A290900 -id:A290900 - cp's OEIS frontend

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

0, 1, 4, 11, 28, 72, 188, 493, 1292, 3383, 8856, 23184, 60696, 158905, 416020, 1089155, 2851444, 7465176, 19544084, 51167077, 133957148, 350704367, 918155952, 2403763488, 6293134512, 16475640049, 43133785636, 112925716859, 295643364940, 774004377960

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).
Note that in A290890, s = (1,2,3,4,...); i.e., A000027(n+1) for n>=0, whereas in A290990, s = (0,1,2,3,4,...); i.e., A000027(n) for n>=0.
Guide to p-INVERT sequences using s = (1,2,3,4,5,...) = A000027:
p(S)                 t(1,2,3,4,5,...)
- S                    A001906
- S^2                  A290890; see A113067 for signed version
- S^3                  A290891
- S^4                  A290892
- S^5                  A290893
- S^6                  A290894
- S^7                  A290895
- S^8                  A290896
- S - S^2              A289780
- S - S^3              A290897
- S - S^4              A290898
- S^2 - S^4            A290899
- S^2 - S^3            A290900
- S^3 - S^4            A290901
- 2S                   A052530;  (1/2)*A052530 = A001353
- 3S                   A290902;  (1/3)*A290902 = A004254
- 4S                   A003319;  (1/4)*A003319 = A001109
- 5S                   A290903;  (1/5)*A290903 = A004187
- 2*S^2                A290904;  (1/2)*A290904 = A290905
- 3*S^2                A290906;  (1/3)*A290906 = A290907
- 4*S^2                A290908;  (1/4)*A290908 = A099486
- 5*S^2                A290909;  (1/5)*A290909 = A290910
- 6*S^2                A290911;  (1/6)*A290911 = A290912
- 7*S^2                A290913;  (1/7)*A290913 = A290914
- 8*S^2                A290915;  (1/8)*A290915 = A290916
(1 - S)^2                A290917
(1 - S)^3                A290918
(1 - S)^4                A290919
(1 - S)^5                A290920
(1 - S)^6                A290921
- S - 2*S^2            A290922
- 2*S - 2*S^2          A290923;  (1/2)*A290923 = A290924
- 3*S - 2*S^2          A290925
(1 - S^2)^2              A290926
(1 - S^2)^3              A290927
(1 - S^3)^2              A290928
(1 - S)(1 - S^2)         A290929
(1 - S^2)(1 - S^4)       A290930
- 3 S + S^2            A291025
- 4 S + S^2            A291026
- 5 S + S^2            A291027
- 6 S + S^2            A291028
- S - S^2 - S^3        A291029
- S - S^2 - S^3 - S^4  A201030
- 3 S + 2 S^3          A291031
- S - S^2 - S^3 + S^4  A291032
- 6 S                  A291033
- 7 S                  A291034
- 8 S                  A291181
- 3 S + 2 S^3          A291031
- 3 S + 2 S^2          A291182
- 4 S + 2 S^3          A291183
- 4 S + 3 S^3          A291184

			(See the examples at A289780.)

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -5, 4, -1)

Cf. A000027, A113067, A289780, A113067 (signed version of same sequence).

z = 60; s = x/(1 - x)^2; p = 1 - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290890 *)

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

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

A372883 Irregular triangle read by rows: T(n,k) is the number of flattened Catalan words of length n with exactly k symmetric peaks, with k >= 0.

Original entry on oeis.org

1, 2, 4, 1, 9, 5, 23, 17, 1, 63, 51, 8, 176, 149, 39, 1, 491, 439, 153, 11, 1362, 1308, 540, 70, 1, 3762, 3912, 1812, 342, 14, 10369, 11671, 5935, 1439, 110, 1, 28559, 34637, 19175, 5541, 645, 17, 78653, 102222, 61302, 20214, 3170, 159, 1, 216638, 300190, 194080, 71242, 13903, 1089, 20

Offset: 1

Stefano Spezia, May 15 2024

			The irregular triangle begins:
     1;
     2;
     4,    1;
     9,    5;
    23,   17,    1;
    63,   51,    8;
   176,  149,   39,   1;
   491,  439,  153,  11;
  1362, 1308,  540,  70,  1;
  3762, 3912, 1812, 342, 14;
  ...
T(4,1) = 5 since there are 5 flattened Catalan words of length 4 with 1 symmetric peak: 0100, 0101, 0010, 0110, and 0121.

Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, Flattened Catalan Words, arXiv:2405.05357 [math.CO], 2024. See pp. 21-22.

Cf. A007051 (row sums), A290900 (2nd column), A369328 (1st column), A371965, A372879, A372884.

T[n_,k_]:=SeriesCoefficient[x(1-x)(1-2x)/(1-5x+8x^2-5x^3-x^2y+2x^3y),{x,0,n},{y,0,k}]; Table[T[n,k],{n,14},{k,0,Floor[(n-1)/2]}]//Flatten

G.f.: x*(1 - x)*(1 - 2*x)/(1 - 5*x + 8*x^2 - 5*x^3 - x^2*y + 2*x^3*y).

Sum_{k>=0} T(n,k) = A007051(n-1).

cp's OEIS Frontend

A290890 p-INVERT of the positive integers, where p(S) = 1 - S^2.

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