A291000 -id:A291000 - cp's OEIS frontend

A291013 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)^3.

0, 3, 6, 15, 36, 85, 198, 456, 1040, 2352, 5280, 11776, 26112, 57600, 126464, 276480, 602112, 1306624, 2826240, 6094848, 13107200, 28114944, 60162048, 128450560, 273678336, 581959680, 1235222528, 2617245696, 5536481280, 11693719552, 24662507520, 51942260736

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..186
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A000012, A033453, A289780, A291000.

[0,3,6] cat [2^(n-6)*(60 +17*n +n^2): n in [3..40]]; // G. C. Greubel, Jun 05 2023

z = 60; s = x/(1-x); p = (1 - s^2)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291013 *)
LinearRecurrence[{6,-12,8}, {0,3,6,15,36,85}, 41] (* G. C. Greubel, Jun 05 2023 *)

concat(0, Vec(x*(3 -12*x +15*x^2 -6*x^3 +x^4)/(1-2*x)^3 + O(x^50))) \\ Colin Barker, Aug 23 2017

[(2^(n-2)*(60 +17*n +n^2) -15*int(n==0) + 9*int(n==1))//16 for n in range(41)] # G. C. Greubel, Jun 05 2023

G.f.: x*(3 - 12*x + 15*x^2 - 6*x^3 + x^4)/(1 - 2*x)^3.
a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3) for n >= 7.
a(n) = 2^(n-6) * (60 + 17*n + n^2) for n>2. - Colin Barker, Aug 23 2017
E.g.f.: -(15/16) + (9/16)*x - x^2/16 + (1/16)*(15 +9*x +x^2)*exp(2*x). - G. C. Greubel, Jun 05 2023

A291014 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

Original entry on oeis.org

0, 0, 2, 6, 12, 23, 48, 105, 228, 486, 1026, 2161, 4548, 9555, 20026, 41874, 87384, 182043, 378648, 786429, 1631120, 3378750, 6990510, 14447045, 29826156, 61516455, 126761190, 260978922, 536870916, 1103567983, 2266788288, 4652881233, 9544371772, 19565962134

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 (6,-15,22,-21,12,-4).

Cf. A000012, A010892, A033453, A099254, A289780, A291000.

R:=PowerSeriesRing(Integers(), 50); [0,0] cat Coefficients(R!( x^2*(2-6*x+6*x^2-3*x^3)/((1-2*x)*(1-x+x^2))^2 )); // G. C. Greubel, Jun 05 2023

z = 60; s = x/(1-x); p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291014 *)
LinearRecurrence[{6,-15,22,-21,12,-4}, {0,0,2,6,12,23}, 50] (* G. C. Greubel, Jun 05 2023 *)

def A291014_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(2-6*x+6*x^2-3*x^3)/((1-2*x)*(1-x+x^2))^2 ).list()
A291014_list(50) # G. C. Greubel, Jun 05 2023

G.f.: x^2*(2 - 6*x + 6*x^2 - 3*x^3)/( (1-2*x)*(1-x+x^2) )^2.
a(n) = 6*a(n-1) - 15*a(n-2) + 22*a(n-3) - 21*a(n-4) + 12*a(n-5) - 4*a(n-6) for n >= 7.
a(n) = (1/9)*( 2^(n-1)*(n + 8) - 3*(A099254(n) - A099254(n-1)) - A010892(n) - 5*A010892(n-1) ). - G. C. Greubel, Jun 05 2023

A291015 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

Original entry on oeis.org

2, 7, 23, 75, 244, 793, 2576, 8366, 27167, 88215, 286439, 930072, 3019941, 9805712, 31838986, 103380599, 335674791, 1089929347, 3538978588, 11490991649, 37311016064, 121148109014, 393365440335, 1277249563655, 4147203285279, 13465884484800, 43723452275981

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,-6,1).

Cf. A000012, A033453, A289780, A291000.

I:=[2,7,23]; [n le 3 select I[n] else 5*Self(n-1) -6*Self(n-2) +Self(n-3): n in [1..50]]; // G. C. Greubel, Jun 06 2023

z = 60; s = x/(1-x); p = (1 - s^3)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291015 *)
LinearRecurrence[{5,-6,1}, {2,7,23}, 50] (* G. C. Greubel, Jun 06 2023 *)

@CachedFunction
def a(n): # a = A291015
    if (n<3): return (2,7,23)[n]
    else: return 5*a(n-1) - 6*a(n-2) + a(n-3)
[a(n) for n in range(51)] # G. C. Greubel, Jun 06 2023

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

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

A291016 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 4*S + S^2.

Original entry on oeis.org

4, 19, 90, 426, 2016, 9540, 45144, 213624, 1010880, 4783536, 22635936, 107114400, 506870784, 2398538304, 11350005120, 53708800896, 254152774656, 1202663842560, 5691066407424, 26930415389184, 127436093890560, 603034071008256, 2853587862706176

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 (6,-6).

Cf. A000012, A033453, A289780, A291000.

