1,1

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 (benoit7848c(AT)orange.fr), 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

M. F. Hasler, Table of n, a(n) for n = 1..124

(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

Sequence in context: A106814 A074669 A010329 this_sequence A146569 A081463 A014884

Adjacent sequences: A034086 A034087 A034088 this_sequence A034090 A034091 A034092

easy,nice,nonn,base

Erich Friedman (erich.friedman(AT)stetson.edu)

Edited, corrected and extended by M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Nov 18 2007