A289780 -id:A289780 - cp's OEIS frontend

A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.

0, 0, 1, 1, 2, 5, 9, 18, 36, 70, 137, 268, 522, 1017, 1980, 3852, 7492, 14568, 28321, 55051, 106999, 207952, 404134, 785366, 1526186, 2965752, 5763103, 11198858, 21761463, 42286357, 82169547, 159668921, 310262351, 602888757, 1171506956, 2276419286

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, -1, -2, -1, 1)

Cf. A000045, A289975, A289780.

z = 60; s = x^3/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,1,2,3,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289976 *)

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

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

A290997 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^6.

Original entry on oeis.org

0, 0, 1, 3, 6, 12, 27, 63, 143, 315, 684, 1479, 3195, 6903, 14932, 32361, 70266, 152775, 332397, 723330, 1573829, 3423444, 7444722, 16185939, 35185779, 76483890, 166253545, 361396431, 785621808, 1707884880, 3712912632, 8071922817, 17548551692, 38150905170

Offset: 0

Clark Kimberling, Aug 22 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 (6,-15,21,-18,9,-1).

Cf. A000012, A033453, A289780, A291000.

R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( x^2*(1-3*x+3*x^2)/(1-6*x+15*x^2-21*x^3 + 18*x^4-9*x^5+x^6) )); // G. C. Greubel, Apr 14 2023

z = 60; s = x/(1-x); p= 1 -s^3 -s^6;
Drop[CoefficientList[Series[s, {x,0,z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x,0,z}], x], 1]  (* A290997 *)
LinearRecurrence[{6,-15,21,-18,9,-1}, {0,0,1,3,6,12}, 40] (* G. C. Greubel, Apr 14 2023 *)

concat(vector(2), Vec(x^2*(1-3*x+3*x^2)/(1-6*x+15*x^2-21*x^3 + 18*x^4-9*x^5+x^6) + O(x^50))) \\ Colin Barker, Aug 22 2017

def A290997_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(1-3*x+3*x^2)/(1-6*x+15*x^2-21*x^3 + 18*x^4-9*x^5+x^6) ).list()
A290997_list(40) # G. C. Greubel, Apr 14 2023

a(n) = 6*a(n-1) - 15*a(n-2) + 21*a(n-3) - 18*a(n-4) + 9*a(n-5) - a(n-6) for n >= 7.

G.f.: x^2*(1 - 3*x + 3*x^2) / (1 - 6*x + 15*x^2 - 21*x^3 + 18*x^4 - 9*x^5 + x^6). - Colin Barker, Aug 22 2017

A289779 p-INVERT of the squares, where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 6, 28, 125, 546, 2371, 10304, 44838, 195209, 849896, 3700025, 16107530, 70121400, 305262325, 1328913506, 5785228011, 25185131956, 109639724218, 477300202625, 2077855302992, 9045633454817, 39378817534750, 171429815189636, 746294159430429, 3248880433597858

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 (7, -16, 22, -12, 5, -1)

Cf. A000290, A033453, A289780.

z = 60; s = (x + x^2)/(1 - x)^3; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000290 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289779 *)

G.f.: (1 - x + 2 x^2 + 3 x^3 - x^4)/(1 - 7 x + 16 x^2 - 22 x^3 + 12 x^4 - 5 x^5 + x^6).

a(n) = 7*a(n-1) - 16*a(n-2) + 22*a(n-3) - 12*a(n-4) + 5*a(n-5) - a(n-6).

A289781 p-INVERT of the positive Fibonacci numbers (A000045), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 3, 9, 27, 80, 237, 701, 2073, 6129, 18120, 53569, 158367, 468181, 1384083, 4091760, 12096453, 35760689, 105719157, 312537041, 923951760, 2731474161, 8075043963, 23872213729, 70573310907, 208635540400, 616788246957, 1823408134821, 5390532719313

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 (3, 1, -3, -1)

Cf. A000045, A289780, A289975, A289976, A289977.

z = 60; s = x/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289781 *)

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

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

A289785 p-INVERT of the (5^n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 7, 48, 325, 2183, 14588, 97161, 645719, 4285240, 28411789, 188257719, 1246893028, 8256349457, 54659946215, 361825274112, 2394939574997, 15851402375719, 104912178457996, 694343294142105, 4595323060281271, 30412598132972936, 201274210714545437

