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Least positive integer k such that n divides (2k)! - (2j)! for some j in [1,k-1].
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%I #15 Nov 20 2024 19:17:51

%S 2,2,3,3,4,3,5,3,4,4,2,3,8,5,4,4,10,4,4,4,5,2,4,3,4,8,6,5,3,4,5,5,4,

%T 10,5,4,20,4,8,4,11,5,10,4,4,4,5,4,6,4,10,8,6,6,4,5,8,3,13,4,8,5,5,5,

%U 8,4,16,10,4,5,7,4,4,20,4,8,7,8,11,4,6,11,22,5,10,10,3,4,5,4,8,4

%N Least positive integer k such that n divides (2k)! - (2j)! for some j in [1,k-1].

%C For a guide to related sequences, see A204892.

%C From _Robert Israel_, Nov 20 2024: (Start)

%C a(n) <= ceil(A002034(n)/2) + 1.

%C The last occurrence of k >= 2 in the sequence is a((2*k)! - 2) = k. (End)

%H Robert Israel, <a href="/A205561/b205561.txt">Table of n, a(n) for n = 1..10000</a>

%e 1 divides (2*2)!-(2*1)! -> k=2, j=1

%e 2 divides (2*2)!-(2*1)! -> k=2, j=1

%e 3 divides (2*3)!-(2*2)! -> k=3, j=2

%e 4 divides (2*3)!-(2*2)! -> k=3, j=2

%e 5 divides (2*4)!-(2*3)! -> k=4, j=3

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

%p S:= {}:

%p x:= 1:

%p for j from 1 do

%p x:=x*2*j*(2*j-1) mod n;

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

%p S:= S union {x}

%p od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Nov 18 2024

%t s = Table[(2n)!, {n, 1, 120}];

%t lk = Table[NestWhile[# + 1 &, 1,

%t Min[Table[Mod[s[[#]] - s[[j]], z], {j, 1, # - 1}]] =!= 0 &], {z, 1, Length[s]}]

%t Table[NestWhile[# + 1 &, 1,

%t Mod[s[[lk[[j]]]] - s[[#]], j] =!= 0 &],

%t {j, 1, Length[lk]}]

%t (* _Peter J. C. Moses_, Jan 27 2012 *)

%Y Cf. A002034, A010050, A204892, A205546, A378188, A378189.

%K nonn

%O 1,1

%A _Clark Kimberling_, Feb 01 2012