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

1,1

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.

Sequence in context: A143721 A333850 A070725 * A022994 A023480 A036743

Adjacent sequences: A249277 A249278 A249279 * A249281 A249282 A249283

Keyword

nonn,base,easy

Author

Felix Fröhlich, Oct 26 2014

Status

approved