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%I #15 Oct 18 2014 09:12:26
%S 1,1,3,2,1,3,4,3,3,1,5,3,1,4,3,4,3,3,6,2,4,5,7,3,2,1,3,4,10,3,5,4,5,3,
%T 4,3,6,6,3,3,1,4,8,5,3,7,11,4,4,2,3,2,8,3,5,4,6,10,9,3,6,5,4,5,1,5,11,
%U 3,7,4,8,3,4,6,3,6,5,3,5,4,3,1,7,4,3,8,10,5,3,3,4
%N A213437 becomes periodic mod n starting at this position.
%F Empirically,
%F A214636(2^n) = (1,2,3,4,4,5,6,6,7,8,8,...) = A004523(n+2) for n>1.
%F A214636(3^n) = 3, A214636(7^n) = 4, A214636(11^n) = 5 for all n>0.
%F A214636(5^n) = A214636(10^n) = (1,2,5,8,11,...) = A016789(n-2) for n>1.
%F A214636(6^n) = (3,3,3,4,4,5,6,6,...) = A214636(2^n) for n>2.
%F A214636(15^n) = (3,3,5,8,11,...) = A214636(5^n) for n>2. - _M. F. Hasler_, Jul 24 2012
%o (PARI) A214636(n, N=199)={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(m-p+1)); return(1))} /* 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 (most probably due to these constraints). */
%Y Cf. A213437, A214635.
%K nonn
%O 1,3
%A _David Applegate_, _M. F. Hasler_ and _N. J. A. Sloane_, Jul 23 2012
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