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%I #31 Sep 01 2021 03:59:42
%S 9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,
%T 1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,
%U 1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1
%N a(n) = gcd(n,9).
%C Start with positive integer n. At each step, either (a) multiply by any positive integer or (b) remove all zeros from the number. a(n) is the smallest number that can be reached by this process. - _David W. Wilson_, Nov 01 2005
%C From _Martin Fuller_, Jul 09 2007: (Start)
%C Also the minimal positive difference between numbers whose digit sum is a multiple of n. Proof:
%C Construction: Pick a positive number that does not end with 9, and has a digit sum n-a(n). To form the lower number, append 9 until the digit sum is a multiple of n. This is always possible since the difference is gcd(n,9). Add a(n) to form the higher number, which will have digit sum n.
%C E.g., n=12: prefix=18, lower=18999, higher=19002, difference=3.
%C Minimality: All numbers are a multiple of a(n) if their digit sum is a multiple of n. Hence the minimal difference is at least a(n). (End)
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,1)
%F a(n) = 1 + 2*[3|n] + 6*[9|n], where [x|y] = 1 when x divides y, 0 otherwise.
%F a(n) = a(n-9).
%F Multiplicative with a(p^e, 9) = gcd(p^e, 9). - _David W. Wilson_ Jun 12 2005
%F G.f.: (-9 - x - x^2 - 3*x^3 - x^4 - x^5 - 3*x^6 - x^7 - x^8) / ((x-1)*(1 + x + x^2)*(x^6 + x^3 + 1)). - _R. J. Mathar_, Apr 04 2011
%F Dirichlet g.f.: (1+2/3^s+6/9^s)*zeta(s). - _R. J. Mathar_, Apr 04 2011
%o (PARI) a(n)=gcd(n,9) \\ _Charles R Greathouse IV_, Oct 07 2015
%o (Python)
%o from math import gcd
%o def a(n): return gcd(n, 9)
%o print([a(n) for n in range(101)]) # _Michael S. Branicky_, Sep 01 2021
%Y Cf. A109004.
%Y Cf. A109007, A109008, A109009, A109010, A109011, A109013, A109014, A109015.
%K nonn,easy,mult
%O 0,1
%A _Mitch Harris_
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