A393486 - OEIS

%I #28 Apr 07 2026 23:03:05

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

%T 243

%N 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).

%C 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).

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

%C 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.

%H Morten Brydensholt, <a href="/A393486/a393486.py.txt">Python program</a>

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

%F 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).

%e Triangle begins:

%e   n\k:  0     1     2     3     4     5

%e   0:    1

%e   1:    3     3

%e   2:    3     4     9

%e   3:  -14    12    -3    27

%e   4:  -45   -56    70   -54    81

%e   5: 1222  -140  -315   387  -297   243

%e 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)).

%o (Python) # See Links

%Y Cf. A394326.

%K sign,tabl,more

%O 0,2

%A _Morten Brydensholt_, Mar 16 2026