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%I #11 Jul 16 2018 16:32:42
%S 19999999999999999999999,28999999999999999999999,
%T 29899999999999999999999,29989999999999999999999,
%U 29998999999999999999999,29999899999999999999999,29999989999999999999999,29999998999999999999999,29999999899999999999999,29999999989999999999999
%N Numbers n with additive persistence = 4.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AdditivePersistence.html">Additive Persistence</a>
%F A031286(a(n)) = 4.
%e Repeatedly taking the sum of digits starting with 19999999999999999999999 gives 199, 19, 10 and 1. There are four steps, so the additive persistence is 4 and 19999999999999999999999 is a member.
%t Take[ Sort@ Flatten[ (FromDigits /@ Permutations@#) & /@ IntegerPartitions[ 199, {23}, Range@ 9]], 10000] (* first 10000 terms, _Giovanni Resta_, May 29 2018 *)
%o (PARI) nb(n) = {my(nba = 0); while (n > 9, n = sumdigits(n); nba++); nba;}
%o isok(n) = nb(n) == 4; \\ _Michel Marcus_, May 29 2018
%Y Cf. A031286.
%Y Cf. Numbers with additive persistence k: A304366 (k=1), A304367 (k=2), A304368 (k=3).
%K nonn,base
%O 1,1
%A _Jaroslav Krizek_, May 28 2018
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