A031286 Additive persistence: number of summations of digits needed to obtain a single digit (the additive digital root).
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 0
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..10000
- Antonios Meimaris, On the additive persistence of a number in base p, Preprint, 2015.
- N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
- Eric Weisstein's World of Mathematics, Additive Persistence
Crossrefs
Programs
-
Maple
read("transforms") ; A031286 := proc(n) local a,nper; nper := n ; a := 0 ; while nper > 9 do nper := digsum(nper) ; a := a+1 ; end do: a ; end proc: seq(A031286(n),n=0..80) ; # R. J. Mathar, Jan 02 2018
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Mathematica
lst = {}; Do[s = 0; While[n > 9, s++; n = Plus @@ IntegerDigits[n]]; AppendTo[lst, s], {n, 0, 98}]; lst (* Arkadiusz Wesolowski, Oct 17 2012 *)
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PARI
dsum(n)=my(s);while(n,s+=n%10;n\=10);s a(n)=my(s);while(n>9,s++;n=dsum(n));s \\ Charles R Greathouse IV, Sep 13 2012
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Python
def A031286(n): ap = 0 while n > 9: n = sum(int(d) for d in str(n)) ap += 1 return ap # Chai Wah Wu, Aug 23 2014
Extensions
Corrected by Reinhard Zumkeller, Feb 05 2009