OFFSET
1,7
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}
For mean instead of median we have A327481.
(End)
LINKS
John Tyler Rascoe, Rows n = 1..100, flattened
EXAMPLE
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
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}]]
PROG
(PARI)
A231147_row(n) = {Vecrev(Vec(numerator((-1+(x+(1/x))^n)/(x+(1/x)-1))))} \\ John Tyler Rascoe, Sep 10 2024
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Clark Kimberling, Nov 05 2013
STATUS
approved