A289780 -id:A289780 - cp's OEIS frontend

A289803 p-INVERT of the even bisection (A001906) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

1, 5, 23, 103, 456, 2009, 8833, 38803, 170399, 748176, 3284833, 14421533, 63314735, 277968871, 1220356440, 5357681369, 23521603225, 103265890987, 453363808127, 1990383615264, 8738295434881, 38363361811637, 168425013526727, 739429075564711, 3246283590352104

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
Rigoberto Flórez, Javier González, Mateo Matijasevick, Cristhian Pardo, José Luis Ramírez, Lina Simbaqueba, and Fabio Velandia, Lattice paths in corridors and cyclic corridors, Contrib. Disc. Math. (2024) Vol. 19. No. 2, 36-55. See p. 11.
Index entries for linear recurrences with constant coefficients, signature (7,-13,7,-1).

Cf. A001906, A289780, A298804.

z = 60; s = x/(1 - 3*x + x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A001906 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289803 *)

G.f.: (1 - 2 x + 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).

A289846 p-INVERT of (1,0,1,0,1,0,1,0,1,...) (A059841), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 2, 4, 9, 18, 39, 80, 170, 353, 744, 1553, 3262, 6824, 14313, 29970, 62823, 131596, 275782, 577777, 1210704, 2536657, 5315210, 11136700, 23334969, 48893202, 102446199, 214654136, 449764562, 942387569, 1974580920, 4137324929, 8668915558, 18163921856

Offset: 0

Clark Kimberling, Aug 14 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, 3, -1, -1)

Cf. A059841, A289780.

z = 60; s = x/(1 - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A059841 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289846 *)

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

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

A289847 p-INVERT of the primes (A000040), where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 11, 53, 253, 1205, 5740, 27336, 130200, 620129, 2953634, 14067934, 67004505, 319137367, 1520027050, 7239773429, 34482491204, 164237487721, 782250685197, 3725800625523, 17745705518523, 84521448139914, 402569240665810, 1917406730442806, 9132462688572345

Offset: 0

Clark Kimberling, Aug 14 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. A000040, A030017 ("INVERT" applied to the primes), A289928.

z = 60; s = Sum[Prime[k] x^k, {k, 1, z}]; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000040 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1](* A289847 *)

A289925 p-INVERT of the lower Wythoff sequence (A000201), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 5, 19, 72, 265, 979, 3618, 13374, 49447, 182807, 675843, 2498594, 9237316, 34150422, 126254366, 466763346, 1725627604, 6379658213, 23585644300, 87196304028, 322365390600, 1191787269208, 4406046481612, 16289186920873, 60221246337260, 222638399818776

Offset: 0

Clark Kimberling, Aug 14 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. A000201, A289926.

z = 60; r = GoldenRatio; s = Sum[Floor[k*r] x^k, {k, 1, z}]; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000201 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1]  (* A289925 *)

A289926 p-INVERT of the upper Wythoff sequence (A001950), where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 13, 71, 376, 1991, 10564, 56051, 297384, 1577797, 8371133, 44413759, 235640987, 1250213362, 6633113651, 35192550325, 186717077925, 990643385291, 5255942989944, 27885853904294, 147950776760552, 784965467407868, 4164701250741605, 22096177765889378

Offset: 0

Clark Kimberling, Aug 14 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. A001950, A289925.

z = 60; r = 1 + GoldenRatio; s = Sum[Floor[k*r] x^k, {k, 1, z}]; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A001950 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1]  (* A289926 *)

A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 7, 12, 23, 41, 77, 140, 258, 470, 861, 1570, 2867, 5225, 9526, 17352, 31607, 57547, 104766, 190684, 347029, 631476, 1148985, 2090427, 3803044, 6918379, 12585209, 22892932, 41641932, 75744383, 137772396, 250592150, 455792833, 829016539

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, -2, 0, -1, -1, 0, 1)

Cf. A000045, A289975, A289780.

