A290890 -id:A290890 - cp's OEIS frontend

A291181 p-INVERT of the positive integers, where p(S) = 1 - 8*S.

8, 80, 792, 7840, 77608, 768240, 7604792, 75279680, 745192008, 7376640400, 73021211992, 722835479520, 7155333583208, 70830500352560, 701149669942392, 6940666199071360, 68705512320771208, 680114457008640720, 6732439057765635992, 66644276120647719200

Offset: 0

Clark Kimberling, Aug 19 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 (10, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 8 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291181 *)
LinearRecurrence[{10,-1},{8,80},30] (* Harvey P. Dale, Jul 31 2023 *)

G.f.: 8/(1 - 10 x + x^2).
a(n) = 10*a(n-1) - a(n-2).
a(n) = 8*A004189(n+1) for n >= 0.

A290891 p-INVERT of the positive integers, where p(S) = 1 - S^3.

Original entry on oeis.org

0, 0, 1, 6, 21, 57, 138, 330, 827, 2175, 5826, 15519, 40836, 106584, 277696, 724968, 1897380, 4972113, 13029534, 34125561, 89336141, 233831262, 612074526, 1602358863, 4195173507, 10983645498, 28756340047, 75285234408, 197097337248, 516002648064

Offset: 0

Clark Kimberling, Aug 15 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 A280890 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, -15, 6, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290891 *)

G.f.: x^2/(1 - 6 x + 15 x^2 - 21 x^3 + 15 x^4 - 6 x^5 + x^6).

a(n) = 6*a(n-1) - 15*a(n-2) + 21*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

A290892 p-INVERT of the positive integers, where p(S) = 1 - S^4.

Original entry on oeis.org

0, 0, 0, 1, 8, 36, 120, 331, 808, 1852, 4248, 10312, 26968, 74012, 204968, 558253, 1483336, 3860588, 9938488, 25570103, 66214096, 172926104, 454504816, 1197527184, 3152221296, 8275051544, 21663395536, 56615219385, 147898879304, 386593228980, 1011521607736

Offset: 0

Clark Kimberling, Aug 15 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 A280890 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, -28, 56, -69, 56, -28, 8, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290892 *)

concat(vector(3), Vec(x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)) + O(x^50))) \\ Colin Barker, Aug 16 2017

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 69*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

G.f.: x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)). - Colin Barker, Aug 16 2017

A290893 p-INVERT of the positive integers, where p(S) = 1 - S^5.

Original entry on oeis.org

0, 0, 0, 0, 1, 10, 55, 220, 715, 2003, 5025, 11650, 25850, 57475, 134883, 345090, 952195, 2722455, 7765010, 21615771, 58293475, 152593575, 390679925, 988851150, 2502813930, 6394182650, 16569837550, 43533891575, 115440190725, 307108317769, 815362167365

Offset: 0

Clark Kimberling, Aug 15 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 A280890 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 (10, -45, 120, -210, 253, -210, 120, -45, 10, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290893 *)

concat(vector(4), Vec(x^4 / ((1 - 3*x + x^2)*(1 - 7*x + 23*x^2 - 44*x^3 + 55*x^4 - 44*x^5 + 23*x^6 - 7*x^7 + x^8)) + O(x^50))) \\ Colin Barker, Aug 16 2017

a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 253*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10).

G.f.: x^4 / ((1 - 3*x + x^2)*(1 - 7*x + 23*x^2 - 44*x^3 + 55*x^4 - 44*x^5 + 23*x^6 - 7*x^7 + x^8)). - Colin Barker, Aug 16 2017

A290894 p-INVERT of the positive integers, where p(S) = 1 - S^6.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 12, 78, 364, 1365, 4368, 12377, 31848, 75882, 170560, 370266, 803712, 1827099, 4531980, 12346791, 35783396, 105681186, 308229948, 873545479, 2392395276, 6336768804, 16309261148, 41095234896, 102361858716, 254804224832, 640481466012

Offset: 0

Clark Kimberling, Aug 15 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 (12, -66, 220, -495, 792, -923, 792, -495, 220, -66, 12, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290894 *)

concat(vector(5), Vec(x^5 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)) + O(x^50))) \\ Colin Barker, Aug 16 2017

a(n) = 12*a(n-1) - 66*a(n-2) + 220*a(n-3) - 495*a(n-4) + 792*a(n-5) - 923*a(n-6) + 792*a(n-7) - 495*a(n-8) + 22*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12).

