A214636 A213437 becomes periodic mod n starting at this position.

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, 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, 3, 7, 4, 8, 3, 4, 6, 3, 6, 5, 3, 5, 4, 3, 1, 7, 4, 3, 8, 10, 5, 3, 3, 4

Offset: 1

Author

David Applegate, M. F. Hasler and N. J. A. Sloane, Jul 23 2012

Keywords

• nonn

Crossrefs

Cf. A213437, A214635.

Programs

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). */

Formula

Empirically,

A214636(2^n) = (1,2,3,4,4,5,6,6,7,8,8,...) = A004523(n+2) for n>1.

A214636(3^n) = 3, A214636(7^n) = 4, A214636(11^n) = 5 for all n>0.

A214636(5^n) = A214636(10^n) = (1,2,5,8,11,...) = A016789(n-2) for n>1.

A214636(6^n) = (3,3,3,4,4,5,6,6,...) = A214636(2^n) for n>2.

A214636(15^n) = (3,3,5,8,11,...) = A214636(5^n) for n>2. - M. F. Hasler, Jul 24 2012