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Triangular array read by rows. For T(n,k) where 1 <= k <= n, start with x = k and repeat the map x -> floor(n/x) + (n mod x) until an x occurs that has already appeared. The number of applications of the map is T(n,k).
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%I #21 Jan 29 2023 04:38:33

%S 1,2,2,2,1,2,2,1,2,2,2,2,1,3,2,2,2,2,3,3,2,2,2,1,1,2,3,2,2,2,3,2,3,4,

%T 3,2,2,2,1,2,1,3,2,3,2,2,2,2,1,2,3,2,3,3,2,2,2,2,3,2,1,3,4,3,3,2,2,2,

%U 2,2,3,2,3,4,3,3,3,2,2,2,2,1,1,3,1,4,2,2,3,3,2,2,2,4,3,3,3,2,3

%N Triangular array read by rows. For T(n,k) where 1 <= k <= n, start with x = k and repeat the map x -> floor(n/x) + (n mod x) until an x occurs that has already appeared. The number of applications of the map is T(n,k).

%C T(n,k) = 1 if k^2 >= n > k and k-1 divides n-k.

%C If A234575(n,k) = j then either T(n,k) = T(n,j)+1 and A357554(n,k) = A357554(n,j), or T(n,k) = T(n,j), A357554(n,k) = k and A357554(n,j) = j.

%H Robert Israel, <a href="/A358052/b358052.txt">Table of n, a(n) for n = 1..10011</a>(rows 1 to 141, flattened)

%e For T(13,2) we have 2 -> floor(13/2) + (13 mod 2) = 7 -> floor(13/7) + (13 mod 7) = 7. That is two applications of the map so T(13,2) = 2.

%e Triangle starts:

%e 1;

%e 2, 2;

%e 2, 1, 2;

%e 2, 1, 2, 2;

%e 2, 2, 1, 3, 2;

%e 2, 2, 2, 3, 3, 2;

%e 2, 2, 1, 1, 2, 3, 2;

%e 2, 2, 3, 2, 3, 4, 3, 2;

%e 2, 2, 1, 2, 1, 3, 2, 3, 2;

%e 2, 2, 2, 1, 2, 3, 2, 3, 3, 2;

%p f:= proc(n, k) local x, S,count;

%p S:= {k};

%p x:= k;

%p for count from 1 do

%p x:= iquo(n, x) + irem(n, x);

%p if member(x, S) then return count fi;

%p S:= S union {x};

%p od

%p end proc:

%p for n from 1 to 20 do seq(f(n, k), k=1..n) od;

%t f[n_, k_] := Module[{x, S, count}, S = {k}; x = k; For[count = 1, True, count++, x = Quotient[n, x] + Mod[n, x]; If[MemberQ[S, x], Return[count]]; S = S~Union~{x}]];

%t Table[f[n, k], {n, 1, 20}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Jan 29 2023, after Maple code *)

%Y Cf. A234575, A357554.

%K nonn,tabl

%O 1,2

%A _J. M. Bergot_ and _Robert Israel_, Oct 27 2022