G.f.: x^5 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)). - Colin Barker, Aug 16 2017

A290895 p-INVERT of the positive integers, where p(S) = 1 - S^7.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 14, 105, 560, 2380, 8568, 27132, 77521, 203518, 497826, 1148126, 2527609, 5401676, 11508168, 25437917, 60978022, 162008098, 468103230, 1409724358, 4259541790, 12617126893, 36241765553, 100599743538, 269998374114, 702694008002

Offset: 0

Clark Kimberling, Aug 15 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 (14, -91, 364, -1001, 2002, -3003, 3433, -3003, 2002, -1001, 364, -91, 14, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^7;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290895 *)

concat(vector(6), Vec(x^6 / ((1 - 3*x + x^2)*(1 - 11*x + 57*x^2 - 182*x^3 + 398*x^4 - 626*x^5 + 727*x^6 - 626*x^7 + 398*x^8 - 182*x^9 + 57*x^10 - 11*x^11 + x^12)) + O(x^50))) \\ Colin Barker, Aug 16 2017

a(n) = 14*a(n-1) - 91*a(n-2) + 364*a(n-3) - 1001*a(n-4) + 2002*a(n-5) - 3003*a(n-6) + 3433*a(n-7) - 3003*a(n-8) + 2002*a(n-9) - 1001*a(n-10) + 364*a(n-11) - 91*a(n-12) + 14*a(n-13) - a(n-14).

G.f.: x^6 / ((1 - 3*x + x^2)*(1 - 11*x + 57*x^2 - 182*x^3 + 398*x^4 - 626*x^5 + 727*x^6 - 626*x^7 + 398*x^8 - 182*x^9 + 57*x^10 - 11*x^11 + x^12)). - Colin Barker, Aug 16 2017

A290896 p-INVERT of the positive integers, where p(S) = 1 - S^8.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 16, 136, 816, 3876, 15504, 54264, 170544, 490315, 1307536, 3269288, 7732144, 17436220, 37819152, 79883544, 167737776, 362063944, 839161648, 2158258904, 6136548496, 18586871324, 57486027952, 176258492200, 527387147664, 1529591016109

Offset: 0

Clark Kimberling, Aug 15 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 (16, -120, 560, -1820, 4368, -8008, 11440, -12869, 11440, -8008, 4368, -1820, 560, -120, 16, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^8;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290896 *)

concat(vector(7), Vec(x^7 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 4*x + 7*x^2 - 4*x^3 + x^4)*(1 - 8*x + 28*x^2 - 56*x^3 + 71*x^4 - 56*x^5 + 28*x^6 - 8*x^7 + x^8)) + O(x^40))) \\ Colin Barker, Aug 16 2017

a(n) = 16*a(n-1) - 120*a(n-2) + 560*a(n-3) - 1820*a(n-4) + 4368*a(n-5) - 8008*a(n-6) + 11440*a(n-7) - 12869*a(n-8) + 11440*a(n-9) - 8008*a(n-10) + 4368*a(n-11) - 1820*a(n-12) + 560*a(n-13) - 120*a(n-14) + 16*a(n-15) - a(n-16).

G.f.: x^7 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 4*x + 7*x^2 - 4*x^3 + x^4)*(1 - 8*x + 28*x^2 - 56*x^3 + 71*x^4 - 56*x^5 + 28*x^6 - 8*x^7 + x^8)). - Colin Barker, Aug 16 2017

A290897 p-INVERT of the positive integers, where p(S) = 1 - S - S^3.

