A256679 A signed triangle of V. I. Arnold for the Springer numbers (A001586).

1, 1, 0, -2, -3, -3, -8, -6, -3, 0, 40, 48, 54, 57, 57, 256, 216, 168, 114, 57, 0, -1952, -2208, -2424, -2592, -2706, -2763, -2763, -17408, -15456, -13248, -10824, -8232, -5526, -2763, 0, 177280, 194688, 210144, 223392, 234216, 242448, 247974, 250737, 250737

Offset: 0

Author

Vladimir Kruchinin, Apr 07 2015

Keywords

• sign
• tabl

Comments

This triangle is denoted R(b) on page 6 of the Arnold reference.

Unsigned version of triangle is triangle of A202818.

First column (m=0) is A000828. - Robert Israel, Apr 08 2015

Examples

			Triangle begins:
    1;
    1,   0;
   -2,  -3,  -3;
   -8,  -6,  -3,   0;
   40,  48,  54,  57, 57;
  256, 216, 168, 114, 57, 0;
		

Links

• V. I. Arnold, The calculus of snakes and the combinatorics of Bernoulli, Euler and Springer numbers of Coxeter groups (in Russian), Uspekhi Mat. nauk., 47 (#1, 1992), 3-45 = Russian Math. Surveys, Vol. 47 (1992), 1-51. (See page 6)

Crossrefs

Cf. A000828, A001586, A002436, A008280, A122045, A202818, A256665.

Programs

Maple

T:= (n,m) -> add(add(4^i*euler(2*i)*binomial(2*k,2*i)*binomial(n-m,2*k-m),i=0..k),k=floor(m/2)..floor(n/2)):
seq(seq(T(n,m),m=0..n),n=0..10); # Robert Israel, Apr 08 2015
# Second program, about 100 times faster than the first for the first 100 rows.
Triangle := proc(len) local s, A, n, k;
A := Array(0..len-1,[1]); lprint(A[0]);
for n from 1 to len-1 do
   if n mod 2 = 1 then s := 0 else
   s := 2^(3*n+1)*(Zeta(0,-n,1/8)-Zeta(0,-n,5/8)) fi;
   A[n] := s;
   for k from n-1 by -1 to 0 do
       s := s + A[k]; A[k] := s od;
   lprint(seq(A[k], k=0..n));
od end:
Triangle(100); # Peter Luschny, Apr 08 2015

Mathematica

T[n_, m_] := Sum[4^i EulerE[2i] Binomial[2k, 2i] Binomial[n-m, 2k-m], {k, Floor[m/2], n/2}, {i, 0, k}];
Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2019 *)

Maxima

T(n,m):=(sum((sum(4^i*euler(2*i)*binomial(2*k,2*i),i,0,k))*binomial(n-m,2*k-m),k,floor(m/2),n/2));

Sage

def triangle(len):
    L = [1]; print(L)
    for n in range(1,len):
        if is_even(n):
            s = 2^(3*n+1)*(hurwitz_zeta(-n,1/8)-hurwitz_zeta(-n,5/8))
        else: s = 0
        L.append(s)
        for k in range(n-1,-1,-1):
            s = s + L[k]; L[k] = s
        print(L)
triangle(7) # Peter Luschny, Apr 08 2015

Formula

E.g.f.: cosh(x+y)/cosh(2*(x+y))*exp(x).

T(n,m) = Sum_{k=floor(m/2)..floor(n/2)} Sum_{i=0..k} 4^i*E(2*i)*C(2*k,2*i)*C(n-m,2*k-m), where E(n) are the Euler secant numbers A122045.