A292479 -id:A292479 - cp's OEIS frontend

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

2, 7, 22, 70, 222, 705, 2238, 7105, 22556, 71608, 227332, 721705, 2291178, 7273743, 23091762, 73308814, 232731578, 738846865, 2345597854, 7446508273, 23640235416, 75050038224, 238259397096, 756395887969, 2401310279090, 7623377054503, 24201736119310

Offset: 0

Clark Kimberling, Sep 04 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).
In the following guide to p-INVERT sequences using s = (1,1,0,0,0,...) = A019590, in some cases t(1,1,0,0,0,...) is a shifted version of the cited sequence:
p(S)                     t(1,1,0,0,0,...)
1 - S                       A000045 (Fibonacci numbers)
1 - S^2                     A094686
1 - S^3                     A115055
1 - S^4                     A291379
1 - S^5                     A281380
1 - S^6                     A281381
1 - 2 S                     A002605
1 - 3 S                     A125145
(1 - S)^2                   A001629
(1 - S)^3                   A001628
(1 - S)^4                   A001629
(1 - S)^5                   A001873
(1 - S)^6                   A001874
1 - S - S^2                 A123392
1 - 2 S - S^2               A291382
1 - S - 2 S^2               A124861
1 - 2 S - S^2               A291383
(1 - 2 S)^2                 A073388
(1 - 3 S)^2                 A291387
(1 - 5 S)^2                 A291389
(1 - 6 S)^2                 A291391
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 3 S)            A291394
(1 - 2 S)(1 - 3 S)          A291395
(1 - S)(1 - 2 S)            A291393
(1 - S)(1 - 2 S)(1 - 3 S)   A291396
1 - S - S^3                 A291397
1 - S^2 - S^3               A291398
1 - S - S^2 - S^3           A186812
1 - S - S^2 - S^3 - S^4     A291399
1 - S^2 - S^4               A291400
1 - S - S^4                 A291401
1 - S^3 - S^4               A291402
1 - 2 S^2 - S^4             A291403
1 - S^2 - 2 S^4             A291404
1 - 2 S^2 - 2 S^4           A291405
1 - S^3 - S^6               A291407
(1 - S)(1 - S^2)            A291408
(1 - S^2)(1 - S)^2          A291409
1 - S - S^2 - 2 S^3         A291410
1 - 2 S - S^2 + S^3         A291411
1 - S - 2 S^2 + S^3         A291412
1 - 3 S + S^2 + S^3         A291413
1 - 2 S + S^3               A291414
1 - 3 S + S^2               A291415
1 - 4 S + S^2               A291416
1 - 4 S + 2 S^2             A291417

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 3, 2, 1)

Cf. A019590, A290890, A291000, A291219, A291728, A292479, A292480.

z = 60; s = x + x^2; p = 1 - 2 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291382 *)

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

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

A292480 p-INVERT of the odd positive integers, where p(S) = 1 - S^2.

Original entry on oeis.org

0, 1, 6, 20, 56, 160, 480, 1456, 4384, 13136, 39360, 118064, 354272, 1062928, 3188736, 9565936, 28697632, 86093264, 258280512, 774841520, 2324523104, 6973567888, 20920705152, 62762119792, 188286360736, 564859074896, 1694577214656, 5083731648560

Offset: 0

Clark Kimberling, Oct 02 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).
In the following guide to p-INVERT sequences using s = (1,3,5,7,9,...) = A005408, in some cases t(1,3,5,7,9,...) is a shifted (or differently indexed) version of the cited sequence:
p(S) *********** t(1,3,5,7,9,...)
- S               A003946
- S^2             A292480
- S^3             (not yet in OEIS)
(1 - S)^2           (not yet in OEIS)
(1 - S)^3           (not yet in OEIS)
- S - S^2         A289786
+ S - S^2         A289484
- S - 2 S^2       A289785
- S - 3 S^2       A289786
- S - 4 S^2       A289787
- S - 5 S^2       A289788
- S - 6 S^2       A289789
- S - 7 S^2       A289790
+ S - 2 S^2       A289791
- S + S^2 - S^3   A289792
+ S - 3 S^2       A289793
- S - S^2 - S^3   A289794

			s = (1,3,5,7,9,...), S(x) = x + 3 x^2 + 5 x^3 + 7 x^4 + ...,
p(S(x)) = 1 - ( x + 3 x^2 + 5 x^3 + 7 x^4 + ...)^2,
1/p(S(x)) = 1 + x^2 + 6 x^3 + 20 x^4 + 56 x^5 + ...
T(x) = (-1 + 1/p(S(x)))/x = x + 6 x^2 + 20 x^3 + 56 x^4 + ...
t(s) = (0,1,2,20,56,...).

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,6)

Cf. A005408, A292479.

I:=[0,1,6,20]; [n le 4 select I[n] else 4*Self(n-1)- 5*Self(n-2)+6*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Oct 03 2017

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292480 *)
Join[{0}, LinearRecurrence[{4, -5, 6}, {1, 6, 20}, 30]] (* Vincenzo Librandi, Oct 03 2017 *)

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

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

A292532 p-INVERT of the squares (A000290), where p(S) = 1 - S^3.

