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A152547 Triangle, read by rows, derived from Pascal's triangle (see g.f. and example for generating methods).

%I A152547 #10 Aug 28 2012 23:30:18
%S A152547 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 A152547 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 A152547 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 A152547 Triangle, read by rows, derived from Pascal's triangle (see g.f. and example for generating methods).
%H A152547 Paul D. Hanna, <a href="/A152547/b152547.txt">Table of rows 0..14 listed as n, a(n) for n = 0..7059</a>
%F A152547 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 A152547 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 A152547 The number of terms in row n is C(n,[n/2]).
%e A152547 Triangle begins:
%e A152547 [1],
%e A152547 [2],
%e A152547 [3,1],
%e A152547 [4,2,2],
%e A152547 [5,3,3,3,1,1],
%e A152547 [6,4,4,4,4,2,2,2,2,2],
%e A152547 [7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1],
%e A152547 [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 A152547 [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 A152547 ...
%e A152547 ILLUSTRATION OF GENERATING METHOD.
%e A152547 Row n is derived from the binomial coefficients in the following way.
%e A152547 Place markers in an array so that the number of contiguous markers
%e A152547 in row k is C(n,k) and then count the markers along columns.
%e A152547 For example, row 6 of this triangle is generated from C(6,k) like so:
%e A152547 ------------------------------------------
%e A152547 1: o - - - - - - - - - - - - - - - - - - -
%e A152547 6: o o o o o o - - - - - - - - - - - - - -
%e A152547 15:o o o o o o o o o o o o o o o - - - - -
%e A152547 20:o o o o o o o o o o o o o o o o o o o o
%e A152547 15:o o o o o o o o o o o o o o o - - - - -
%e A152547 6: o o o o o o - - - - - - - - - - - - - -
%e A152547 1: o - - - - - - - - - - - - - - - - - - -
%e A152547 ------------------------------------------
%e A152547 Counting the markers along the columns gives row 6 of this triangle:
%e A152547 [7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1].
%e A152547 Continuing in this way generates all the rows of this triangle.
%e A152547 ...
%e A152547 Number of repeated terms in each row of this triangle forms A008315:
%e A152547 1;
%e A152547 1;
%e A152547 1, 1;
%e A152547 1, 2;
%e A152547 1, 3, 2;
%e A152547 1, 4, 5;
%e A152547 1, 5, 9, 5;
%e A152547 1, 6, 14, 14;
%e A152547 1, 7, 20, 28, 14;...
%o A152547 (PARI) {T(n,k)=polcoeff(sum(j=0,n,(x^binomial(n,j) - 1)/(x-1)),k)}
%o A152547 for(n=0,10, for(k=0, binomial(n,n\2)-1, print1(T(n,k),","));print(""))
%Y A152547 Cf. A152548 (row squared sums), A008315; A152545.
%K A152547 nonn,tabf
%O A152547 0,2
%A A152547 _Paul D. Hanna_, Dec 14 2008