A291408 -id:A291408 - 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.

A275439 Sum of the asymmetry degrees of all compositions of n with parts in {1,2}.

Original entry on oeis.org

0, 0, 0, 2, 2, 6, 12, 22, 42, 78, 140, 252, 448, 788, 1380, 2402, 4158, 7170, 12316, 21082, 35982, 61246, 103992, 176184, 297888, 502728, 846984, 1424738, 2393114, 4014270, 6725196, 11253694, 18810930, 31410894, 52400132, 87335604, 145438624, 242001692

Offset: 0

Emeric Deutsch, Aug 16 2016

The asymmetry degree of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the asymmetry degree of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).

A sequence is palindromic if and only if its asymmetry degree is 0.

			a(5) = 6 because the compositions of 5 with parts in {1,2} are 122, 212, 221, 1112, 1121, 1211, 2111, and 11111 and the sum of their asymmetry degrees is 1 + 0 + 1 + 1 + 1 + 1 + 1 + 0 = 6.

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

Cf. A000045, A275438, A291408.

f := n -> combinat:-fibonacci(n):
a := n -> (n+1)/2-(3/2)*floor((n+2)/3)+(3/5)*(n+1)*f(n)-(1/10)*(2*n+5)*f(n+1):
seq(a(n), n = 0..40);
# alternative program:
g := 2*z^3/((1+z+z^2)*(1-z-z^2)^2):
gser := series(g, z=0, 45):
seq(coeff(gser, z, n), n = 0..40);

Join[{0}, Table[Total@ Map[Total, Map[BitXor[Take[# - 1, Ceiling[Length[#]/2]], Reverse@ Take[# - 1, -Ceiling[Length[#]/2]]] &,
Flatten[Map[Permutations, DeleteCases[IntegerPartitions@ n, {a_, _} /; a > 2]], 1]]], {n, 30}]] (* Michael De Vlieger, Aug 17 2016 *)

concat(vector(3), Vec(2*x^3/((1+x+x^2)*(1-x-x^2)^2) + O(x^50))) \\ Colin Barker, Aug 28 2016

G.f.: g(z) = 2*z^3/((1+z+z^2)*(1-z-z^2)^2). In the more general situation of compositions into a[1]=1} z^(a[j]), we have g(z) = (F(z)^2-F(z^2))/((1+F(z))*(1-F(z))^2).
a(n) = (n+1)/2-(3/2)*floor((n+2)/3)+(3/5)*(n+1)*f(n)-(1/10)*(2*n+5)*f(n+1), where f(j) = A000045(j) are the Fibonacci numbers.
a(n) = Sum_{k>=0} k*A275438(n,k).
a(n) = 2*A291408(n-4) for n>=4. - Alois P. Heinz, May 20 2025

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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