A353515 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.

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, 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, 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

Offset

1,4

Comments

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.

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.

Links

Antti Karttunen, Table of n, a(n) for n = 1..16383

Index entries for sequences computed from indices in prime factorization

Formula

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))).

a(p) = 1 for all primes p.

a(n) < A099307(n), unless A099307(n) = 0. [I.e., for all n in A099308]

Example

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.
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.
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.
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.

Prog

(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
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));

Crossrefs

Cf. A003415, A003961, A048103, A051903, A099307, A099308, A099309, A100716, A349905.

Cf. also A327969 and A343221, A344027, A353484.

Sequence in context: A074695 A069098 A126241 * A019777 A337515 A090885

Adjacent sequences: A353512 A353513 A353514 * A353516 A353517 A353518

Keyword

nonn

Author

Antti Karttunen, Apr 23 2022

Status

approved