A377384 a(n) is the number of iterations that n requires to reach a noninteger or a factorial number under the map x -> x / f(x), where f(k) = A034968(k) is the sum of digits in the factorial-base representation of k; a(n) = 0 if n is a factorial number.

0, 0, 1, 1, 1, 0, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 0, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset

1,8

Comments

The factorial numbers are fixed points of the map, since f(k!) = 1 for all k >= 0. Therefore they are arbitrarily assigned the value a(k!) = 0.

Each number n starts a chain of a(n) integers: n, n/f(n), (n/f(n))/f(n/f(n)), ..., of them the first a(n)-1 integers are factorial-base Niven numbers (A118363).

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

a(n) = 0 if and only if n is in A000142 (by definition).

a(n) = 1 if and only if n is in A286607.

a(n) >= 2 if and only if n is in A118363 \ A000142 (i.e., n is a factorial-base Niven number that is not a factorial number).

a(n) >= 3 if and only if n is in A377385 \ A000142.

a(n) >= 4 if and only if n is in A377386 \ A000142.

a(n) < A000005(n).

Example

a(8) = 2 since 8/f(8) = 4 and 4/f(4) = 2 is a factorial number that is reached after 2 iterations.
a(27) = 3 since 27/f(27) = 9, 9/f(9) = 3 and 3/f(3) = 3/2 is a noninteger that is reached after 3 iterations.

Mathematica

fdigsum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s += r; m++]; s]; a[n_] := a[n] = Module[{s = fdigsum[n]}, If[s == 1, 0, If[!Divisible[n, s], 1, 1 + a[n/s]]]]; Array[a, 100]

Prog

(PARI) fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s; }
a(n) = {my(f = fdigsum(n)); if(f == 1, 0, if(n % f, 1, 1 + a(n/f))); }
(Python)
def f(n, p=2): return n if n<p else f(n//p, p+1) + n%p
def a(n): return 0 if (fn:=f(n)) == 1 else (1 if n%fn else 1 + a(n//fn))
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Oct 27 2024

Crossrefs

Cf. A000005, A000142, A034968, A118363, A286607, A377385, A377386.

Analogous sequences: A376615 (binary), A377208 (Zeckendorf).

Sequence in context: A156839 A172299 A025453 * A094215 A329040 A296978

Adjacent sequences: A377378 A377379 A377380 * A377385 A377386 A377387

Keyword

nonn,easy,base

Author

Amiram Eldar, Oct 27 2024

Status

approved