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A268317 Irregular triangle read by rows: T(n,k) gives the columns sum in the table Fib(n+1) X Fib(n), where k = 1..Fib(n), and 1's are assigned to cells on the longest diagonal path.

%I A268317 #11 Mar 03 2016 15:57:19
%S A268317 0,1,2,2,2,2,3,2,2,3,2,3,2,2,3,2,3,3,2,3,2,2,3,2,3,3,2,3,2,3,3,2,3,2,
%T A268317 2,3,2,3,3,2,3,2,3,3,2,3,3,2,3,2,3,3,2,3,2,2,3,2,3,3,2,3,2,3,3,2,3,3,
%U A268317 2,3,2,3,3,2,3,2,3,3,2,3,3,2,3,2,3,3,2,3,2,2,3,2,3,3,2,3,2,3,3,2,3
%N A268317 Irregular triangle read by rows: T(n,k) gives the columns sum in the table Fib(n+1) X Fib(n), where k = 1..Fib(n), and 1's are assigned to cells on the longest diagonal path.
%C A268317 Inspired by sun flower spirals which come in Fib(i) and Fib(i+1) numbers in opposite directions. The present case of the Fib(n+1) X Fib(n) table has the following properties:
%C A268317 (i) Columns sum create the present irregular triangle.
%C A268317 (ii) Rows sum create the irregular triangle A268318.
%C A268317 (iii) The row sum of each of these irregular triangles is conjectured to be A000071.
%C A268317 (iv) The first differences of the sequence of half of the voids (0's) are conjectured to give A191797.
%C A268317 See illustrations in the links.
%H A268317 Kival Ngaokrajang, <a href="/A268317/a268317.pdf">Illustration of initial terms</a>, <a href="/A268317/a268317_1.pdf">Sun flower spirals</a>
%e A268317 Irregular triangle begins:
%e A268317 1
%e A268317 2
%e A268317 2  2
%e A268317 2  3  2
%e A268317 2  3  2  3  2
%e A268317 2  3  2  3  3  2  3  2
%e A268317 2  3  2  3  3  2  3  2  3  3  2  3  2
%e A268317 ...
%o A268317 (Small Basic)
%o A268317 TextWindow.Write("0, 1, 2, 2, 2, 2, 3, 2, ")
%o A268317 t[4][1] = 2
%o A268317 t[4][2] = 3
%o A268317 t[4][3] = 2
%o A268317 k[3] = 2
%o A268317 k[4] = 3
%o A268317 For n = 5 To 12
%o A268317   k[n] = k[n-1]+k[n-2]
%o A268317   c = math.Ceiling(k[n]/2)
%o A268317   i1 = 1
%o A268317   For j = 1 To k[n]
%o A268317     If Math.Remainder(k[n],2)<>0 Then
%o A268317       If j > c then
%o A268317         t[n][j] = t[n][j-2*i1]
%o A268317         i1 = i1 + 1
%o A268317       Else
%o A268317         t[n][j] = t[n-1][j]
%o A268317       EndIf
%o A268317     Else
%o A268317       If j <= c then
%o A268317         t[n][j] = t[n-1][j]
%o A268317       Else
%o A268317         if j = c+1 Then
%o A268317           t[n][j] = t[n][j-1]
%o A268317         else
%o A268317         t[n][j] = t[n][j-(2*i1+1)]
%o A268317         i1 = i 1+ 1
%o A268317         endif
%o A268317       EndIf
%o A268317     EndIf
%o A268317     TextWindow.Write(t[n][j]+", ")
%o A268317   EndFor
%o A268317 EndFor
%Y A268317 Cf. A000071, A191797, A268318.
%K A268317 nonn,base,tabf
%O A268317 0,3
%A A268317 _Kival Ngaokrajang_, Feb 01 2016