A275439 -id:A275439 - cp's OEIS frontend

A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

1, 3, 6, 11, 21, 39, 70, 126, 224, 394, 690, 1201, 2079, 3585, 6158, 10541, 17991, 30623, 51996, 88092, 148944, 251364, 423492, 712369, 1196557, 2007135, 3362598, 5626847, 9405465, 15705447, 26200066, 43667802, 72719312, 121000846, 201185334, 334265089

Offset: 0

Clark Kimberling, Sep 07 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 A291382 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 (1,2,1,-2,-3,-1)

Cf. A019590, A275439, A291382.

a:=[1,3,6,11,21,39];;
for n in [7..10^2] do a[n]:=a[n-1]+2*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]- a[n-6]; od; a; # Muniru A Asiru, Sep 10 2017

z = 60; s = x + x^2; p = (1 - s)(1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291408 *)

G.f.: -(((1 + x) (-1 - x + 2 x^3 + x^4))/((-1 + x + x^2)^2 (1 + x + x^2))).
a(n) = a(n-1) + 2*a(n-2) + a(n-3) - 2*a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.
a(n) = (1/2) * A275439(n+4). - Alois P. Heinz, May 20 2025

A275438 Triangle read by rows: T(n,k) is the number of compositions of n with parts in {1,2} having asymmetry degree equal to k (n>=0; 0<=k<=floor(n/3)).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 2, 2, 6, 5, 4, 4, 3, 14, 4, 8, 10, 16, 5, 30, 12, 8, 13, 20, 48, 8, 8, 60, 36, 40, 21, 40, 124, 32, 16, 13, 116, 88, 144, 16, 34, 76, 292, 112, 96, 21, 218, 204, 432, 80, 32, 55, 142, 648, 320, 400, 32, 34, 402, 444, 1160, 320, 224

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).
Number of entries in row n is 1 + floor(n/3).
Sum of entries in row n is A000045(n+1) (Fibonacci).
T(n,0) = A053602(n+1) (= number of palindromic compositions of n with parts in {1,2}).
Sum_{k>=0} k*T(n,k) = A275439(n).

			Row 4 is [3,2] because the compositions of 4 with parts in {1,2} are 22, 112, 121, 211, and 1111, having asymmetry degrees 0, 1, 0, 1, 0, respectively.
Triangle starts:
  1;
  1;
  2;
  1,2;
  3,2;
  2,6;
  5,4,4.

Krithnaswami Alladi and V. E. Hoggatt, Jr. Compositions with Ones and Twos, Fibonacci Quarterly, 13 (1975), 233-239.
V. E. Hoggatt, Jr. and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.

Cf. A000045, A053602, A275439.

G:=(1+z+z^2)/(1-z^2-2*t*z^3-z^4): Gser:=simplify(series(G,z=0,25)): for n from 0 to 20 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 20 do seq(coeff(P[n],t,j),j=0..degree(P[n])) end do; # yields sequence in triangular form

Join[{{1}}, Table[BinCounts[#, {0, 1 + Floor[n/3], 1}] &@ 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, 17}]] // Flatten (* Michael De Vlieger, Aug 17 2016 *)

G.f.: G(t,z) = (1+z+z^2)/(1-z^2-2*t*z^3-z^4). In the more general situation of compositions into a[1]=1} z^(a[j]), we have G(t,z) = (1+F(z))/(1-F(z^2)-t*(F(z)^2-F(z^2))). In particular, for t=0 we obtain Theorem 1.2 of the Hoggatt et al. reference.

cp's OEIS Frontend

A291408 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

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