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A093094
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"Products into digits": start with a(1)=2, a(2)=2; adjoin digits of product of a(k) and a(k+1) for k from 1 to infinity.
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5
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2, 2, 4, 8, 3, 2, 2, 4, 6, 4, 8, 2, 4, 2, 4, 3, 2, 1, 6, 8, 8, 8, 1, 2, 6, 2, 6, 4, 8, 6, 4, 6, 4, 8, 2, 1, 2, 1, 2, 1, 2, 2, 4, 3, 2, 4, 8, 2, 4, 2, 4, 2, 4, 3, 2, 1, 6, 2, 2, 2, 2, 2, 2, 4, 8, 1, 2, 6, 8, 3, 2, 1, 6, 8, 8, 8, 8, 8, 1, 2, 6, 2, 6, 1, 2, 4, 4, 4, 4, 4, 8, 3, 2, 8, 2, 1, 2, 4, 8, 2, 4
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OFFSET
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1,1
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COMMENTS
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Only the digits 1,2,3,4,6,8 occur, infinitely often. The sequence is not periodic. Around a(800) there are many 8's.
Proof that sequence is not periodic:
Let us assume that somewhere in the sequence there is a subsequence of 3 adjacent 8': ...,8,8,8,....(which is true).
Then we know that in the following there will be the subsequence ...,6,4,6,4.. (i.e. 8x8, 8x8) again, there will be somewhere ...,2,4,2,4,2,4,... (i.e. 6x4, 4x6, 6x4) and finally ...,8,8,8,8,8,...
Analogously, starting from 8,8,8,8 we obtain 6,4,6,4,6,4 then 2,4,2,4,2,4,2,4,2,4 and finally 8,8,8,8,8,8,8,8,8.
Generalizing, if somewhere appears a run of k>2 8's, then in some future position will appear a run of at least 4*k-7 8's (where since k>2, 4*k-7>k).
So the sequence will contain arbitrary long runs of 8's, without being constantly equal to 8, thus it cannot be periodic. (End)
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LINKS
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EXAMPLE
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a(3)=a(1)*a(2), a(4)=a(2)*a(3), a(5)=first digit of (a(3)*a(4)), a(6)=2nd digit of (a(3)*a(4)), a(9)=a(6)*a(7)
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PROG
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(Haskell)
a093094 n = a093094_list !! (n-1)
a093094_list = f [2, 2] where
f (u : vs@(v : _)) = u : f (vs ++
if w < 10 then [w] else uncurry ((. return) . (:)) $ divMod w 10)
where w = u * v
(Python)
from itertools import islice
from collections import deque
def agen(): # generator of terms
a = deque([2, 2])
while True:
a.extend(list(map(int, str(a[0]*a[1]))))
yield a.popleft()
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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