Original entry on oeis.org

1, 3, 9, 29, 95, 307, 976, 3073, 9645, 30283, 95207, 299625, 943363, 2970320, 9351621, 29439359, 92671625, 291715157, 918275995, 2890621063, 9099375792, 28643956245, 90168412937, 283841284899, 893503898503, 2812659866565, 8853968158791, 27871395427616

Offset: 0

Clark Kimberling, Aug 15 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 (7, -19, 27, -19, 7, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290897 *)

Vec((1 - 4*x + 7*x^2 - 4*x^3 + x^4) / ((1 - 3*x + 4*x^2 - x^3)*(1 - 4*x + 3*x^2 - x^3)) + O(x^30)) \\ Colin Barker, Aug 16 2017

a(n) = 7*a(n-1) - 19*a(n-2) + 27*a(n-3) - 19*a(n-4) + 7*a(n-5) - a(n-6).

G.f.: (1 - 4*x + 7*x^2 - 4*x^3 + x^4) / ((1 - 3*x + 4*x^2 - x^3)*(1 - 4*x + 3*x^2 - x^3)). - Colin Barker, Aug 16 2017

A290898 p-INVERT of the positive integers, where p(S) = 1 - S - S^4.

Original entry on oeis.org

1, 3, 8, 22, 65, 203, 647, 2053, 6423, 19811, 60490, 183750, 557551, 1693921, 5157224, 15731043, 48041589, 146785994, 448475954, 1369853581, 4182850121, 12769287055, 38976737437, 118967979141, 363132913719, 1108463577238, 3383732698880, 10329587789993

Offset: 0

Clark Kimberling, Aug 17 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 (9, -34, 71, -89, 71, -34, 9, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290898 *)

Vec((1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4) / (1 - 9*x + 34*x^2 - 71*x^3 + 89*x^4 - 71*x^5 + 34*x^6 - 9*x^7 + x^8) + O(x^40)) \\ Colin Barker, Aug 18 2017

a(n) = 9*a(n-1) - 34*a(n-2) + 71*a(n-3) - 89*a(n-4) + 71*a(n-5) - 34*a(n-6) + 9*a(n-7) - a(n-8).

G.f.: (1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4) / (1 - 9*x + 34*x^2 - 71*x^3 + 89*x^4 - 71*x^5 + 34*x^6 - 9*x^7 + x^8). - Colin Barker, Aug 18 2017

A290899 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^4.

Original entry on oeis.org

0, 1, 4, 12, 36, 110, 332, 983, 2876, 8380, 24428, 71357, 208868, 612178, 1795228, 5264684, 15436060, 45248195, 132616392, 388652536, 1138993032, 3338020181, 9782903524, 28671786116, 84032220964, 246284956558, 721820483900, 2115530739035, 6200240318564

Offset: 0

Clark Kimberling, Aug 17 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 (8, -27, 52, -63, 52, -27, 8, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290899 *)

concat(0, Vec(x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8) + O(x^40))) \\ Colin Barker, Aug 18 2017

a(n) = 8*a(n-1) - 27*a(n-2) + 52*a(n-3) - 63*a(n-4) + 52*a(n-5) - 27*a(n-6) + 8*a(n-7) - a(n-8).

G.f.: x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8). - Colin Barker, Aug 18 2017

cp's OEIS Frontend

A291181 p-INVERT of the positive integers, where p(S) = 1 - 8*S.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A290891 p-INVERT of the positive integers, where p(S) = 1 - S^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

A290892 p-INVERT of the positive integers, where p(S) = 1 - S^4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290893 p-INVERT of the positive integers, where p(S) = 1 - S^5.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290894 p-INVERT of the positive integers, where p(S) = 1 - S^6.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290895 p-INVERT of the positive integers, where p(S) = 1 - S^7.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290896 p-INVERT of the positive integers, where p(S) = 1 - S^8.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A290897 p-INVERT of the positive integers, where p(S) = 1 - S - S^3.

Views

Author

Keywords