A289788 -id:A289788 - cp's OEIS frontend

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

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.

A289787 p-INVERT of the even positive integers (A005843), where p(S) = 1 - S - S^2.

Original entry on oeis.org

2, 12, 62, 312, 1570, 7908, 39838, 200688, 1010978, 5092860, 25655582, 129241512, 651061762, 3279762132, 16521995710, 83230530528, 419278719938, 2112141348588, 10640036959358, 53599815453720, 270012240337762, 1360202629711812, 6852101192007262

Offset: 0

Clark Kimberling, Aug 10 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 (6, -6, 6, -1)

Cf. A005843, A289780, A289788.

z = 60; s = 2*x/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005843 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289787 *)
u/2 (* A289788 *)

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

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

cp's OEIS Frontend

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

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