OFFSET
1,3
COMMENTS
det(M_N) = -(2*N-3)!! where M_N is the N X N matrix with M_N(i,j) = T(i+1,j) for 1<=i,j<=N (T(n,m)=0 for m>n-1); (-1)!!=1 by convention.
From Sah Deepak Kumar Sureshprasad, Jun 23 2026: (Start)
The companion construction H2(u)=V(u)-I(u) yields S(n,m)=(-1)^(n+m)(n-2)!(n-1-m)/m!, for n>=2 and 0<=m<=n-2.
Equivalently, S(n,m)=(-1)^(n+m)A132159(n-2,m), where S(n,m) are the coefficients of R(n,v) in H2^(n)(v)=[R(n,v)exp(v)+Q2(n,v)]/v^n, with Q2(n,v)=(-1)^(n-1)(n-2)!(v+n-1), V'(u)=I(u), and I(u)=Integral_{t=0..u}(exp(t)-1)/t dt.
Moreover, |S(n,m)|=A007318(n-2,m)(n-1-m)!, so diagonal normalization yields Pascal's triangle A007318, while A395459 yields the Lucas triangle A029635.
Extending S(n,m) beyond its triangular domain 0<=m<=n-2 to the off-triangle region m>=n gives rational values in Q\Z satisfying Sum_{k>=1} Sum_{n>=2} S(n,n+k)=exp(-1)=-H(-1), where H(u)=V(u)+I(u).
Thus the companion constructions H(u)=V(u)+I(u) and H2(u)=V(u)-I(u) give rise to coefficient triangles whose diagonal normalizations yield the Lucas and Pascal triangles, respectively, while their off-triangle double sums are cross-linked through the boundary values H2(-1) and -H(-1) at u=-1. (End)
REFERENCES
L. B. W. Jolley, Summation of Series, Dover, 1961, Eq. 302.
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of triangle, flattened).
Sah Deepak Kumar Sureshprasad, The n-th Derivative of H at Integer Points: Integer Structure, Closed Forms, and a Derangement Identity, Zenodo, 2026.
Sah Deepak Kumar Sureshprasad, The Coefficient triangle
Z.-W. Sun and D. Xu, A Combinatorial Identity with Applications to Labeled Rooted Forests, arXiv:1007.1339 [math.CO], 2010.
FORMULA
Recurrence: T(n,m) = T(n-1,m-1) - (n-1-m)*T(n-1,m); T(1,0) = 1, T(n,n-1) = 2 for n >= 2, 0 <= m <= n-2.
G.f. for row n (n>=2): Sum_{m>=0} (-1)^(n+m-1)*(n-2)!*(n+m-1)/m! * x^m = (-1)^(n-1)*(n-2)!*(n-1-x)*exp(-x), where the coefficients (-1)^(n+m-1)*(n-2)!*(n+m-1)/m! are integers (= T(n,m)) for 0 <= m <= n-1, and lie in Q\Z for m >= n.
For k >= 1, the off triangle values (lying in Q\Z) satisfy Sum_{n>=2} T(n,n+k) = (-1)^(k-1)*(k+2)/(k*(k+1)!), with double sum Sum_{k>=1} Sum_{n>=2} T(n,n+k) = 2*Ein(1) - exp(-1) = H2(-1) where H2(u) = V(u) - I(u).
G.f. for right edge (d=0): Sum_{n>=2} T(n,n-1)*x^n = 2*x^2/(1-x).
G.f. for diagonal d (d>=1): Sum_{n>=d+1} T(n,n-1-d)*x^n = (-1)^d * d! * x^(d+1)*(1+x)/(1-x)^(d+1).
Antidiagonal sums T(s) = Sum_{n+m=s, n>=2, 0<=m<=n-1} T(n,m) satisfy T(s) = (-1)^(s-1)*(s-1)*sigma(s) + 2*[s % 2 != 0], where sigma(s) = (s-2)*sigma(s-1) - sigma(s-2) + (1+(-1)^s), sigma(2)=sigma(3)=1.
