OFFSET
1,4
COMMENTS
More precisely, the sequence a(n) is created from the Eulerian numbers E(3,r;k(1),k(2),k(3)) of the third order ordered by the following relation: E(3,r;k(1),k(2),k(3)) < E(3,s;l(1),l(2),l(3)) if r < s or r = s and simultaneously one of the following three conditions holds true: k(1) < l(1) or k(1) = l(1) and k(2) < l(2) or k(1) = l(1) and k(2) = l(2) and k(3) < l(3) for any r,s,k(i),l(i) in N, k(1)+k(2)+k(3)=2*r+1, each k(i) <= r, l(1)+l(2)+l(3)=2*s+1, and each l(i) =< s.
We see that the Eulerian numbers E(3,r;k(1),k(2),k(3)) are defined for all r,k(i) in N, k(i) <= r, satisfying the condition k(1)+k(2)+k(3)=2*r+1.
If at least one k(i) is equal to one we set E(3,r;k(1),k(2),k(3)) = 1. For the other cases we define E(3,r+1;k(1),k(2),k(3)) = k(1)*E'(3,r;k(1),k(2)-1,k(3)-1) + k(2)*E'(3,r;k(1)-1,k(2),k(3)-1) + k(3)*E'(3,r;k(1)-1,k(2)-1,k(3)), where the respective E'() is equal to the respective E(), whenever the last number could be defined, and is equal to 0 for the other case. We note that E(3,r;k(1),k(2),k(3)) = E(3,r;k(p(1)),k(p(2)),k(p(3))) for any permutation p of the set {1,2,3}.
We note that if a function x(t) satisfies the following differential equation: x'(t) = A*(x(t)-a)*(x(t)-b)*(x(t)-c), where A,a,b,c are the fixed complex numbers and A is nonzero, then for every positive integer r we obtain the identity:
d^r(x(t))/dt = A^r*sum{over all positive integers k(i) <= r, i=1,2,3, such that k(1)+k(2)+k(3)=2*r+1} E(3,r;k(1),k(2),k(3))*(x(t)-a)^(k(1))*(x(t)-b)^(k(2))*(x(t)-c)^(k(3)) - this identity is the real source of creating the Eulerian numbers of the third order.
For the Author and his colleagues (E. Hetmaniok and D. Slota) from the Rzadkowski's paper comes the inspiration for creating the above Eulerian numbers of the third order and their generalizations (called the multiindices Eulerian numbers).
REFERENCES
R. Witula, E. Hetmaniok and D. Slota, Proposition of the generalisation of Eulerian numbers and their applications, submitted (2012).
LINKS
G. Rzadkowski, Derivatives and Eulerian Numbers, Amer. Math. Monthly, 115 (2008), 458-460.
EXAMPLE
We have E(3,1;1,1,1)=1, E(3,2;1,2,2)=1, E(3,3;1,3,3)=1,
E(3,3;2,2,3) = 2*2*E(3,2;1,2,2) = 4, E(3,4;1,4,4)=1,
E(3,4;2,3,4) = 2*E(3,3;2,2,3) + 3*E(3,3;1,3,3) = 11,
E(3,4;3,3,3) = 3*3*E(3,3;2,2,3) = 36, E(3,5;1,5,5)=1,
E(3,5;2,4,5) = 2*E(3,4;2,3,4) + 4*E(3,4;1,4,4) = 26,
E(3,5;3,3,5) = 2*3*E(3,4;2,3,4) = 66, E(3,5;3,4,4) =
3*E(3,4;3,3,3) + 2*4*E(3,4;2,3,4) = 196, E(3,6;1,6,6)=1,
E(3,6;2,5,6) = 2*E(3,5;2,4,5) + 5*E(3,5;1,5,5) = 57,
E(3,6;3,4,6) = 3*E(3,5;3,3,5) + 4*E(3,5;2,4,5) = 302,
E(3,6;3,5,5) = 3*E(3,5;3,4,4) + 2*5*E(3,5;2,4,5) = 848,
E(3,6;4,4,5) = 2*4*E(3,5;3,4,4) + 5*E(3,5;3,3,5) = 1898.
CROSSREFS
KEYWORD
nonn
AUTHOR
Roman Witula, Sep 24 2012
STATUS
approved