A351256 Maximal exponent in the prime factorization of A351255(n).

0, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 1, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 3, 3, 4, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6

Offset

1,5

Comments

The largest digit in the primorial base representation of A328116(n).

The scatter plot gives some sense of how it is harder to eventually reach zero by iterating A003415, when starting from a number with large exponent(s) in its prime factorization. See also A099308.

Links

Antti Karttunen, Table of n, a(n) for n = 1..105368 (computed for all 19-smooth terms of A351255, and also for A276086(9699690) = 23)

Formula

a(n) = A328114(A328116(n)) = A051903(A351255(n)).

For all n, a(n) < A351257(n). [See A351258 for the differences].

Example

A328116(27) = 50, and A049345(50) = "1310", where the largest primorial base digit is 3, therefore a(27) = 3. Equally, A351255(27) = 2625 = 3 * 5^3 * 7, thus A051903(2625) = 3 and a(27) = 3.

Prog

(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));
A051903(n) = if((1==n), 0, vecmax(factor(n)[, 2]));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A099307(n) = { my(s=1); while(n>1, n = A003415checked(n); s++); if(n, s, 0); };
for(n=0, 2^9, u=A276086(n); c = A099307(u); if(c>0, print1(A051903(u), ", ")));

Crossrefs

Cf. A003415, A049345, A051903, A099308, A276086, A328114, A328116, A351255, A351257, A351258.

Sequence in context: A238013 A303940 A280134 * A143488 A201159 A356225

Adjacent sequences: A351253 A351254 A351255 * A351257 A351258 A351259

Keyword

nonn

Author

Antti Karttunen, Feb 11 2022

Status

approved