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A249280 Repeatedly apply 'Reverse and add' to n. a(n) gives the number of steps needed to reach a sum containing each digit from 0 to 9 at least once.

%I A249280 #28 Aug 20 2022 18:04:21
%S A249280 38,37,35,36,54,34,45,35,48,53,52,33,51,44,32,34,50,47,43,52,33,51,44,
%T A249280 32,34,50,47,43,42,33,51,44,32,34,50,47,43,42,31,51,44,32,34,50,47,43,
%U A249280 42,31,41,44,32,34,50,47,43,42,31,41,33,32,34,50,47,43,42
%N A249280 Repeatedly apply 'Reverse and add' to n. a(n) gives the number of steps needed to reach a sum containing each digit from 0 to 9 at least once.
%C A249280 Conjecture 1: a(n) exists for all n.
%C A249280 Conjecture 2: There exists an upper bound c such that a(n) < c for all n.
%C A249280 The conjectures seem highly likely, especially since a(n) = 0 for almost all n. A lower bound for c is a(1418993) = 73. (Checked to 10^9.) - _Charles R Greathouse IV_, Oct 28 2014
%H A249280 Charles R Greathouse IV, <a href="/A249280/b249280.txt">Table of n, a(n) for n = 1..10000</a>
%H A249280 Felix Fröhlich, <a href="/A249280/a249280.txt">C++ program for this sequence</a>
%t A249280 Table[Length[NestWhileList[#+IntegerReverse[#]&,n,Min[DigitCount[#]] == 0&]]-1,{n,70}] (* _Harvey P. Dale_, Aug 20 2022 *)
%o A249280 (PARI) fromdigits(v,b=10)=subst(Pol(v),'x,b) \\ needed for gp < 2.63 or so
%o A249280 A056964(n)=fromdigits(Vecrev(digits(n)))+n
%o A249280 ispan(n)=#Set(digits(n))==10
%o A249280 a(n)=my(k); while(!ispan(n), n=A056964(n); k++); k \\ _Charles R Greathouse IV_, Oct 28 2014
%Y A249280 Cf. A056964.
%K A249280 nonn,base,easy
%O A249280 1,1
%A A249280 _Felix Fröhlich_, Oct 26 2014