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A069862 Smallest k such that n divides the concatenation of numbers from (n+1) to (n+k), where (n+1) is on the most significant side. 10

%I #21 Oct 22 2014 11:23:41

%S 1,2,2,2,5,2,9,4,8,10,10,8,22,16,5,4,2,8,3,20,20,10,17,12,25,22,26,16,

%T 25,20,110,20,11,2,20,8,998,52,38,20,60,20,4,32,35,42,50,20,96,50,2,

%U 96,93,26,10,20,3,50,44,20,46,40,45,40,50,32,86,32,17,20,75,72,26,926,50

%N Smallest k such that n divides the concatenation of numbers from (n+1) to (n+k), where (n+1) is on the most significant side.

%C Minimum number of consecutive subsequent integers after n that must be concatenated together in ascending order such that n divides the concatenated term.

%C Concatenation always begins at n+1. Note that multiples of 11 seems to require more terms than any other number. 385 requires 9860. 451 requires 100270 terms be concatenated together into a 495,000 digit number. - Chuck Seggelin (barkeep(AT)plastereddragon.com), Oct 29 2003; corrected by _Chai Wah Wu_, Oct 19 2014

%H Chai Wah Wu, <a href="/A069862/b069862.txt">Table of n, a(n) for n = 1..10000</a>

%H C. Seggelin, <a href="http://www.plastereddragon.com/maths/catcon.htm">Concatenation of Consecutive Integers</a>.

%e a(7) = 9 as 7 divides 8910111213141516 the concatenation of numbers from 8(= 7+1) to 16 (= 7+9).

%e a(5) = 5 because 5 will divide the number formed by concatenating the 5 integers after 5 in ascending order (i.e. 678910). a(385) = 9860 because 385 will divide the concatenation of 386,387,388,...,10245.

%p c[1] := 1:for n from 2 to 172 do k := 1:g := (n+k) mod n:while(true) do k := k+1:b := convert(n+k,base,10):g := (g*10^nops(b)+n+k) mod n: if((g mod n)=0) then c[n] := k:break:fi:od:od:seq(c[l],l=1..172);

%t f[n_] := Block[{k = n + 1}, d = k; While[ !IntegerQ[d/n], k++; d = d*10^Floor[Log[10, k] + 1] + k]; k - n]; Table[ f[n], {n, 1, 75}] (* _Robert G. Wilson v_, Nov 04 2003 *)

%o (Python)

%o def A069862(n):

%o ....nk, kr, r = n+1, 1, 1 if n > 1 else 0

%o ....while r:

%o ........nk += 1

%o ........kr = (kr + 1) % n

%o ........r = (r*(10**len(str(nk)) % n)+kr) % n

%o ....return nk-n # _Chai Wah Wu_, Oct 20 2014

%Y Cf. A069860, A069861, A088797, A088799, A088868, A088870, A088872, A088885.

%Y Records are in A088947, A088343.

%K base,nonn

%O 1,2

%A _Amarnath Murthy_, Apr 18 2002

%E More terms from _Sascha Kurz_, Jan 28 2003

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)