A351259 First noncomposite number reached when iterating the map x -> x', when starting from x = A351255(n). Here x' is the arithmetic derivative of x, A003415.

1, 2, 3, 5, 5, 7, 5, 7, 31, 7, 41, 71, 191, 2711, 7, 5, 7, 41, 103, 59, 71, 271, 71, 1031, 2887, 439, 5, 5, 7, 631, 251, 401, 3491, 1031, 1319, 17747, 9733, 1931, 16319, 77351, 131, 5, 419, 7079, 22343, 971, 5981, 6861581, 419, 18731, 11903, 33937, 7079, 15287, 15287, 6143, 6944111, 1415651, 11, 13, 5, 61, 103, 401, 631

Offset: 1

Author

Antti Karttunen, Feb 11 2022

Keywords

• nonn
• look

Comments

For the initial 105367 19-smooth terms of A351255, the last 7 occurs here at a(54796), with A351255(54796) = 289993286583 = 3^2 * 7 * 11 * 13^2 * 19^5, and the last 5 occurs here at a(65777), with A351255(65777) = 391899820830375516750 = 2 * 3^2 * 5^3 * 7^3 * 13^3 * 17^3 * 19^6, already a moderately high starting value, in whose vicinity most ending primes for successful iterations are much larger. This observation motivates a conjecture: Even from large numbers with high exponents in their prime factorization it is sometimes possible to reach a small prime. Compare to the conjecture 8 in Ufnarovski & Åhlander paper.

Examples

			From A351255(27) = 2625 it takes 12 iterations of the map x -> A003415(x) to reach zero: 2625 -> 2825 -> 1155 -> 886 -> 445 -> 94 -> 49 -> 14 -> 9 -> 6 -> 5 -> 1 -> 0. Two steps before the final zero is the first and only prime on the path, 5, therefore a(27) = 5.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..12878 (computed for all 17-smooth terms of A351255)
• Antti Karttunen, 105368 initial terms, without indices (computed for all 19-smooth terms of A351255, and also for A276086(9699690) = 23)
• Victor Ufnarovski and Bo Åhlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003, #03.3.4.

Crossrefs

Cf. A000040, A003415, A276086, A327975, A327977, A351078, A351255, A351257, A351261.

Programs

PARI

A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i,2]>=f[i,1],return(0), s += f[i, 2]/f[i, 1])); (n*s));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A351078(n) = { while(n>1&&!isprime(n), n = A003415checked(n)); (n); };
for(n=0, 2^9, u=A276086(n); p = A351078(u); if(p>0,print1(p, ", ")));

Formula

a(n) = A351078(A351255(n)).

a(1) = 1, and for n > 1, a(n) = A003415^[A351257(n)-2](A351255(n)). [This means: take the (A351257(n)-2)-th arithmetic derivative of A351255(n)].