A256679 - OEIS

%I #31 Mar 07 2020 12:08:48

%S 1,1,0,-2,-3,-3,-8,-6,-3,0,40,48,54,57,57,256,216,168,114,57,0,-1952,

%T -2208,-2424,-2592,-2706,-2763,-2763,-17408,-15456,-13248,-10824,

%U -8232,-5526,-2763,0,177280,194688,210144,223392,234216,242448,247974,250737,250737

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

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

%C Unsigned version of triangle is triangle of A202818.

%C First column (m=0) is A000828. - _Robert Israel_, Apr 08 2015

%H V. I. Arnold, <a href="http://mi.mathnet.ru/eng/umn4470">The calculus of snakes and the combinatorics of Bernoulli, Euler and Springer numbers of Coxeter groups (in Russian)</a>, Uspekhi Mat. nauk., 47 (#1, 1992), 3-45 = Russian Math. Surveys, Vol. 47 (1992), 1-51. (See page 6)

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

%F 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.

%e     1;

%e     1,   0;

%e    -2,  -3,  -3;

%e    -8,  -6,  -3,   0;

%e    40,  48,  54,  57, 57;

%e   256, 216, 168, 114, 57, 0;

%p 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)):

%p seq(seq(T(n,m),m=0..n),n=0..10); # _Robert Israel_, Apr 08 2015

%p # Second program, about 100 times faster than the first for the first 100 rows.

%p Triangle := proc(len) local s, A, n, k;

%p A := Array(0..len-1,[1]); lprint(A[0]);

%p for n from 1 to len-1 do

%p    if n mod 2 = 1 then s := 0 else

%p    s := 2^(3*n+1)*(Zeta(0,-n,1/8)-Zeta(0,-n,5/8)) fi;

%p    A[n] := s;

%p    for k from n-1 by -1 to 0 do

%p        s := s + A[k]; A[k] := s od;

%p    lprint(seq(A[k], k=0..n));

%p od end:

%p Triangle(100); # _Peter Luschny_, Apr 08 2015

%t 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}];

%t Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 12 2019 *)

%o (Maxima)

%o 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));

%o (Sage)

%o def triangle(len):

%o     L = [1]; print(L)

%o     for n in range(1,len):

%o         if is_even(n):

%o             s = 2^(3*n+1)*(hurwitz_zeta(-n,1/8)-hurwitz_zeta(-n,5/8))

%o         else: s = 0

%o         L.append(s)

%o         for k in range(n-1,-1,-1):

%o             s = s + L[k]; L[k] = s

%o         print(L)

%o triangle(7) # _Peter Luschny_, Apr 08 2015

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

%K sign,tabl

%O 0,4

%A _Vladimir Kruchinin_, Apr 07 2015