A351079 a(n) is the largest term encountered on the path from n to 0 when iterating the map x -> x', or 0 if 0 cannot be reached from n (or if n is 0). Here x' is the arithmetic derivative of x, A003415.

0, 1, 2, 3, 0, 5, 6, 7, 0, 9, 10, 11, 0, 13, 14, 0, 0, 17, 21, 19, 0, 21, 22, 23, 0, 25, 0, 0, 0, 29, 31, 31, 0, 33, 34, 0, 0, 37, 38, 0, 0, 41, 42, 43, 0, 0, 46, 47, 0, 49, 0, 0, 0, 53, 0, 0, 0, 57, 58, 59, 0, 61, 62, 0, 0, 65, 66, 67, 0, 0, 70, 71, 0, 73, 0, 0, 0, 77, 78, 79, 0, 0, 82, 83, 0, 85, 0, 0, 0, 89

Offset: 0

Author

Antti Karttunen, Feb 11 2022

Keywords

• nonn

Comments

Question: Is there any good upper bound for ratio a(n)/n? See also comments in A351261.

Examples

			For n = 15, if we iterate with A003415, we get a path 15 -> 8 -> 12 -> 16 -> 32 -> 80 -> 176 -> 368 -> ..., where the terms just keep on growing without ever reaching zero, therefore a(15) = 0.
For n = 18, its path down to zero, when iterating A003415 is: 18 -> 21 -> 10 -> 7 -> 1 -> 0, and the largest term is 21, therefore a(18) = 21.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537

Crossrefs

Cf. A003415, A099307, A099309, A351078, A351261 [= a(A351255(n))].

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));
A351079(n) = { my(m=n); while(n>1, n = A003415checked(n); m = max(m,n)); if(n,m); };

Formula

For n > 0, a(n) = 0 if A099307(n) = 0, otherwise a(n) = max(n, a(A003415(n))).

a(0) = 0 and a(A099309(n)) = 0 for all n.