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A001155
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Describe the previous term! (method A - initial term is 0).
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20
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0, 10, 1110, 3110, 132110, 1113122110, 311311222110, 13211321322110, 1113122113121113222110, 31131122211311123113322110, 132113213221133112132123222110, 11131221131211132221232112111312111213322110, 31131122211311123113321112131221123113111231121123222110
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OFFSET
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1,2
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COMMENTS
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Method A = 'frequency' followed by 'digit'-indication.
a(n), A001140, A001141, A001143, A001145, A001151 and A001154 are all identical apart from the last digit of each term (the seed). This is because digits other than 1, 2 and 3 never arise elsewhere in the terms (other than at the end of each of them) of look-and-say sequences of this type (as is mentioned by Carmine Suriano in A006751). - Chayim Lowen, Jul 16 2015
a(n+1) - a(n) is divisible by 10^5 for n > 5. - Altug Alkan, Dec 04 2015
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REFERENCES
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S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 452-455.
I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 4.
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LINKS
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EXAMPLE
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The term after 3110 is obtained by saying "one 3, two 1's, one 0", which gives 132110.
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MATHEMATICA
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PROG
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(PARI) A001155(n, a=0)={ while(n--, my(c=1); for(j=2, #a=Vec(Str(a)), if( a[j-1]==a[j], a[j-1]=""; c++, a[j-1]=Str(c, a[j-1]); c=1)); a[#a]=Str(c, a[#a]); a=concat(a)); a } \\ M. F. Hasler, Jun 30 2011
(Python)
from itertools import accumulate, groupby, repeat
def summarize(n, _): return int("".join(str(len(list(g)))+k for k, g in groupby(str(n))))
def aupton(terms): return list(accumulate(repeat(0, terms), summarize))
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CROSSREFS
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KEYWORD
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nonn,base,easy,nice
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AUTHOR
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STATUS
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approved
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