|
%I #5 Jun 10 2012 12:32:14
%S 1,1,1,1,2,1,2,2,1,3,2,2,1,4,3,3,2,1,5,4,4,3,3,1,1,7,5,5,5,4,3,3,1,1,
%T 9,7,7,7,5,5,5,4,3,1,1,1,12,9,9,9,8,7,7,7,5,4,4,4,1,1,1,1,16,12,12,12,
%U 12,9,9,9,8,8,8,7,5,4,4,4,1,1,1,1,1,21,16,16,16,16,13,12,12,12,12,12,11,9,8
%N Padovan-Fibonacci triangle, read by rows, where the first column equals the Padovan spiral numbers (A134816), while the row sums equal the Fibonacci numbers (A000045).
%C The number of terms in each row equal the Padovan spiral numbers (A134816, with offset).
%H Paul D. Hanna, <a href="/A152545/b152545.txt">Table of n, a(n) for n = 0..261</a>
%F G.f. for row n: Sum_{k=0..A000931(n+5)-1} (x^{T(n-1,k)+T(n-2,k)} - 1)/(x-1) = Sum_{k=0..A000931(n+6)-1} T(n,k)*x^k for n>1 with T(0,0)=T(1,0)=1, where A000931 is the Padovan sequence.
%e Triangle begins:
%e [1],
%e [1],
%e [1,1],
%e [2,1],
%e [2,2,1],
%e [3,2,2,1],
%e [4,3,3,2,1],
%e [5,4,4,3,3,1,1],
%e [7,5,5,5,4,3,3,1,1],
%e [9,7,7,7,5,5,5,4,3,1,1,1],
%e [12,9,9,9,8,7,7,7,5,4,4,4,1,1,1,1],
%e [16,12,12,12,12,9,9,9,8,8,8,7,5,4,4,4,1,1,1,1,1],
%e [21,16,16,16,16,13,12,12,12,12,12,11,9,8,8,8,5,5,5,5,4,1,1,1,1,1,1,1],
%e [28,21,21,21,21,20,16,16,16,16,16,16,13,13,12,12,12,12,11,11,8,6,5,5,5,5,5,5,1,1,1,1,1,1,1,1,1],
%e [37,28,28,28,28,28,22,21,21,21,21,21,20,20,20,20,18,16,16,16,14,13,12,12,12,12,12,11,6,6,6,6,6,5,5,5,5,1,1,1,1,1,1,1,1,1,1,1,1],
%e [49,37,37,37,37,37,33,28,28,28,28,28,28,28,28,28,27,22,22,21,21,21,20,20,20,20,20,18,17,17,16,16,14,12,12,12,12,7,6,6,6,6,6,6,6,6,6,6,5,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
%e ...
%e ILLUSTRATION OF RECURRENCE.
%e Start out with row 0 and row 1 consisting of a single '1'.
%e To obtain any given row of this irregular triangle, first
%e sum the prior two rows term-by-term; for rows 7 and 8 we get:
%e [5,4,4,3,3,1,1] + [7,5,5,5,4,3,3,1,1] = [12,9,9,8,7,4,4,1,1].
%e Place markers in an array so that the number of contiguous markers
%e in each row correspond to the term-by-term sums like so:
%e --------------------------
%e 12:o o o o o o o o o o o o
%e 9: o o o o o o o o o - - -
%e 9: o o o o o o o o o - - -
%e 8: o o o o o o o o - - - -
%e 7: o o o o o o o - - - - -
%e 4: o o o o - - - - - - - -
%e 4: o o o o - - - - - - - -
%e 1: o - - - - - - - - - - -
%e 1: o - - - - - - - - - - -
%e --------------------------
%e Then count the markers by columns to obtain the desired row;
%e here, the number of markers in each column yields row 9:
%e [9,7,7,7,5,5,5,4,3,1,1,1].
%e Continuing in this way generates all the rows of this triangle.
%o (PARI) {T(n,k)=local(G000931=(1-x^2)/(1-x^2-x^3+x*O(x^(n+6))));if(n<0,0,if(n<2&k==0,1, polcoeff(sum(j=0,polcoeff(G000931,n+5)-1,(x^(T(n-1,j)+T(n-2,j)) - 1)/(x-1)),k) ))};
%o /* To print, use Padovan g.f. to get the number of terms in row n: */
%o for(n=0,10,for(k=0,polcoeff((1-x^2)/(1-x^2-x^3+x*O(x^(n+6))),n+6)-1,print1(T(n,k),","));print(""))
%Y Cf. A134816, A000045, A000931; A152546 (row squared sums).
%K nonn,tabf
%O 0,5
%A _Paul D. Hanna_, Dec 13 2008
|
