A246426 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.
1, 2, 12, 35, 266, 2486, 28781, 409929, 6824303, 130615525, 2846762052, 69481280019, 1875107560799, 55466043249309, 1791220240413582, 62561482660495103, 2351822439519905841, 94645194753881075173, 4053272062427790574310, 185031225489554592624010
Offset: 2
Examples
154 + missingdigit(154, 10) = 154 + 2+3+6+7+8+9 = 189. The sequence looks like this in bases 2, 3 and 10: 0 + missingdigit(0,2) = 0 + 1 = 1. missingdigit(1,2) = 0. (1 step) 0 + missingdigit(0,3) = 0 + (1+2) = 10. 10 + missingdigit(10,3) = 10 + 2 = 12. missingdigit(12,3) = 0. (2 steps) 0 + missingdigit(0,10) = 0 + 45 = 45. 45 + missingdigit(45) = 45 + 36 = 81. 81 + missingdigit(81) = 81 + 36 = 117. 117 + missingdigit(117) = 117 + 37 = 154. 154 + missingdigit(154) = 154 + 35 = 189. After 6824303 steps, the sequence reaches 123456978 and stops, because missingdigit(123456978) = 0.
Crossrefs
Cf. A216407.
Programs
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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,", "); ); }
Extensions
a(13)-a(21) from Hiroaki Yamanouchi, Aug 02 2015
Comments