z = 60; s = x^4/(1 - x - x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,0,1,2,3,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A289977 *)
LinearRecurrence[{2,1,-2,0,-1,-1,0,1},{0,0,0,1,1,2,3,7},40] (* Harvey P. Dale, Jul 14 2018 *)

concat(vector(3), Vec(x^3*(1 - x)*(1 - x^2 - x^3) / (1 - 2*x - x^2 + 2*x^3 + x^5 + x^6 - x^8) + O(x^50))) \\ Colin Barker, Aug 24 2017

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

G.f.: x^3*(1 - x)*(1 - x^2 - x^3) / (1 - 2*x - x^2 + 2*x^3 + x^5 + x^6 - x^8). - Colin Barker, Aug 24 2017

A290992 p-INVERT of (0,0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by two zeros, where p(S) = 1 - S - S^2.

Original entry on oeis.org

0, 0, 0, 1, 2, 3, 4, 7, 14, 27, 48, 82, 140, 242, 420, 726, 1250, 2153, 3720, 6446, 11184, 19408, 33676, 58431, 101378, 175861, 304988, 528800, 916714, 1589091, 2754612, 4775074, 8277754, 14350253, 24878304, 43131381, 74777890, 129645147, 224770632

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.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,0,-2,1,0,1).

Cf. A000027, A033453, A289780, A290890, A290990, A290991.

R:=PowerSeriesRing(Integers(), 60); [0,0,0] cat Coefficients(R!( x^3*(1-2*x+x^2+x^4)/(1-4*x+6*x^2-4*x^3+2*x^5-x^6-x^8) )); // G. C. Greubel, Apr 12 2023

z = 60; s = x^4/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* 0,0,0,1,2,3,4,5,... *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290992 *)

concat(vector(3), Vec(x^3*(1 - 2*x + x^2 + x^4) / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^5 - x^6 - x^8) + O(x^50))) \\ Colin Barker, Aug 24 2017

def f(x): return x^3*(1-2*x+x^2+x^4)/(1-4*x+6*x^2-4*x^3+2*x^5-x^6-x^8)
def A290992_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( f(x) ).list()
A290992_list(60) # G. C. Greubel, Apr 12 2023

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

G.f.: x^3*(1 - 2*x + x^2 + x^4) / (1 - 4*x + 6*x^2 - 4*x^3 + 2*x^5 - x^6 - x^8). - Colin Barker, Aug 24 2017

A290993 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^6.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 6, 21, 56, 126, 252, 463, 804, 1365, 2366, 4368, 8736, 18565, 40410, 87381, 184604, 379050, 758100, 1486675, 2884776, 5592405, 10919090, 21572460, 43144920, 87087001, 176565486, 357913941, 723002336, 1453179126, 2906358252, 5791193143

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 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,-15,20,-15,6).

Cf. A000012, A033453, A192080, A289780, A290991, A290992, A291000.

Sequences of the form x^(m-1)/((1-x)^m - x^m): A000079 (m=1), A131577 (m=2), A024495 (m=3), A000749 (m=4), A139761 (m=5), this sequence (m=6), A290994 (m=7), A290995 (m=8).

a:=[0,0,0,0,1];;  for n in [6..35] do a[n]:=6*a[n-1]-15*a[n-2]+20*a[n-3]-15*a[n-4]+6*a[n-5]; od; Concatenation([0],a); # Muniru A Asiru, Oct 23 2018

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0,0] cat Coefficients(R!( x^5/((1-x)^6 - x^6) )); // G. C. Greubel, Apr 11 2023

seq(coeff(series(x^5/((1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Oct 23 2018

z = 60; s = x/(1 - x); p = 1 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290993 *)

concat(vector(5), Vec(x^5 / ((1 - 2*x)*(1 - x + x^2)*(1 - 3*x + 3*x^2)) + O(x^50))) \\ Colin Barker, Aug 24 2017

def A290993_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^5/((1-x)^6 - x^6) ).list()
A290993_list(60) # G. C. Greubel, Apr 11 2023

