login
A034089
Numbers that are proper divisors of the number you get by rotating digits right once.
6
102564, 128205, 142857, 153846, 179487, 205128, 230769, 102564102564, 128205128205, 142857142857, 153846153846, 179487179487, 205128205128, 230769230769, 1012658227848, 1139240506329, 102564102564102564
OFFSET
1,1
COMMENTS
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).
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
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
Numbers comprising multiple copies of a single digit, e.g., 111111, are not permitted. - Harvey P. Dale, Mar 08 2013
From Emmanuel Vantieghem, Oct 25 2015: (Start)
Subsequence of A245680.
Every element of the sequence is a multiple of 3.
The leading digit of every element is < 5.
(End)
LINKS
PROG
(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
CROSSREFS
Sequence in context: A241789 A010329 A184567 * A146569 A081463 A321149
KEYWORD
easy,nice,nonn,base
EXTENSIONS
Edited, corrected and extended by M. F. Hasler, Nov 18 2007
STATUS
approved