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.

38, 37, 35, 36, 54, 34, 45, 35, 48, 53, 52, 33, 51, 44, 32, 34, 50, 47, 43, 52, 33, 51, 44, 32, 34, 50, 47, 43, 42, 33, 51, 44, 32, 34, 50, 47, 43, 42, 31, 51, 44, 32, 34, 50, 47, 43, 42, 31, 41, 44, 32, 34, 50, 47, 43, 42, 31, 41, 33, 32, 34, 50, 47, 43, 42

Offset: 1

Author

Felix Fröhlich, Oct 26 2014

Keywords

• nonn
• base
• easy

Comments

Conjecture 1: a(n) exists for all n.

Conjecture 2: There exists an upper bound c such that a(n) < c for all n.

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

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Felix Fröhlich, C++ program for this sequence

Crossrefs

Cf. A056964.

Programs

Mathematica

Table[Length[NestWhileList[#+IntegerReverse[#]&,n,Min[DigitCount[#]] == 0&]]-1,{n,70}] (* Harvey P. Dale, Aug 20 2022 *)

PARI

fromdigits(v,b=10)=subst(Pol(v),'x,b) \\ needed for gp < 2.63 or so
A056964(n)=fromdigits(Vecrev(digits(n)))+n
ispan(n)=#Set(digits(n))==10
a(n)=my(k); while(!ispan(n), n=A056964(n); k++); k \\ Charles R Greathouse IV, Oct 28 2014