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