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%I #11 Mar 03 2016 15:57:19
%S 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 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 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 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 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 (i) Columns sum create the present irregular triangle.
%C (ii) Rows sum create the irregular triangle A268318.
%C (iii) The row sum 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.
%H Kival Ngaokrajang, <a href="/A268317/a268317.pdf">Illustration of initial terms</a>, <a href="/A268317/a268317_1.pdf">Sun flower spirals</a>
%e Irregular triangle begins:
%e 1
%e 2
%e 2 2
%e 2 3 2
%e 2 3 2 3 2
%e 2 3 2 3 3 2 3 2
%e 2 3 2 3 3 2 3 2 3 3 2 3 2
%e ...
%o (Small Basic)
%o TextWindow.Write("0, 1, 2, 2, 2, 2, 3, 2, ")
%o t[4][1] = 2
%o t[4][2] = 3
%o t[4][3] = 2
%o k[3] = 2
%o k[4] = 3
%o For n = 5 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 = i 1+ 1
%o endif
%o EndIf
%o EndIf
%o TextWindow.Write(t[n][j]+", ")
%o EndFor
%o EndFor
%Y Cf. A000071, A191797, A268318.
%K nonn,base,tabf
%O 0,3
%A _Kival Ngaokrajang_, Feb 01 2016
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