A175764 Number of iterations of the mapping k->f(k) to reach one of 2, 5, or 29, starting with k=n, and with f(k)=(k^2+4)/d, where d is the next-to-largest divisor of k^2+4, or -1 if the sequence never reaches one of the required values.

1, 0, 9, 1, 0, 1, 2, 1, 1, 1, 1, 1, 8, 1, 2, 1, 4, 1, 1, 1, 1, 1, 9, 1, 5, 1, 3, 1, 0, 1, 1, 1, 3, 1, 2, 1, 6, 1, 1, 1, 1, 1, 5, 1, 2, 1, 10, 1, 1, 1, 1, 1, 1, 1, 9, 1, 10, 1, 1, 1, 1, 1, 1, 1, 2, 1, 6, 1, 1, 1, 1, 1, 10, 1, 9, 1, 5, 1, 1, 1, 1, 1, 2, 1, 2, 1, 6, 1, 1, 1, 1, 1, 5, 1, 2, 1, 3, 1, 1, 1, 1, 1, 11

Offset: 1

Author

John W. Layman, Aug 30 2010

Keywords

• nonn

Comments

It appears that the sequence always reaches 2, 5, or 29 for any initial value n. Is this easy to prove?

It appears that a(n) is 1 whenever n>29 and n mod 10 is one of {0,1,2,4,6,8,9}. This has been verified to n=5000. Also, it appears that a(n) is 9 whenever n mod 130 is one of {3,23,55,75,107,127}. This has also been verified to n=5000. Are these conjectures easy to prove?

From Antti Karttunen, May 19 2021: (Start)

Note that the first four terms of iteration 47017 -> 2210598293 -> 4886744813014513853 -> 23880274867524255960728999629928905613 are all primes (see also A231120), but then (4+(23880274867524255960728999629928905613^2)) is composite, and its smallest prime divisor is 1946761. Actually, a(23880274867524255960728999629928905613) = 2, thus a(47017) = 5.

(End)

Examples

			For n=3, we have 3 -> (3^2+4)/d = 13/1 -> (13^2+4)/d = 173/1 -> (173^2+4)/d = 29933/809 = 37, since the divisors of 29933 are {1,37,809,29933}. Continuing, we get the orbit {3,13,173,37,1373,1217,97,9413,89,5,29,5,29,...}, showing that 5 is reached after 9 steps, after which the orbit is periodic {...,5,29,5,29,...}. Thus a(3)=9.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..47016

Crossrefs

Cf. A032742, A076423, A087717, A231120.

Programs

PARI

A175764(n) = if(2==n||5==n||29==n,0,1+A175764(f(n)));
f(k) = { my(u=(4+(k^2)), ds=divisors(u)); (u/ds[#ds-1]); };
\\ Alternatively, "f" could be defined as:
f(k) = { my(u=(4+(k^2))); (u/A032742(u)); };
A032742(n) = if(1==n||isprime(n),1,forprime(p=2,n,if(!(n%p),return(n/p)))); \\ And not requiring full factorization when this is used. - Antti Karttunen, May 19 2021