A357610 Start with x = 3 and repeat the map x -> floor(n/x) + (n mod x) until an x occurs that has already appeared, then that is a(n).
1, 2, 3, 2, 3, 3, 3, 4, 3, 4, 3, 3, 5, 6, 3, 6, 5, 3, 7, 5, 3, 8, 7, 3, 9, 6, 3, 10, 9, 3, 11, 8, 3, 12, 5, 3, 13, 10, 3, 14, 9, 3, 15, 11, 3, 16, 11, 3, 17, 5, 3, 18, 5, 3, 19, 12, 3, 20, 11, 3, 21, 14, 3, 22, 5, 3, 23, 8, 3, 24, 15, 3, 25, 14, 3, 26, 17, 3, 27, 5, 3, 28, 11, 3, 29, 18, 3, 30, 9
Offset: 1
Examples
a(4) = 2 because we have 3 -> floor(4/3) + (4 mod 3) = 2 -> floor(4/2) + (4 mod 2) = 2.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A357554.
Programs
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Maple
g:= proc(n, k) local x, S; S:= {k}; x:= k; do x:= iquo(n, x) + irem(n, x); if member(x, S) then return x fi; S:= S union {x}; od end proc: seq(g(n,3), n=1..100);
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Mathematica
T[n_, k_] := Module[{x, S}, S = {k}; x = k; While[True, x = Total@QuotientRemainder[n, x]; If[MemberQ[S, x], Return[x]]; S = S~Union~{x}]]; a[n_] := T[n, 3]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 17 2022, after Maple code *)
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Python
def a(n): seen, x = set(), 3 while x not in seen: seen.add(x); q, r = divmod(n, x); x = q + r return x print([a(n) for n in range(1, 90)]) # Michael S. Branicky, Oct 06 2022~
Formula
a(n) = 3 if n is divisible by 3.
a(3*k+1) = k+1.
a(15*k+5) = 5 for k >= 1.
a(15*k+11) = 3*k+3.
a(255*k+137) = 15*k+9 for k >= 1.
Comments