A294658 Number of steps required to reach either 5 or 13, starting with n, when iterating the map A125256: x -> smallest odd prime divisor of n^2+1; or a(n) = -1 in case 5 is never reached.

1, 1, 2, 0, 2, 1, 1, 4, 3, 5, 1, 0, 2, 2, 2, 1, 1, 4, 3, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 1, 1, 1, 2, 3, 1, 1, 2, 4, 3, 1, 1, 2, 2, 2, 1, 1, 4, 1, 4, 1, 1, 2, 2, 2, 1, 1, 5, 1, 3, 1, 1, 2, 3, 2, 1, 1, 4, 2, 2, 1, 1, 2, 2, 1, 1

Offset: 2

Author

M. F. Hasler, Nov 06 2017

Keywords

• nonn

Comments

The orbit or trajectory under A125256 appears to end in the cycle 5 -> 13 -> 5 -> etc. for any initial value n.

Sequence A294656 gives the size of the complete orbit of n under the map A125256, including the two elements 5 and 13 of the terminating cycle. Thus a(n) is 2 less than A294656(n) for all n. This is confirmed by careful examination of special cases - assuming, of course, that all trajectories end in the cycle (5, 13).

Examples

			For n = 1 the map A125256 is not defined.
a(2) = 1 because under A125256, 2 -> 2^2+1 = 5 (= its smallest odd prime factor), so 5 is reached after just a(2) = 1 iteration of this map.
a(3) = 1 because A125256(3) = 5, least odd prime factor of 3^2+1 = 10 = 2*5, so here again 5 is reached after just a(2) = 1 iteration of A125256.
a(4) = 2 because A125256(4) = 4^2 + 1 = 17, and A125256(17) = 5 = least odd prime factor of 17^2 + 1 = 289 + 1 = 2*5*29, so 5 is reached after a(4) = 2 iterations of A125256.
a(5) = a(13) = 0 because for these initial values 5 and 13, no iteration is needed until either 5 or 13 is reached.
		

Links

• Ray Chandler, Table of n, a(n) for n = 2..20001

Crossrefs

Cf. A125256, A294656, A294657.

Programs

PARI

A294658(n)=for(k=0,oo,bittest(8224,n)&&return(k);n=A125256(n)) \\ 8224 = 2^5 + 2^13. One could add 2^0 + 2^1 = 3 to avoid an error message for initial values 0 and 1, for which A125256 is not defined.

Formula

a(n) = A294656(n) - 2.