A304373 Numbers n with additive persistence = 4.

19999999999999999999999, 28999999999999999999999, 29899999999999999999999, 29989999999999999999999, 29998999999999999999999, 29999899999999999999999, 29999989999999999999999, 29999998999999999999999, 29999999899999999999999, 29999999989999999999999

Offset: 1

Author

Jaroslav Krizek, May 28 2018

Keywords

• nonn
• base

Examples

			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.
		

Links

• Eric Weisstein's World of Mathematics, Additive Persistence

Crossrefs

Cf. A031286.

Cf. Numbers with additive persistence k: A304366 (k=1), A304367 (k=2), A304368 (k=3).

Programs

Mathematica

Take[ Sort@ Flatten[ (FromDigits /@ Permutations@#) & /@ IntegerPartitions[ 199, {23}, Range@ 9]], 10000] (* first 10000 terms, Giovanni Resta, May 29 2018 *)

PARI

nb(n) = {my(nba = 0); while (n > 9, n = sumdigits(n); nba++); nba;}
isok(n) = nb(n) == 4; \\ Michel Marcus, May 29 2018

Formula

A031286(a(n)) = 4.