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Number of steps in base n after which all digits > 0 are present in x(i) where x(i) = x(i-1) + sum of missing digits of x(i-1) with x(0) = 0.
0

%I #14 Aug 03 2015 04:07:18

%S 1,2,12,35,266,2486,28781,409929,6824303,130615525,2846762052,

%T 69481280019,1875107560799,55466043249309,1791220240413582,

%U 62561482660495103,2351822439519905841,94645194753881075173,4053272062427790574310,185031225489554592624010

%N Number of steps in base n after which all digits > 0 are present in x(i) where x(i) = x(i-1) + sum of missing digits of x(i-1) with x(0) = 0.

%C Let x(i) = x(i-1) + A216407(x(i-1)), with x(0) = 0. a(n) is the smallest i such that A216407(x(i)) = 0.

%e 154 + missingdigit(154, 10) = 154 + 2+3+6+7+8+9 = 189.

%e The sequence looks like this in bases 2, 3 and 10:

%e 0 + missingdigit(0,2) = 0 + 1 = 1.

%e missingdigit(1,2) = 0. (1 step)

%e 0 + missingdigit(0,3) = 0 + (1+2) = 10.

%e 10 + missingdigit(10,3) = 10 + 2 = 12.

%e missingdigit(12,3) = 0. (2 steps)

%e 0 + missingdigit(0,10) = 0 + 45 = 45.

%e 45 + missingdigit(45) = 45 + 36 = 81.

%e 81 + missingdigit(81) = 81 + 36 = 117.

%e 117 + missingdigit(117) = 117 + 37 = 154.

%e 154 + missingdigit(154) = 154 + 35 = 189.

%e After 6824303 steps, the sequence reaches 123456978 and stops, because missingdigit(123456978) = 0.

%o (PARI) { bmx=12; for(b=2,bmx, n=0; miss=1; c=0; all=b*(b-1)/2; while(miss>0, d=vecsort(digits(n,b),,8); miss=all-sum(i=1,#d,d[i]); n+=miss; c++; ); print1(c-1,", "); ); }

%Y Cf. A216407.

%K nonn,base

%O 2,2

%A _Anthony Sand_, Aug 26 2014

%E a(13)-a(21) from _Hiroaki Yamanouchi_, Aug 02 2015