A231147 Array of coefficients of numerator polynomials of the rational function p(n, x + 1/x), where p(n,x) = (x^n - 1)/(x - 1).

1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 4, 3, 4, 1, 1, 1, 1, 5, 4, 9, 4, 5, 1, 1, 1, 1, 6, 5, 14, 9, 14, 5, 6, 1, 1, 1, 1, 7, 6, 20, 14, 29, 14, 20, 6, 7, 1, 1, 1, 1, 8, 7, 27, 20, 49, 29, 49, 20, 27, 7, 8, 1, 1, 1, 1, 9, 8, 35, 27, 76, 49, 99, 49, 76, 27, 35, 8, 9

Offset: 1

Author

Clark Kimberling, Nov 05 2013

Keywords

• nonn
• tabf
• easy

Comments

From Gus Wiseman, Mar 19 2023: (Start)

Also appears to be the number of nonempty subsets of {1,...,n} with median k, where k ranges from 1 to n in steps of 1/2, and the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). For example, row n = 5 counts the following subsets:

{1} {1,2} {2} {1,4} {3} {2,5} {4} {4,5} {5}

{1,3} {2,3} {1,5} {3,4} {3,5}

{1,2,3} {1,2,3,4} {2,4} {1,3,4,5} {1,4,5}

{1,2,4} {1,2,3,5} {1,3,4} {2,3,4,5} {2,4,5}

{1,2,5} {1,3,5} {3,4,5}

{2,3,4}

{2,3,5}

{1,2,4,5}

{1,2,3,4,5}

Central diagonals T(n,(n+1)/2) appear to be A100066 (bisection A006134).

For mean instead of median we have A327481.

For partitions instead of subsets we have A359893, full steps A359901.

Central diagonals T(n,n/2) are A361801 (bisection A079309).

(End)

Examples

			Triangle begins:
  1
  1  1  1
  1  1  3  1  1
  1  1  4  3  4  1  1
  1  1  5  4  9  4  5  1  1
  1  1  6  5 14  9 14  5  6  1  1
  1  1  7  6 20 14 29 14 20  6  7  1  1
  1  1  8  7 27 20 49 29 49 20 27  7  8  1  1
  1  1  9  8 35 27 76 49 99 49 76 27 35  8  9  1  1
First 3 polynomials: 1, 1 + x + x^2, 1 + x + 3*x^2 + x^3 + x^4
		

Links

• John Tyler Rascoe, Rows n = 1..100, flattened

Crossrefs

Cf. A231148.

Row sums are 2^n-1 = A000225(n).

Row lengths are 2n-1 = A005408(n-1).

Removing every other column appears to give A013580.

Cf. A006134, A007318, A008289, A079309, A325347, A327481, A359907, A360005.

Programs

Mathematica

z = 60; p[n_, x_] := p[x] = (x^n - 1)/(x - 1); Table[p[n, x], {n, 1, z/4}]; f1[n_, x_] := f1[n, x] = Numerator[Factor[p[n, x] /. x -> x + 1/x]]; Table[Expand[f1[n, x]], {n, 0, z/4}]
Flatten[Table[CoefficientList[f1[n, x], x], {n, 1, z/4}]]

PARI

A231147_row(n) = {Vecrev(Vec(numerator((-1+(x+(1/x))^n)/(x+(1/x)-1))))} \\ John Tyler Rascoe, Sep 10 2024