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%I #30 Nov 01 2015 05:03:52
%S 102564,128205,142857,153846,179487,205128,230769,102564102564,
%T 128205128205,142857142857,153846153846,179487179487,205128205128,
%U 230769230769,1012658227848,1139240506329,102564102564102564
%N Numbers that are proper divisors of the number you get by rotating digits right once.
%C Let p(q) denote the period of the fraction q; then sequence is generated by p(i / (10k-1)), k=2,3,4,5,6,7,8,9; k <= i <= 9 and the concatenations of those periods, e.g., p(7/39)=a(5) p(2/19)=a(17).
%C Example if k=5: p((5+2)/49)=142857 which is in the sequence as the concatenations 142857142857, 142857142857142857, 142857142857142857142857, etc. - _Benoit Cloitre_, Feb 02 2002
%C The i in p(i / (10k-1)) is the last digit of the period, while k is equal to the ratio (right-rotated of p)/p. Thus no concatenation of any different such p's can be in the sequence. There are 8*9/2 = 36 terms which are not concatenation of previous terms, the last one being a(124) = 1525423728813559322033898305084745762711864406779661016949 with 58 digits. The term a(3)=p(7/49) is the only period of length (6) different from the length (42) of the other terms corresponding to the same value of k. - _M. F. Hasler_, Nov 18 2007
%C Numbers comprising multiple copies of a single digit, e.g., 111111, are not permitted. - _Harvey P. Dale_, Mar 08 2013
%C From _Emmanuel Vantieghem_, Oct 25 2015: (Start)
%C Subsequence of A245680.
%C Every element of the sequence is a multiple of 3.
%C The leading digit of every element is < 5.
%C (End)
%H M. F. Hasler, <a href="/A034089/b034089.txt">Table of n, a(n) for n = 1..124</a>
%o (PARI) period(p,q,S=[])=until(setsearch(S,p),S=setunion(S,[p]);p=10*p%q);S=[];until(p==S[1],S=concat(S,p);p=10*p%q);S*10\q /* print list of periods, right-rotated and ratio */ rotquo(n,d)={d=divrem(n,10);d[1]+=d[2]*10^#Str(d[1]);[n,d[1],d[1]/n]} for(k=2,9,for(i=k,9,print1( i/(10*k-1),"\t",rotquo(sum(j=1,#p=period(i,k*10-1),p[j]*10^(#p-j))))) /* build the sequence up to the greatest period */ A034089()={local(S=[],p); for(k=2,9,for(i=k,9,S=concat(S,sum(j=1,#p=period(i,k*10-1),p[j]*10^(#p-j))))); S=vecsort(S); for(i=1,#S, for(c=2,58\p=#Str(S[i]), S=concat(S,S[i]*(10^(c*p)-1)/(10^p-1)) )); vecsort(S)} \\ _M. F. Hasler_, Nov 18 2007
%K easy,nice,nonn,base
%O 1,1
%A _Erich Friedman_
%E Edited, corrected and extended by _M. F. Hasler_, Nov 18 2007
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