%I #26 Apr 23 2022 18:13:22
%S 0,1,1,4,1,2,1,7,3,2,1,6,1,4,7,8,1,4,1,6,3,2,1,7,3,6,7,8,1,2,1,10,5,2,
%T 6,5,1,4,4,6,1,2,1,6,5,4,1,9,4,5,5,8,1,6,8,8,3,2,1,4,1,6,5,10,5,2,1,5,
%U 4,2,1,8,1,5,6,6,5,2,1,9,7,2,1,4,3,5,7,8,1,5,7,7,3,5,4,9,1,6,7,6,1,3,1,7,2
%N The length of the shortest path from n to 1 when using the transitions x -> A003415(x) and x -> A003961(x), or -1 if no 1 can ever be reached from n.
%C This is a variant of A327969 that seems to be less in need of an escape clause. Note that enough prime shifts with A003961 will eventually transform every term of A100716 (which is a subsequence of A099309) to a term of A048103, and that A051903(A003961(n)) = A051903(n). See also the array A344027.
%C Records 0, 1, 4, 7, 8, 10, 12, 13, 14, 15, 16, 19, ... occur at 1, 2, 4, 8, 16, 32, 128, 256, 768, 1024, 2048, 4096, ..., etc.
%H Antti Karttunen, <a href="/A353515/b353515.txt">Table of n, a(n) for n = 1..16383</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F a(1) = 0, a(p^p) = 1 + a(A003961(p^p)) for primes p, and for other numbers, a(n) = 1 + min(a(A003415(n)), a(A003961(n))).
%F a(p) = 1 for all primes p.
%F a(n) < A099307(n), unless A099307(n) = 0. [I.e., for all n in A099308]
%e From n = 4, we can reach 1 with just four steps as A003961(4) = 9, A003415(9) = 6, A003415(6) = 5 and A003415(5) = 1, and because there are no shorter paths we have a(4) = 4.
%e From n = 8, we can reach 1 with seven steps, as A003415(8) = 12, A003961(12) = 45, A003415(45) = 39, A003961(39) = 85, A003415(85) = 22, A003415(22) = 13, A003415(13) = 1, and because there are no shorter paths we have a(8) = 7.
%e For n = 15, as A003415(15) = 8, we know that a(15) is at most a(8)+1, i.e., 8. But we can do better, as A003961(15) = 35, A003961(35) = 77, A003415(77) = 18, A003415(18) = 21, A003415(21) = 10, A003415(10) = 7, A003415(7) = 1, and because there are no shorter paths we have a(15) = 7.
%e From n = 49, we can reach 1 in four steps, as A003961(49) = 121, A003415(121) = 22, A003415(22) = 13, A003415(13) = 1. Note that this is less than A099307(49)-1, as it would take 5 steps to reach 1 if using the arithmetic derivative only, 49 -> 14 -> 9 -> 6 -> 5 -> 1.
%o (PARI)
%o A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
%o A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
%o A353515(n) = if(1==n,0,my(xs=Set([n]),newxs,a,b,u); for(k=1,oo, newxs=Set([]); for(i=1,#xs,u = xs[i]; a = A003415(u); if(1==a, return(k)); if(isprime(a), return(k+1)); b = A003961(u); newxs = setunion([a],newxs); newxs = setunion([b],newxs)); xs = newxs));
%Y Cf. A003415, A003961, A048103, A051903, A099307, A099308, A099309, A100716, A349905.
%Y Cf. also A327969 and A343221, A344027, A353484.
%K nonn
%O 1,4
%A _Antti Karttunen_, Apr 23 2022