cp's OEIS Frontend

6, 31, 146, 652, 2816, 11896, 49496, 203752, 832376, 3381736, 13683896, 55206952, 222242936, 893219176, 3585623096, 14380739752, 57637717496, 230895178216, 924613703096, 3701553914152, 14815513224056, 59289946122856, 237243465219896, 949224905162152

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 (9,-26,24).

[(2^n-16*3^n+27*4^n)/2: n in [0..40]]; // G. C. Greubel, Apr 27 2023

z = 60; s = x/(1-x); p = (1-s)*(1-2*s)*(1-3*s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291002 *)
LinearRecurrence[{9,-26,24}, {6,31,146}, 41] (* G. C. Greubel, Apr 27 2023 *)

[(2^n-16*3^n+27*4^n)/2 for n in range(41)] # G. C. Greubel, Apr 27 2023

G.f.: (6 - 23*x + 23*x^2)/(1 - 9*x + 26*x^2 - 24*x^3).
a(n) = 9*a(n-1) - 26*a(n-2) + 24*a(n-3) for n >= 4.
a(n) = (2^n - 16*3^n + 27*4^n) / 2. - Colin Barker, Aug 23 2017
E.g.f.: (1/2)*(exp(2*x) - 16*exp(3*x) + 27*exp(4*x)). - G. C. Greubel, Apr 27 2023

A291003 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)(1 - 2S)(1 - 3S)(1 - 4*S).

10, 75, 490, 2956, 16944, 93800, 506600, 2687256, 14064904, 72873880, 374671560, 1914880856, 9741440264, 49378177560, 249583291720, 1258711575256, 6336814854024, 31857331730840, 159980377179080, 802678826106456, 4024508089842184, 20167014882109720

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..999
Index entries for linear recurrences with constant coefficients, signature (14,-71,154,-120).

[(-2^n +48*3^n -243*4^n +256*5^n)/6: n in [0..40]]; // G. C. Greubel, May 23 2023

z = 60; s = x/(1-x); p = (1 - s)(1 - 2 s)(1 - 3 s)(1 - 4 s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291003 *)
LinearRecurrence[{14,-71,154,-120}, {10,75,490,2956}, 41] (* G. C. Greubel, May 23 2023 *)

[(-2^n +48*3^n -243*4^n +256*5^n)//6 for n in range(41)] # G. C. Greubel, May 23 2023

G.f.: (10 - 65*x + 150*x^2 - 119*x^3)/(1 - 14*x + 71*x^2 - 154*x^3 + 120*x^4).
a(n) = 14*a(n-1) - 71*a(n-2) + 154*a(n-3) - 120*a(n-4) for n >= 5.
a(n) = (-2^n + 16*3^(1+n) - 243*4^n + 256*5^n) / 6. - Colin Barker, Aug 23 2017

A291004 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 3*S)^2.

6, 33, 168, 816, 3840, 17664, 79872, 356352, 1572864, 6881280, 29884416, 128974848, 553648128, 2365587456, 10066329600, 42681237504, 180388626432, 760209211392, 3195455668224, 13400297963520, 56075093016576, 234195976716288, 976366325465088, 4063794976260096

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,-16).

[3*(4^(n-1)*(3*n+8)): n in [0..30]]; // Vincenzo Librandi, Aug 27 2017

z = 60; s = x/(1-x); p = (1 - 3 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291004 *)
LinearRecurrence[{8, -16}, {6, 33}, 25] (* Vincenzo Librandi, Aug 27 2017 *)

[3*4^n*(3*n+8)//4 for n in range(41)] # G. C. Greubel, Jun 01 2023

G.f.: 3*(2 - 5*x)/(1 - 4*x)^2.
a(n) = 8*a(n-1) - 16*a(n-2) for n >= 3.
a(n) = 3*A006234(n+3) for n >= 0.
a(n) = 3 * 4^(n-1) * (3*n+8). - Colin Barker, Aug 23 2017

A291005 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 2 S - 2 S^3.

2, 6, 20, 68, 230, 774, 2600, 8732, 29330, 98526, 330980, 1111868, 3735110, 12547374, 42150440, 141596132, 475664450, 1597901526, 5367837140, 18032197268, 60575633990, 203491974774, 683591422280, 2296391457932, 7714277207570, 25914602943726, 87055031555300

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,-7,5).

Cf. A000012, A289780, A291000, A291337.

I:=[2,6,20]; [n le 3 select I[n] else 5*Self(n-1)-7*Self(n-2)+5*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Aug 27 2017

z = 60; s = 1 - 2 s - 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291005 *)
u / 2  (* A291337 *)
LinearRecurrence[{5, -7, 5}, {2, 6, 20}, 30] (* Vincenzo Librandi, Aug 27 2017 *)

def A291005_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 2*(1-2*x+2*x^2)/(1-5*x+7*x^2-5*x^3) ).list()
A291005_list(30) # G. C. Greubel, Jun 01 2023

G.f.: 2*(1 - 2*x + 2*x^2)/(1 - 5*x + 7*x^2 - 5*x^3).
a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) for n >= 4.
a(n) = 2*A291337(n) for n >= 0.

