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A093086
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"Fibonacci in digits": start with a(0)=0, a(1)=1; repeatedly adjoin the digits of the sum of the next two terms.
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15
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0, 1, 1, 2, 3, 5, 8, 1, 3, 9, 4, 1, 2, 1, 3, 5, 3, 3, 4, 8, 8, 6, 7, 1, 2, 1, 6, 1, 4, 1, 3, 8, 3, 3, 7, 7, 5, 5, 4, 1, 1, 1, 1, 6, 1, 0, 1, 4, 1, 2, 1, 0, 9, 5, 2, 2, 2, 7, 7, 1, 1, 5, 5, 3, 3, 1, 9, 1, 4, 7, 4, 4, 9, 1, 4, 8, 2, 6, 1, 0, 8, 6, 4, 1, 0, 1, 0, 5, 1, 1, 1, 1, 8, 1, 3, 1, 0, 5, 1, 2, 1, 0, 8, 7, 1, 8
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OFFSET
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0,4
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COMMENTS
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Formally, define strings of digits S_i as follows. S_0={0}, S_1={0,1}. For n >= 1, let S_n={t_0, t_1, ..., t_z}. Then S_{n+1} is obtained by adjoining the digits of t_{n-1}+t_n to S_n. The sequence gives the limiting string S_oo.
All digits appear infinitely often, although the sequence is not periodic.
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LINKS
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EXAMPLE
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After S_6 = {0,1,1,2,3,5,8} we have 5+8 = 13, so we get
S_7 = {0,1,1,2,3,5,8,1,3}. Then 8+1 = 9, so we get
S_8 = {0,1,1,2,3,5,8,1,3,9}. Then 1+3 = 4, so we get
S_9 = {0,1,1,2,3,5,8,1,3,9,4}, and so on.
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MAPLE
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with(linalg): A:=matrix(1, 2, [0, 1]): for n from 1 to 100 do if A[1, n]+A[1, n+1]<10 then A:=concat(A, matrix(1, 1, A[1, n]+A[1, n+1])) else A:=concat(A, matrix(1, 2, [1, A[1, n]+A[1, n+1]-10])) fi od: matrix(A); # Emeric Deutsch, May 31 2005
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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