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A268317
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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.
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3
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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, 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, 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
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OFFSET
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0,3
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COMMENTS
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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:
(i) Columns sum create the present irregular triangle.
(ii) Rows sum create the irregular triangle A268318.
(iii) The row sum of each of these irregular triangles is conjectured to be A000071.
(iv) The first differences of the sequence of half of the voids (0's) are conjectured to give A191797.
See illustrations in the links.
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LINKS
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EXAMPLE
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Irregular triangle begins:
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
...
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PROG
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(Small Basic)
TextWindow.Write("0, 1, 2, 2, 2, 2, 3, 2, ")
t[4][1] = 2
t[4][2] = 3
t[4][3] = 2
k[3] = 2
k[4] = 3
For n = 5 To 12
k[n] = k[n-1]+k[n-2]
c = math.Ceiling(k[n]/2)
i1 = 1
For j = 1 To k[n]
If Math.Remainder(k[n], 2)<>0 Then
If j > c then
t[n][j] = t[n][j-2*i1]
i1 = i1 + 1
Else
t[n][j] = t[n-1][j]
EndIf
Else
If j <= c then
t[n][j] = t[n-1][j]
Else
if j = c+1 Then
t[n][j] = t[n][j-1]
else
t[n][j] = t[n][j-(2*i1+1)]
i1 = i 1+ 1
endif
EndIf
EndIf
TextWindow.Write(t[n][j]+", ")
EndFor
EndFor
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CROSSREFS
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KEYWORD
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nonn,base,tabf
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AUTHOR
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STATUS
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approved
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