A291006 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

1, 3, 9, 27, 80, 235, 688, 2013, 5891, 17244, 50482, 147791, 432672, 1266680, 3708288, 10856241, 31782309, 93044665, 272394011, 797450348, 2334585333, 6834643282, 20008841823, 58577124509, 171488162320, 502042223184, 1469759722591, 4302812676894

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..999
Index entries for linear recurrences with constant coefficients, signature (5,-8,6,-1).

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-2*x+2*x^2)/(1-5*x+8*x^2-6*x^3+x^4) )); // G. C. Greubel, Jun 01 2023

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]  (* A291006 *)
LinearRecurrence[{5,-8,6,-1}, {1,3,9,27}, 41] (* G. C. Greubel, Jun 01 2023 *)

def A291006_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-2*x+2*x^2)/(1-5*x+8*x^2-6*x^3+x^4) ).list()
A291006_list(40) # G. C. Greubel, Jun 01 2023

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

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

A291007 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 - S^5.

1, 3, 9, 27, 81, 242, 720, 2137, 6337, 18789, 55715, 165232, 490058, 1453493, 4311025, 12786359, 37923789, 112480082, 333610072, 989469949, 2934716101, 8704215281, 25816251319, 76569665176, 227101665034, 673571786617, 1997779058053, 5925309279179

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,-7,2).

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1 -3*x +4*x^2 -2*x^3 +x^4)/(1 -6*x +13*x^2 -14*x^3 +7*x^4 -2*x^5) )); // G. C. Greubel, Jun 01 2023

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]  (* A291007 *)
LinearRecurrence[{6,-13,14,-7,2},{1,3,9,27,81},30] (* Harvey P. Dale, Apr 07 2019 *)

Vec((1 -3*x +4*x^2 -2*x^3 +x^4)/(1 -6*x +13*x^2 -14*x^3 +7*x^4 - 2*x^5) + O(x^30)) \\ Colin Barker, Aug 23 2017

def A291007_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1 -3*x +4*x^2 -2*x^3 +x^4)/(1 -6*x +13*x^2 -14*x^3 +7*x^4 -2*x^5) ).list()
A291007_list(40) # G. C. Greubel, Jun 01 2023

a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 7*a(n-4) + 2*a(n-5) for n >= 6.

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 - 2*x^5). - Colin Barker, Aug 23 2017

A291009 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)(1 - 3S).

4, 17, 70, 284, 1144, 4592, 18400, 73664, 294784, 1179392, 4718080, 18873344, 75495424, 301985792, 1207951360, 4831821824, 19327320064, 77309345792, 309237514240, 1236950319104, 4947801800704, 19791208251392, 79164835102720, 316659344605184, 1266637386809344

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,-8).

[2^(n-1)*(9*2^n -1): n in [0..40]]; // G. C. Greubel, Jun 04 2023

z = 60; s = x/(1-x); p = (1-s)(1-3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291009 *)
LinearRecurrence[{6,-8}, {4,17}, 40] (* G. C. Greubel, Jun 04 2023 *)

Vec((4-7*x)/((1-2*x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017

A291009=BinaryRecurrenceSequence(6,-8,4,17)
[A291009(n) for n in range(41)] # G. C. Greubel, Jun 04 2023

a(n) = 6*a(n-1) - 8*a(n-2) for n >= 3.
From Colin Barker, Aug 23 2017: (Start)
G.f.: (4 - 7*x) / ((1 - 2*x)*(1 - 4*x)).
a(n) = 2^(n-1) * (9*2^n - 1).
(End)

A291010 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - 2S)(1 - 3*S).

5, 24, 108, 468, 1980, 8244, 33948, 138708, 563580, 2280564, 9200988, 37040148, 148869180, 597602484, 2396787228, 9606280788, 38482518780, 154102262004, 616925608668, 2469252116628, 9881657512380, 39540577187124, 158204150161308, 632942124883668

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,-12).

[36*(4^(n-1)-3^(n-2)): n in [0..40]]; // G. C. Greubel, Jun 04 2023

z = 60; s = x/(1-x); p = (1-2s)(1-3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* this sequence *)
LinearRecurrence[{7,-12}, {5,24}, 40] (* G. C. Greubel, Jun 04 2023 *)

Vec((5-11*x)/((1-3*x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017

A291010=BinaryRecurrenceSequence(7,-12,5,24)
[A291010(n) for n in range(41)] # G. C. Greubel, Jun 04 2023

G.f.: (5 - 11*x)/(1 - 7*x + 12*x^2).
a(n) = 7*a(n-1) - 12*a(n-2) for n >= 3.
a(n) = 9*4^n - 4*3^n. - Colin Barker, Aug 23 2017
E.g.f.: 9*exp(4*x) - 4*exp(3*x). - G. C. Greubel, Jun 04 2023

A291011 p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S)^2 * (1 - 2*S).

4, 15, 52, 172, 552, 1736, 5384, 16536, 50440, 153112, 463176, 1397720, 4210568, 12668568, 38083528, 114414424, 343587336, 1031482904, 3095956040, 9291013848, 27879595144, 83652416920, 250985562312, 753015407192, 2259167856392, 6777755227416, 20333785775944

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..999
Index entries for linear recurrences with constant coefficients, signature (7,-16,12).