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 (11, -29)

Cf. A000351, A289780.

z = 60; s = x/(1 - 5*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000351 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289785 *)

Vec(x*(1 - 4*x) / (1 - 11*x + 29*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

G.f.: (1 - 4 x)/(1 - 11 x + 29 x^2).
a(n) = 11*a(n-1) - 29*a(n-2).
a(n) = (2^(-n-1)*((11-sqrt(5))^(n+1)*(-7+2*sqrt(5)) + (11+sqrt(5))^(n+1)*(7+2*sqrt(5)))) / (29*sqrt(5)). - Colin Barker, Aug 11 2017
a(n) = A081575(n+1)-4*A081575(n). - R. J. Mathar, Jul 08 2022

A289786 p-INVERT of the odd positive integers (A005408), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 5, 20, 77, 291, 1098, 4149, 15689, 59332, 224369, 848447, 3208370, 12132345, 45878109, 173486772, 656035301, 2480778763, 9380993978, 35473960589, 134143768193, 507260826084, 1918192318185, 7253589435975, 27429241169378, 103722891648049, 392225150722037

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 (5, -6, 5, 1)

Cf. A005408, A289780.

z = 60; s = x (1 + x)/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289786 *)
LinearRecurrence[{5,-6,5,1},{1,5,20,77},30] (* Harvey P. Dale, May 06 2018 *)

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

A289788 a(n) = (1/2)*A289787(n).

Original entry on oeis.org

1, 6, 31, 156, 785, 3954, 19919, 100344, 505489, 2546430, 12827791, 64620756, 325530881, 1639881066, 8260997855, 41615265264, 209639359969, 1056070674294, 5320018479679, 26799907726860, 135006120168881, 680101314855906, 3426050596003631, 17258932500172776

Offset: 0

Clark Kimberling, Aug 10 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -6, 6, -1)

Cf. A289780, A289787.

z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)
u/2 (* A289788 *)

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

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

A289789 p-INVERT of A016777, where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 6, 26, 111, 460, 1905, 7910, 32880, 136675, 568050, 2360825, 9811650, 40777750, 169474875, 704348000, 2927312625, 12166086250, 50562982500, 210142784375, 873366003750, 3629761440625, 15085506018750, 62696266831250, 260569441284375, 1082942209562500

Offset: 0

Clark Kimberling, Aug 11 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, 5, 5)

Cf. A016777, A289780.

z = 60; s = x (1 + 2*x)/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A016777 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289789 *)

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

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

A289790 p-INVERT of A016789, where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 13, 72, 385, 2056, 11000, 58872, 315065, 1686086, 9023167, 48287964, 258415702, 1382925814, 7400803253, 39605804028, 211952630117, 1134276112400, 6070140759292, 32484690838716, 173843602765153, 930333564584074, 4978731041147699, 26643951936925764

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 (6, -5, 8, 1)

Cf. A016789, A289780.

z = 60; s = x (2 + x)/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A016789 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289790 *)

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

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

A289795 p-INVERT of (3n), where p(S) = 1 - S - S^2.

Original entry on oeis.org

3, 24, 162, 1083, 7260, 48681, 326406, 2188536, 14674041, 98388840, 659693103, 4423214952, 29657473194, 198852130383, 1333295304660, 8939689838877, 59940250397646, 401896898269128, 2694702070258437, 18067865859946320, 121144292846335179, 812267469938047224

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, -3, 7, -1)

Cf. A008585, A289780, A289796.

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.: 3 (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).

cp's OEIS Frontend

A289976 p-INVERT of (0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by two zeros, where p(S) = 1 - S - S^2.

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A290997 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^3 - S^6.

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A289779 p-INVERT of the squares, where p(S) = 1 - S - S^2.

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A289781 p-INVERT of the positive Fibonacci numbers (A000045), where p(S) = 1 - S - S^2.

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A289785 p-INVERT of the (5^n), where p(S) = 1 - S - S^2.

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A289786 p-INVERT of the odd positive integers (A005408), where p(S) = 1 - S - S^2.

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A289788 a(n) = (1/2)*A289787(n).

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A289789 p-INVERT of A016777, where p(S) = 1 - S - S^2.

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