%I #9 Mar 03 2016 16:10:06
%S 0,1,1,1,1,2,1,1,2,1,2,1,1,2,1,2,2,1,2,1,1,2,1,2,2,1,2,1,2,2,1,2,1,1,
%T 2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,1,1,2,1,2,2,1,2,1,2,2,1,2,2,1,
%U 2,1,2,2,1,2,1,2,2,1,2,2,1,2,1,2,2,1,2,1,1,2,1,2,2,1,2,1,2,2,1,2,2
%N Irregular triangle read by rows: T(n,k) gives the row sums in the table Fib(n+1) X Fib(n), where k = 1..Fib(n+1), and 1's are assigned to cells on the longest diagonal path.
%C Inspired by sun flower spirals which come in Fib(i) and Fib(i+1) numbers in opposite directions. The present Fib(n+1) X Fib(n) table has the following properties:
%C (i) Columns sum create the irregular triangle A268317.
%C (ii) Rows sum create the present irregular triangle.
%C (iii) The row sums of each of these irregular triangles is conjectured to be A000071.
%C (iv) The first differences of the sequence of half of the voids (0's) are conjectured to give A191797.
%C See illustrations in the links of A268317.
%e Irregular triangle begins:
%e 0
%e 1
%e 1 1
%e 1 2 1
%e 1 2 1 2 1
%e 1 2 1 2 2 1 2 1
%e 1 2 1 2 2 1 2 1 2 2 1 2 1
%e 1 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 1
%e ...
%o (Small Basic)
%o TextWindow.Write("0, 1, 1, 1, 1, 2, 1, ")
%o t[3][1] = 1
%o t[3][2] = 2
%o t[3][3] = 1
%o k[2] = 2
%o k[3] = 3
%o For n = 4 To 12
%o k[n] = k[n-1] + k[n-2]
%o c = math.Ceiling(k[n]/2)
%o i1 = 1
%o For j = 1 To k[n]
%o If Math.Remainder(k[n],2)<>0 Then
%o If j > c then
%o t[n][j] = t[n][j-2*i1]
%o i1 = i1 + 1
%o Else
%o t[n][j] = t[n-1][j]
%o EndIf
%o Else
%o If j <= c then
%o t[n][j] = t[n-1][j]
%o Else
%o if j = c+1 Then
%o t[n][j] = t[n][j-1]
%o else
%o t[n][j] = t[n][j-(2*i1+1)]
%o i1 = i1 + 1
%o endif
%o EndIf
%o EndIf
%o TextWindow.Write(t[n][j]+", ")
%o EndFor
%o EndFor
%Y Cf. A000071, A191797, A268317.
%K nonn,base,tabf
%O 0,6
%A _Kival Ngaokrajang_, Feb 01 2016
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