A022470 Describe the previous term! (method B - initial term is 2).
2, 21, 2111, 2113, 211231, 2112213111, 211222113113, 21122312311231, 2112223111213112213111, 21122331132111311222113113, 211222321231211331122312311231, 21122331211121311121123212223111213112213111, 21122232112113211131132112213121112331132111311222113113
Offset: 1
Examples
E.g., the term after 2113 is obtained by saying "2 once, 1 twice, 3 once", which gives 211231.
Links
- Peter J. C. Moses, Table of n, a(n) for n = 1..23
Programs
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Mathematica
a[1] = 2; a[n_] := a[n] = FromDigits[Flatten[{First[#], Length[#]} & /@ Split[IntegerDigits[a[n - 1]]]]]; Map[a,Range[1, 23]] (* Peter J. C. Moses, Mar 22 2013 *)
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Python
from itertools import accumulate, groupby, repeat def summarize(n, _): return int("".join(k+str(len(list(g))) for k, g in groupby(str(n)))) def aupton(nn): return list(accumulate(repeat(2, nn), summarize)) print(aupton(13)) # Michael S. Branicky, Feb 21 2021
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