OFFSET
1,5
FORMULA
Empirically:
A214635(7^n) = (1,6,42,294,...) = 6*7^(n-2) for n>1.
A214635(11^n) = (1,20,220,2420,...) = 20*11^(n-2) for n>1. - M. F. Hasler, Jul 24 2012
PROG
(PARI) A214635(n, N=99)={my(a=[Mod(1, n)]); for(n=1, N-1, a=concat(a, a[n]+(a[n]+1)*prod(k=1, n-1, a[k]))); for(p=1, N\3, forstep(m=N, p+1, -1, a[m]==a[m-p]&next; 3*m>N&next(2); return(p)); return(p))} /* the 2nd optional parameter must be taken large enough, at least 3 times the period length and starting position. The script returns zero if the period is not found (probably due to these constraints). */
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved