0,5

Consider the k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].

An element of CP(n,k) is a n-tupel T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,..,n^k.

We count members T of CP(n,k) which satisfy some condition delta(T_t),

so delta(.) is an indicator function which attains values of 1 or 0

depending on whether T_t is to be counted or not; the summation

sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.

For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j = i_(j+1):

T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j = i_(j+1)).

The test on i_j > i_(j+1) generates A158498. One gets the Pascal triangle A007318

if the indicator function tests whether for any two i_j, i_(j+1) in T_t one has i_j >= i_(j+1).

Use of other indicator functions can also calculate the Bell numbers A000110, A000045 or A000108.

Thomas Wieder, Home Page.

Thomas Wieder, (Old) Home Page

Columns: T(n,2)=A002378(n-1). T(n,3)=A011379(n-1). T(n,5)=A101362(n-1).

Rows: T(2,k) = A040000(k). T(3,k) = A003945(k), T(4,k)=A003946(k), T(5,k)=A003947(k), T(6,k)=A003948(k).

T(n,k) = (n-1)^(k-1)+(n-1)^k = n*A079901(n-1,k-1), k>0.

sum_{k=0..n} T(n,k) = (n*(n-1)^n-2)/(n-2), n>2.

The triangle begins

1

1 1

1 2 2

1 3 6 12

1 4 12 36 108

1 5 20 80 320 1280

1 6 30 150 750 3750 18750

1 7 42 252 1512 9072 54432 326592

1 8 56 392 2744 19208 134456 941192 6588344

T(3,3)=12 counts the triples (1,2,1), (1,2,3), (1,3,1), (1,3,2), (2,1,2),

(2,1,3), (2,3,1), (2,3,2), (3,1,2), (3,1,3), (3,2,1), (3,2,3) out of a total of 3^3=27 triples in the CP(3,3).

A007318, A158498.

Sequence in context: A110858 A008279 A056043 this_sequence A110564 A007441 A111933

Adjacent sequences: A158494 A158495 A158496 this_sequence A158498 A158499 A158500

nonn,tabl

Thomas Wieder (thomas.wieder(AT)t-online.de), Mar 20 2009

Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 31 2009