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%I #8 Jun 15 2013 23:41:45
%S 1,3,2,2,5,2,6,2,7,2,2,2,2,11,2,12,2,2,2,2,2,17,2,18,2,2,2,2,2,2,24,2,
%T 25,2,26,2,27,2,28,2,2,2,2,2,2,2,2,2,2,38,2,39,2,2,2,2,2,2,2,2,2,2,2,
%U 50,2,51,2,52,2,53,2,54,2,55,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,71,2,72,2,2,2
%N a(n) = length of n-th run of consecutive numbers in A136120.
%H Klaus Brockhaus, <a href="/A139420/b139420.txt">Table of n, a(n) for n=1..395</a>
%e A136120 starts 1, 3, 4, 5, 9, 10, 15, ..., so a(1) = 1, a(2) = 3, a(3) = 2.
%t f[seq_] := Module[{s = seq, n1, n2}, n++; n1 = s[[n]] + n; If[n1 <= len, n2 = Min[n - 1 + 2*s[[n]], len]; len -= n2 - n1 + 1; Drop[s, {n1, n2}], s]]; n = 0; len = 29800; Length /@Split[FixedPoint[f, Range[len]], #2 - #1 == 1 &] (* _Jayanta Basu_, Jun 15 2013 - using the Mma program in A136120 *)
%Y Cf. A136120, A139418, A139419.
%K nonn
%O 1,2
%A _Klaus Brockhaus_, Apr 22 2008
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