Original entry on oeis.org

0, 0, 1, 12, 75, 329, 1158, 3606, 10971, 35601, 126168, 467541, 1722714, 6173070, 21563906, 74452230, 257613930, 899546303, 3166966692, 11185908147, 39459021883, 138761604786, 486746839758, 1705955898935, 5982257083623, 20999661326520, 73772324787965

Offset: 0

Clark Kimberling, Oct 04 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 A292479 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, -36, 85, -123, 129, -83, 36, -9, 1)

Cf. A000290, A292479.

z = 60; s = x (x + 1)/(1 - x)^3; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292532 *)
LinearRecurrence[{9,-36,85,-123,129,-83,36,-9,1},{0,0,1,12,75,329,1158,3606,10971},30] (* Harvey P. Dale, Sep 27 2023 *)

G.f.: -((x^2 (1 + x)^3)/((-1 + 4 x - 2 x^2 + x^3) (1 - 5 x + 14 x^2 - 18 x^3 + 18 x^4 - 7 x^5 + x^6))).

a(n) = 9*a(n-1) - 36*a(n-2) + 85*a(n-3) - 123*a(n-4) + 129*a(n-5) - 83*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n >= 10.

A292533 p-INVERT of the squares (A000290), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 11, 46, 187, 748, 2948, 11480, 44273, 169374, 643601, 2431526, 9140616, 34212350, 127563959, 474022478, 1756118055, 6488228880, 23912815820, 87935847700, 322713694333, 1182114988606, 4322734288413, 15782353895178, 57537481431056, 209479529802682

Offset: 0

Clark Kimberling, Oct 04 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 A292479 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, -20, 18, -12, 4, -1)

Cf. A000290, A292479.

z = 60; s = x (x + 1)/(1 - x)^3; p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292533 *)

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

a(n) = 8*a(n-1) - 20*a(n-2) + 18*a(n-3) - 12*a(n-4) + 4*a(n-5) - a(n-6) for n >= 7.

A292534 p-INVERT of the squares (A000290), where p(S) = 1 + S - S^2.

Original entry on oeis.org

-1, -2, 4, 21, 30, 11, 80, 622, 2055, 4584, 10711, 34354, 115480, 341213, 934750, 2640483, 7874188, 23564242, 68738591, 198108496, 575654335, 1688669686, 4951141372, 14443935957, 42064267934, 122731975243, 358682023576, 1047906654118, 3058580566407

Offset: 0

Clark Kimberling, Oct 04 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 A292479 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, -12, 22, -16, 7, -1)

Cf. A000290, A292479.

z = 60; s = x (x + 1)/(1 - x)^3; 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]  (* A292534 *)

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

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

A292535 p-INVERT of the squares (A000290), where p(S) = 1 + S - 2 S^2.

Original entry on oeis.org

1, 7, 38, 189, 909, 4368, 21093, 102051, 493702, 2387689, 11546425, 55837024, 270025769, 1305841103, 6315023830, 30539305893, 147687325509, 714212301776, 3453913488845, 16703042620715, 80775512179990, 390628431683601, 1889069687788593, 9135495517900480

Offset: 0

Clark Kimberling, Oct 04 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 A292479 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, -15, 24, -11, 5, -1)

Cf. A000290, A292479.

z = 60; s = x (x + 1)/(1 - x)^3; p = 1 + s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292535 *)

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

a(n) = 7*a(n-1) - 15*a(n-2) + 24*a(n-3) - 11*a(n-4) + 5*a(n-5) - a(n-6) for n >= 7.

A292536 p-INVERT of the squares (A000290), where p(S) = 1 + S - 3 S^2.

Original entry on oeis.org

1, 8, 48, 255, 1310, 6773, 35260, 183740, 956765, 4980320, 25924725, 134956612, 702554244, 3657326875, 19039098206, 99112598721, 515954630808, 2685927132776, 13982245762937, 72787973059648, 378915453775913, 1972536332660240, 10268516498713448

Offset: 0

Clark Kimberling, Oct 04 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 A292479 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, -14, 26, -10, 5, -1)

Cf. A000290, A292479.

z = 60; s = x (x + 1)/(1 - x)^3; p = 1 + s - 3 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292536 *)

G.f.: -(((1 + x) (-1 - 6 x^2 + x^3))/(1 - 7 x + 14 x^2 - 26 x^3 + 10 x^4 - 5 x^5 + x^6)).

a(n) = 7*a(n-1) - 14*a(n-2) + 26*a(n-3) - 10*a(n-4) + 5*a(n-5) - a(n-6) for n >= 7.

cp's OEIS Frontend

A291382 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2.

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A292480 p-INVERT of the odd positive integers, where p(S) = 1 - S^2.

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A292532 p-INVERT of the squares (A000290), where p(S) = 1 - S^3.

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A292533 p-INVERT of the squares (A000290), where p(S) = (1 - S)^2.

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A292534 p-INVERT of the squares (A000290), where p(S) = 1 + S - S^2.

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A292535 p-INVERT of the squares (A000290), where p(S) = 1 + S - 2 S^2.

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A292536 p-INVERT of the squares (A000290), where p(S) = 1 + S - 3 S^2.

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