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A175764
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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.
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1
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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
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OFFSET
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1,3
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COMMENTS
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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?
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.
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LINKS
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EXAMPLE
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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.
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PROG
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(PARI)
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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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