A249280 - OEIS

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

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OFFSET

1,1

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

MATHEMATICA

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

PROG

(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

CROSSREFS

Cf. A056964.