A152547 - OEIS

%I #10 Aug 28 2012 23:30:18

%S 1,2,3,1,4,2,2,5,3,3,3,1,1,6,4,4,4,4,2,2,2,2,2,7,5,5,5,5,5,3,3,3,3,3,

%T 3,3,3,3,1,1,1,1,1,8,6,6,6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2,2,2,2,

%U 2,2,2,2,2,2,2,2,2,2,9,7,7,7,7,7,7,7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5

%N Triangle, read by rows, derived from Pascal's triangle (see g.f. and example for generating methods).

%H Paul D. Hanna, <a href="/A152547/b152547.txt">Table of rows 0..14 listed as n, a(n) for n = 0..7059</a>

%F G.f. of row n: Sum_{k=0..n} (x^binomial(n,k) - 1)/(x-1) = Sum_{k=0..binomial(n,n\2)-1} T(n,k)*x^k.

%F A152548(n) = Sum_{k=0..C(n,[n/2])-1} T(n,k)^2 = Sum_{k=0..[(n+1)/2]} C(n+1, k)*(n+1-2k)^3/(n+1).

%e The number of terms in row n is C(n,[n/2]).

%e Triangle begins:

%e [1],

%e [2],

%e [3,1],

%e [4,2,2],

%e [5,3,3,3,1,1],

%e [6,4,4,4,4,2,2,2,2,2],

%e [7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1],

%e [8,6,6,6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2],

%e [9,7,7,7,7,7,7,7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1],

%e ...

%e ILLUSTRATION OF GENERATING METHOD.

%e Row n is derived from the binomial coefficients in the following way.

%e Place markers in an array so that the number of contiguous markers

%e in row k is C(n,k) and then count the markers along columns.

%e For example, row 6 of this triangle is generated from C(6,k) like so:

%e ------------------------------------------

%e 1: o - - - - - - - - - - - - - - - - - - -

%e 6: o o o o o o - - - - - - - - - - - - - -

%e 15:o o o o o o o o o o o o o o o - - - - -

%e 20:o o o o o o o o o o o o o o o o o o o o

%e 15:o o o o o o o o o o o o o o o - - - - -

%e 6: o o o o o o - - - - - - - - - - - - - -

%e 1: o - - - - - - - - - - - - - - - - - - -

%e ------------------------------------------

%e Counting the markers along the columns gives row 6 of this triangle:

%e [7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1].

%e Continuing in this way generates all the rows of this triangle.

%e ...

%e Number of repeated terms in each row of this triangle forms A008315:

%e 1;

%e 1;

%e 1, 1;

%e 1, 2;

%e 1, 3, 2;

%e 1, 4, 5;

%e 1, 5, 9, 5;

%e 1, 6, 14, 14;

%e 1, 7, 20, 28, 14;...

%o (PARI) {T(n,k)=polcoeff(sum(j=0,n,(x^binomial(n,j) - 1)/(x-1)),k)}

%o for(n=0,10, for(k=0, binomial(n,n\2)-1, print1(T(n,k),","));print(""))

%Y Cf. A152548 (row squared sums), A008315; A152545.

%K nonn,tabf

%O 0,2

%A _Paul D. Hanna_, Dec 14 2008