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) for n>5. Corrected by Colin Barker, Aug 24 2017
G.f.: x^5 / ((1 - 2*x)*(1 - x + x^2)*(1 - 3*x + 3*x^2)). - Colin Barker, Aug 24 2017
a(n) = A192080(n-5) for n > 5. - Georg Fischer, Oct 23 2018
G.f.: x^5/((1-x)^6 - x^6). - G. C. Greubel, Apr 11 2023

A290994 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^7.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 7, 28, 84, 210, 462, 924, 1717, 3017, 5110, 8568, 14756, 27132, 54264, 116281, 257775, 572264, 1246784, 2641366, 5430530, 10861060, 21242341, 40927033, 78354346, 150402700, 291693136, 574274008, 1148548016, 2326683921, 4749439975

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 (7,-21,35,-35,21,-7,2).

Cf. A000012, A033453, A049017, A289780, A291000.

Sequences of the form x^(m-1)/((1-x)^m - x^m): A000079 (m=1), A131577 (m=2), A024495 (m=3), A000749 (m=4), A139761 (m=5), A290993 (m=6), this sequence (m=7), A290995 (m=8).

R:=PowerSeriesRing(Integers(), 60); [0,0,0,0,0,0] cat Coefficients(R!( x^6/((1-x)^7 - x^7) )); // G. C. Greubel, Apr 11 2023

z = 60; s = x/(1 - x); p = 1 - s^7;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290994 *)

concat(vector(6), Vec(x^6 / ((1 - 2*x)*(1 - 5*x + 11*x^2 - 13*x^3 + 9*x^4 - 3*x^5 + x^6)) + O(x^50))) \\ Colin Barker, Aug 22 2017

def A290994_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^6/((1-x)^7 - x^7) ).list()
A290994_list(60) # G. C. Greubel, Apr 11 2023

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + 2*a(n-7) for n >= 8.
G.f.: x^6 / ((1 - 2*x)*(1 - 5*x + 11*x^2 - 13*x^3 + 9*x^4 - 3*x^5 + x^6)). - Colin Barker, Aug 22 2017
a(n) = A049017(n-6) for n > 5. - Georg Fischer, Oct 23 2018
G.f.: x^6/((1-x)^7 - x^7). - G. C. Greubel, Apr 11 2023

A291008 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 7*S^2.

Original entry on oeis.org

0, 7, 14, 70, 224, 868, 3080, 11368, 41216, 150640, 548576, 2000992, 7293440, 26592832, 96946304, 353449600, 1288577024, 4697851648, 17127165440, 62441440768, 227645874176, 829940392960, 3025756030976, 11031154419712, 40216845025280, 146620616568832

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 (2,6).

Cf. A000012, A083099, A289780, A291000.

[n le 2 select 7*(n-1) else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jun 01 2023

z = 60; s = x/(1 - x); p = 1 - s^7;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291008 *)
LinearRecurrence[{2,6}, {0,7}, 40] (* G. C. Greubel, Jun 01 2023 *)

A291008=BinaryRecurrenceSequence(2,6,0,7)
[A291008(n) for n in range(41)] # G. C. Greubel, Jun 01 2023

G.f.: 7*x/(1 - 2*x - 6*x^2).
a(n) = 2*a(n-1) + 6*a(n-2) for n >= 3.
a(n) = 7*A083099(n).
a(n) = (sqrt(7)*((1+sqrt(7))^n - (1-sqrt(7))^n)) / 2. - Colin Barker, Aug 23 2017
a(n) = 7*i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - G. C. Greubel, Jun 01 2023

cp's OEIS Frontend

A289803 p-INVERT of the even bisection (A001906) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A289846 p-INVERT of (1,0,1,0,1,0,1,0,1,...) (A059841), where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A289847 p-INVERT of the primes (A000040), where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

A289925 p-INVERT of the lower Wythoff sequence (A000201), where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

A289926 p-INVERT of the upper Wythoff sequence (A001950), where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

A289977 p-INVERT of (0,0,0,1,2,3,5,8,...), the Fibonacci numbers preceded by three zeros, where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290992 p-INVERT of (0,0,0,1,2,3,4,5,...), the nonnegative integers A000027 preceded by two zeros, where p(S) = 1 - S - S^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A290993 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S^6.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple