A351259 - OEIS

%I #18 Feb 12 2022 17:47:25

%S 1,2,3,5,5,7,5,7,31,7,41,71,191,2711,7,5,7,41,103,59,71,271,71,1031,

%T 2887,439,5,5,7,631,251,401,3491,1031,1319,17747,9733,1931,16319,

%U 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

%N 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.

%C 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.

%H Antti Karttunen, <a href="/A351259/b351259.txt">Table of n, a(n) for n = 1..12878</a> (computed for all 17-smooth terms of A351255)

%H Antti Karttunen, <a href="/A351259/a351259.txt">105368 initial terms, without indices</a> (computed for all 19-smooth terms of A351255, and also for A276086(9699690) = 23)

%H Victor Ufnarovski and Bo Åhlander, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.html">How to Differentiate a Number</a>, J. Integer Seqs., Vol. 6, 2003, #03.3.4.

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

%F 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)].

%e 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.

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

%o A351078(n) = { while(n>1&&!isprime(n), n = A003415checked(n)); (n); };

%o for(n=0, 2^9, u=A276086(n); p = A351078(u); if(p>0,print1(p, ", ")));

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

%K nonn,look

%O 1,2

%A _Antti Karttunen_, Feb 11 2022