4*x^9 + x^12). - Colin Barker, Aug 13 2017

cp's OEIS Frontend

A289801 p-INVERT of the tetrahedral numbers (A000292), where p(S) = 1 - S - S^2.

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A289802 p-INVERT of the quarter-squares (A002620), where p(S) = 1 - S - S^2.

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A289804 p-INVERT of the even bisection (A001519) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

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A289805 p-INVERT of A103889, where p(S) = 1 - S - S^2.

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A289806 p-INVERT of (1,1,2,2,3,3,...) (A008619), where p(S) = 1 - S - S^2.

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A289807 p-INVERT of (1,2,2,3,3,4,4,...) (A080513), where p(S) = 1 - S - S^2.

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A289809 p-INVERT of (1,2,1,3,1,4,1,5,...) (A133622), where p(S) = 1 - S - S^2.

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A289810 p-INVERT of A081696, where p(S) = 1 - S - S^2.

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A289843 p-INVERT of (1,0,2,0,3,0,4,0,5,...) (A027656), where p(S) = 1 - S - S^2.