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%I #34 Aug 28 2015 17:48:15
%S 5,8,9,18,21,24,26,27,36,44,45,50,54,60,62,63,72,80,81,86,90,108,116,
%T 117,126,132,134,135,140,144,152,153,162,170,171,180,200,204,206,207,
%U 210,216,224,225,230,234,240,242,243,252,260,261,264,270,306,312,314
%N Numbers n such that n is divisible by (10^k - digitsum(n)), where k equals the number of digits of digitsum(n).
%C This sequence is infinite because all numbers with a digitsum equal to 9 are part of this sequence.
%H Pieter Post, <a href="/A260348/b260348.txt">Table of n, a(n) for n = 1..12089</a>
%e a(1) = 5, because 5 divided by (10 - 5) equals 1.
%e a(7) = 26, because digitsum(26) = 8 and 26 divided by (10 - 8) equals 13.
%e a(20) = 86, the first member of this sequence where digitsum(n) >= 10. Digitsum(86) = 14, so k = 10^2 - 14 = 86, so 86 is a member of this sequence.
%t fQ[n_] := Block[{d = Total@ IntegerDigits@ n, k}, k = IntegerLength@ d;
%t Divisible[n, 10^k - d]]; Select[Range@ 314, fQ] (* or *)
%t Select[Range@ 314, Divisible[#, (10^(Floor[Log[10, Total@ IntegerDigits@ #]] + 1) - Total@ IntegerDigits@ #)] &] (* _Michael De Vlieger_, Aug 05 2015 *)
%o (Python)
%o def sod(n,m):
%o ....kk = 0
%o ....while n > 0:
%o ........kk= kk+(n%m)
%o ........n =int(n//m)
%o ....return kk
%o for c in range (1, 10**6):
%o ....k=len(str(sod(c,10)))
%o ....kl=10**k-sod(c,10)
%o ....if c%kl==0:
%o ........print (c)
%o (PARI) isok(n)=my(sd = sumdigits(n), nsd = #digits(sd)); n % (10^nsd - sd) == 0; \\ _Michel Marcus_, Aug 05 2015
%Y Cf. A005349, A007953, A113315.
%K nonn,base,less
%O 1,1
%A _Pieter Post_, Jul 23 2015
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