|
%I #14 Jan 11 2021 13:20:18
%S 1,11,31,4113,612314,8112332416,1113253342618,151528344153628,
%T 1817210364454648,102118211310455661768,3028110212311475962788,
%U 50331142143124851064711819,704111621731641051165713829,905011821931841251468714839,1105712022132141451569718869
%N a(n) contains the nonzero frequencies f(d) of digits d=0 .. 9 in all terms up to a(n-1) in concatenated form sorted with respect to d: f(0)//0//f(1)//1//...//f(9)//9. Initial term a(1)=1.
%H J. Mulder, <a href="/A177360/b177360.txt">Table of n, a(n) for n = 1..50000</a>
%e One; one one; three ones; four ones, one three; six ones, two threes, one four; eight ones, one two, three threes, two fours, one six; eleven ones, three twos, five threes, three fours, two sixes, one eight; etc.
%t lst = {Join[{e, 1}, Array[e &, 8]]}; Do[With[{k = Last@lst}, AppendTo[lst, ((k /. e -> 0) + With[{l = StringJoin @@ ToString /@ k}, Table[If[k[[i + 1]] =!= e, 1, 0] + StringCount[l, ToString[i]], {i, 0, 9}]]) /. {0 -> e}]], {1000}] lst = Prepend[ StringJoin @@ MapIndexed[ If[ # =!= e, ToString@# <> ToString[ #2[[1]] - 1], ""] &, # ] & /@ lst, "1"]; (* Jasper Mulder (jasper.mulder(AT)planet.nl), Jun 04 2010 *)
%o (Python)
%o def aupton(nn):
%o alst, last_str = [1], "1"
%o dig_counts = [0 for i in range(10)]
%o for n in range(2, nn+1):
%o nxt = []
%o for d in "0123456789":
%o if d in last_str: dig_counts[int(d)] += last_str.count(d)
%o if dig_counts[int(d)] > 0: nxt += [dig_counts[int(d)], int(d)]
%o nxt_str = "".join(map(str, nxt))
%o alst.append(int(nxt_str)); last_str = nxt_str
%o return alst
%o print(aupton(15)) # _Michael S. Branicky_, Jan 11 2021
%Y Cf. A060857, A177359 - A177368
%K easy,nonn,base
%O 1,2
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, May 10 2010
%E Terms corrected using values in b-file. - _N. J. A. Sloane_, Oct 05 2010
|
