A393486 Triangle read by rows: T(n,k) (0 <= k <= n) such that b(e,e-n) = 2^(e-2n) * Sum_{k=0..n} T(n,k)*binomial(e,k), arising from excess-tail polynomials of row-word inversion clusters on standard Young tableaux of shape (m,m,m).

1, 3, 3, 3, 4, 9, -14, 12, -3, 27, -45, -56, 70, -54, 81, 1222, -140, -315, 387, -297, 243

Offset

0,2

Comments

Triangle arises from the row-word inversion statistic on standard Young tableaux of shape (m,m,m). Connected defect clusters of span s and weight w = 3s-2+e have counts d_s(e) that are polynomial in s of degree e for s >= s_0(e). The binomial decomposition p_e(s) = Sum_{k=0..e} b(e,k)*binomial(s-2,k) yields integer coefficients b(e,k), and the diagonal entries satisfy b(e,e-n) = 2^(e-2n) * Sum_{k=0..n} T(n,k)*binomial(e,k).

The 2-power factor is 2^(e-2n), not 2^(e-n); this specific choice renders all T(n,k) integral.

The triangle is not a standard Stirling transform of 3^n: all ansatze of the form Sum_{m=k..n} (-1)^(n-m)*binomial(n,m)*3^m*S(m,k) using Stirling numbers of either kind have been tested and falsified.

Links

Table of n, a(n) for n=0..20.

Morten Brydensholt, Python program

Formula

T(n,n) = 3^n (verified n = 0..5).

b(e,e-n) = 2^(e-2n) * Sum_{k=0..n} T(n,k)*binomial(e,k), where b(e,k) are the binomial basis coefficients of the excess-tail polynomials (see A394326).

Example

Triangle begins:
  n\k:  0     1     2     3     4     5
  0:    1
  1:    3     3
  2:    3     4     9
  3:  -14    12    -3    27
  4:  -45   -56    70   -54    81
  5: 1222  -140  -315   387  -297   243
For n=2: T(2,0)=3, T(2,1)=4, T(2,2)=9=3^2. This gives b(e,e-2) = 2^(e-4)*(3*binomial(e,0) + 4*binomial(e,1) + 9*binomial(e,2)).

Prog

(Python) # See Links

Crossrefs

Cf. A394326.

Sequence in context: A108688 A266688 A347139 * A285286 A123371 A011277

Adjacent sequences: A393483 A393484 A393485 * A393487 A393488 A393489

Keyword

sign,tabl,more

Author

Morten Brydensholt, Mar 16 2026

Status

approved