I:=[4,19]; [n le 2 select I[n] else 6*(Self(n-1)-Self(n-2)): n in [1..40]]; // G. C. Greubel, Jun 06 2023

z = 60; s = x/(1-x); p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291016 *)
LinearRecurrence[{6,-6}, {4,19}, 40] (* G. C. Greubel, Jun 06 2023 *)

A291016=BinaryRecurrenceSequence(6,-6,4,19)
[A291016(n) for n in range(51)] # G. C. Greubel, Jun 06 2023

G.f.: (4 - 5*x)/(1 - 6*x + 6*x^2).
a(n) = 6*a(n-1) - 6*a(n-2) n >= 3.
From G. C. Greubel, Jun 06 2023: (Start)
a(n) = ((3+sqrt(3))^(n+4) - (3-sqrt(3))^(n+4))/(72*sqrt(3)).
a(n) = 6^(n/2)*(4*ChebyshevU(n, sqrt(3/2)) - (5/sqrt(6))*ChebyshevU(n - 1, sqrt(3/2))).
E.g.f.: exp(3*x)*(4*cosh(sqrt(3)*x) + (7/sqrt(3))*sinh(sqrt(3)*x)). (End)

A291017 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 5 S + S^2.

Original entry on oeis.org

5, 29, 168, 973, 5635, 32634, 188993, 1094513, 6338640, 36708889, 212591743, 1231179978, 7130117645, 41292563669, 239137122168, 1384911909493, 8020423511275, 46448581212474, 268997103908393, 1557839658871433, 9021897884741280, 52248407581088929

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 (7, -7)

Cf. A000012, A289780, A291000.

z = 60; s = x/(1 - x); p = 1 - 5 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291017 *)

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

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

A291018 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6 S + S^2.

Original entry on oeis.org

6, 41, 280, 1912, 13056, 89152, 608768, 4156928, 28385280, 193826816, 1323532288, 9037643776, 61712891904, 421401985024, 2877512744960, 19648886079488, 134170986676224, 916176804773888, 6256046544781312, 42718957920059392, 291703291002224640

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 (8, -8)

Cf. A000012, A289780, A291000.

z = 60; s = x/(1 - x); p = 1 - 6 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291018 *)

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

a(n) = 8*a(n-1) - 8*a(n-2) n >= 3.

A291019 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.

Original entry on oeis.org

1, 3, 9, 25, 68, 185, 504, 1373, 3739, 10180, 27714, 75445, 205376, 559064, 1521840, 4142609, 11276581, 30695881, 83556891, 227449066, 619135745, 1685339900, 4587637263, 12487934387, 33993205996, 92532358762, 251880840375, 685640764594, 1866371634554

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, -8, 6, -3)

Cf. A000012, A289780, A291000.

z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 + s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291019 *)

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

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

A291020 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.

Original entry on oeis.org

1, 3, 9, 27, 79, 228, 656, 1889, 5445, 15701, 45275, 130544, 376388, 1085199, 3128841, 9021083, 26009635, 74991112, 216214692, 623391005, 1797363157, 5182163781, 14941232871, 43078615236, 124204414928, 358106605227, 1032494220505, 2976890957419

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 (6, -13, 14, -7)

Cf. A000012, A289780, A291000.

z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 - s^4 + s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291020 *)

G.f.: -((-1 + 3 x - 4 x^2 + 2 x^3 + x^4)/(1 - 6 x + 13 x^2 - 14 x^3 + 7 x^4) ).

a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 7*a(n-4) for n >= 6.

A291021 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4 + S^5.

Original entry on oeis.org

1, 3, 9, 25, 67, 178, 472, 1249, 3297, 8685, 22843, 60014, 157540, 413289, 1083693, 2840521, 7443331, 19500394, 51079696, 133782385, 350354841, 917456901, 2402365387, 6290338310, 16470047644, 43122600825, 112903347237, 295598625697, 773914899475

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 (6, -13, 14, -9, 2)

Cf. A000012, A289780, A291000.

z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 + s^4 + s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291021 *)

G.f.: (-1 + 3 x - 4 x^2 + 4 x^3 - x^4)/(-1 + 6 x - 13 x^2 + 14 x^3 - 9 x^4 + 2 x^5).

a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 9*a(n-4) + 2*a(n-5) for n >= 5.

A291023 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 3 S^2 + 2 S^3.

Original entry on oeis.org

0, 3, 4, 12, 24, 56, 120, 264, 568, 1224, 2616, 5576, 11832, 25032, 52792, 111048, 233016, 487880, 1019448, 2126280, 4427320, 9204168, 19107384, 39612872, 82021944, 169636296, 350457400, 723284424, 1491308088, 3072094664, 6323146296, 13004206536, 26724240952

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..1000
Index entries for linear recurrences with constant coefficients, signature (3, 0, -4)

Cf. A000012, A289780, A291000.

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]  (* A291023 *)

concat(0, Vec(x*(3 - 5*x) / ((1 + x)*(1 - 2*x)^2) + O(x^40))) \\ Colin Barker, Aug 24 2017

G.f.: (3 x - 5 x^2)/((1 + x) (-1 + 2 x)^2).
a(n) = 3*a(n-1) - 4*a(n-3) for n >= 4.
a(n) = (16*((-1)^(1+n) + 2^n) + 3*2^n*n) / 18. - Colin Barker, Aug 24 2017

cp's OEIS Frontend

A291013 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)^3.

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A291014 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

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A291015 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^3)^2.

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A291016 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 4*S + S^2.

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A291017 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 5 S + S^2.

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A291018 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6 S + S^2.

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A291019 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.

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