T(s) = (-1)^(s-1) * (s-1) * A003470(s-2) + 2 * [s % 2 != 0].
Even antidiagonal sums (for even s >= 2) |T(s)| = (s-1) * A003470(s-2).
Odd antidiagonal sums (for odd s >= 3) T(s) = 2 * A370704((s-1)/2).
Central elements: |T(2*n,n)| = (n-1)! * A051960(n-1) (even rows) and |T(2*n+1,n)| = 3*A000407(n-1) (odd rows), for n >= 1.
T(n,m) = (-1)^(n-1) * A293600(n,m) * (n-1-m)! (with the convention T(1,0)=1).
Partial sums: S(k,n) = Sum_{m=0..k} T(n,m) = (-1)^(n-1) * (n-2)! * (A000166(k)*n - A212291(k)) / k!, for k >= 1, n >= k+1.
|Sum_{m=0..1} T(n,m)| = A000142(n-2), n >= 2.
|Sum_{m=0..2} T(n,m)| = A001710(n-1), n >= 3.
|Sum_{m=0..3} T(n,m)| = A006157(n-4), n >= 4.
From Sah Deepak Kumar Sureshprasad, Jun 03 2026: (Start)
T(n,m) = (-1)^(n+m-1) * A029635(n,m) * (n-1-m)! (with the convention T(1,0)=2).
T(n,m) are the coefficients of P(n,v) in H(n,v) = [P(n,v)*exp(v) + Q(n,v)]/v^n, where H(u) = V(u) + I(u), V'(u) = I(u), I(u) = Integral_{t=0..u} (exp(t)-1)/t dt, and Q(n,v) = (-1)^(n-1)*(n-2)!*(v-n+1). (End)
EXAMPLE
Triangle begins (n=1..9):
n\m | 0 1 2 3 4 5 6 7 8
----+------------------------------------------------------
1 | 1
2 | -1 2
3 | 2 -3 2
4 | -6 8 -5 2
5 | 24 -30 18 -7 2
6 | -120 144 -84 32 -9 2
7 | 720 -840 480 -180 50 -11 2
8 | -5040 5760 -3240 1200 -330 72 -13 2
9 | 40320 -45360 25200 -9240 2520 -546 98 -15 2
MATHEMATICA
A395459[n_, m_] := If[n == 1, 1, (-1)^(n+m-1)*(n-2)!*(n+m-1)/m!];
Table[A395459[n, m], {n, 10}, {m, 0, n-1}] (* Paolo Xausa, May 20 2026 *)
PROG
(Python)
from math import factorial
def T(n, m):
sign = (-1)**(n+m-1)
return sign * factorial(n-2) * (n+m-1) // factorial(m)
for n in range(2, 10):
print([T(n, m) for m in range(n)])
CROSSREFS
Cf. A007395 (diagonal d=0), A005408 (diagonal d=1), A000290 (diagonal d=2 scaled by 2! or A001105), A000330 (diagonal d=3 scaled by 3! or A055112), A002415 (diagonal d=4 scaled by 4!), A005585 (diagonal d=5 scaled by 5!), A040977 (diagonal d=6 scaled by 6!), A050486 (diagonal d=7 scaled by 7!), A053347 (diagonal d=8 scaled by 8!), A054333 (diagonal d=9 scaled by 9!), A054334 (diagonal d=10 scaled by 10!), A057788 (diagonal d=11 scaled by 11!), A266561 (diagonal d=12 scaled by 12!).
Cf. A105926 (row sums absolute for v=1), A392965 (alternating row sums absolute for v=-1), A000312 (for v=n-1).
Cf. A001147 (determinant of triangle).
KEYWORD
AUTHOR
Sah Deepak Kumar Sureshprasad, Apr 23 2026
STATUS
approved