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A152545 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).

%I A152545 #5 Jun 10 2012 12:32:14
%S A152545 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 A152545 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 A152545 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 A152545 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 A152545 The number of terms in each row equal the Padovan spiral numbers (A134816, with offset).
%H A152545 Paul D. Hanna, <a href="/A152545/b152545.txt">Table of n, a(n) for n = 0..261</a>
%F A152545 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 A152545 Triangle begins:
%e A152545 [1],
%e A152545 [1],
%e A152545 [1,1],
%e A152545 [2,1],
%e A152545 [2,2,1],
%e A152545 [3,2,2,1],
%e A152545 [4,3,3,2,1],
%e A152545 [5,4,4,3,3,1,1],
%e A152545 [7,5,5,5,4,3,3,1,1],
%e A152545 [9,7,7,7,5,5,5,4,3,1,1,1],
%e A152545 [12,9,9,9,8,7,7,7,5,4,4,4,1,1,1,1],
%e A152545 [16,12,12,12,12,9,9,9,8,8,8,7,5,4,4,4,1,1,1,1,1],
%e A152545 [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 A152545 [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 A152545 [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 A152545 [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 A152545 ...
%e A152545 ILLUSTRATION OF RECURRENCE.
%e A152545 Start out with row 0 and row 1 consisting of a single '1'.
%e A152545 To obtain any given row of this irregular triangle, first
%e A152545 sum the prior two rows term-by-term; for rows 7 and 8 we get:
%e A152545 [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 A152545 Place markers in an array so that the number of contiguous markers
%e A152545 in each row correspond to the term-by-term sums like so:
%e A152545 --------------------------
%e A152545 12:o o o o o o o o o o o o
%e A152545 9: o o o o o o o o o - - -
%e A152545 9: o o o o o o o o o - - -
%e A152545 8: o o o o o o o o - - - -
%e A152545 7: o o o o o o o - - - - -
%e A152545 4: o o o o - - - - - - - -
%e A152545 4: o o o o - - - - - - - -
%e A152545 1: o - - - - - - - - - - -
%e A152545 1: o - - - - - - - - - - -
%e A152545 --------------------------
%e A152545 Then count the markers by columns to obtain the desired row;
%e A152545 here, the number of markers in each column yields row 9:
%e A152545 [9,7,7,7,5,5,5,4,3,1,1,1].
%e A152545 Continuing in this way generates all the rows of this triangle.
%o A152545 (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 A152545 /* To print, use Padovan g.f. to get the number of terms in row n: */
%o A152545 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 A152545 Cf. A134816, A000045, A000931; A152546 (row squared sums).
%K A152545 nonn,tabf
%O A152545 0,5
%A A152545 _Paul D. Hanna_, Dec 13 2008