A001155 Describe the previous term! (method A - initial term is 0).

0, 10, 1110, 3110, 132110, 1113122110, 311311222110, 13211321322110, 1113122113121113222110, 31131122211311123113322110, 132113213221133112132123222110, 11131221131211132221232112111312111213322110, 31131122211311123113321112131221123113111231121123222110

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• base
• easy
• nice

Comments

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

Examples

			The term after 3110 is obtained by saying "one 3, two 1's, one 0", which gives 132110.
		

References

• S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 452-455.
• I. Vardi, Computational Recreations in Mathematica. Addison-Wesley, Redwood City, CA, 1991, p. 4.

Links

• T. D. Noe, Table of n, a(n) for n=1..20
• J. H. Conway, The weird and wonderful chemistry of audioactive decay, in T. M. Cover and Gopinath, eds., Open Problems in Communication and Computation, Springer, NY 1987, pp. 173-188.
• S. R. Finch, Conway's Constant [Broken link]
• S. R. Finch, Conway's Constant [From the Wayback Machine]

Crossrefs

Cf. A005150, A006751, A006715, A001140, A001141, A001143, A001145, A001151, A001154.

Cf. A036058.

Programs

Mathematica

A001155[1] := 0; A001155[n_] := A001155[n] = FromDigits[Flatten[{Length[#], First[#]}&/@Split[IntegerDigits[A001155[n-1]]]]]; Map[A001155,Range[15]] (* Peter J. C. Moses, Mar 21 2013 *)

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))
print(aupton(11)) # Michael S. Branicky, Jun 28 2022