A268318 - OEIS

%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