A351256 - OEIS

%I #14 Feb 12 2022 17:46:55

%S 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,

%T 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,

%U 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

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

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

%C 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.

%H Antti Karttunen, <a href="/A351256/b351256.txt">Table of n, a(n) for n = 1..105368</a> (computed for all 19-smooth terms of A351255, and also for A276086(9699690) = 23)

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

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

%e 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.

%o (PARI)

%o 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));

%o A051903(n) = if((1==n),0,vecmax(factor(n)[, 2]));

%o A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };

%o A099307(n) = { my(s=1); while(n>1, n = A003415checked(n); s++); if(n,s,0); };

%o for(n=0, 2^9, u=A276086(n); c = A099307(u); if(c>0,print1(A051903(u), ", ")));

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

%K nonn

%O 1,5

%A _Antti Karttunen_, Feb 11 2022