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a(n) = smallest m such that A187824(m) = n, or -1 if A187824 never takes the value n.
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%I #50 May 17 2024 09:56:10

%S -1,-1,-1,2,3,4,5,29,41,55,71,881,791,9360,10009,1079,30239,-1,246960,

%T -1,636481,1360800,3160079,-1,2162161

%N a(n) = smallest m such that A187824(m) = n, or -1 if A187824 never takes the value n.

%C a(17) = -1. Proof: If x mod 9 and x mod 12 are both in {-1, 0, 1} then so is x mod 18. So if x is a number which is congruent to -1, 0 or 1 mod k for k=1..17, then also x mod 18 is congruent to -1, 0 or 1. So there is no x such that A187824(x) = 17. QED

%C From _M. F. Hasler_, Dec 30 2012 and Dec 31 2012: (Start)

%C Similarly, a(19) = -1. Indeed, if x == 0, 1 or -1 (mod 15) and (mod 12), then also (mod 60). [Proof: Write x = 15*(4k+d)+e, |e| < 2, then d = 1, 2, 3 all give impossible x (mod 12).] Therefore A187824 cannot have the value 19 (nor 29, nor 59).

%C Also, a(23) = -1, because x == 0, 1 or -1 (mod 8) and (mod 12) implies the same (mod 24). [To see this, write x = 12*(2k+d)+e, |e| < 2, then d = 1 gives impossible x (mod 8).] Therefore A187824 cannot have the value 23.

%C From A220891 one may deduce the values for n = 26, 28, 31, 36, 40, 42, 46, 48, 52, 58, 60, 61 to be a(n) = 39412801, 107881201, 3625549201, 170918748000, 2355997644001, 237662810985599, 4614209634434399, 7522575180120001, 362645725505263201, 10684484093105222399, 442709913651892286399, 5205240636387758366399. (End)

%C _Don Reble_ shows that a(n) > -1 iff n + 1 is either 12, 2p, 3p or p^k > 3, where p is a prime, k >= 1. - _M. F. Hasler_, Mar 17 2020

%H Robert Israel, <a href="/A220890/b220890.txt">Table of n, a(n) for n = 0..70</a>

%H Don Reble, <a href="/A187824/a187824.pdf">Division gets rough: OEIS A187824 and A220890</a>.

%p N:= 70: # maximum m

%p V[0]:= -1: V[1]:= -1: V[2]:= -1:

%p S[3]:= {$0..5}: M[3]:= 6:

%p # M[m] is the lcm of 1..m

%p # S[m] is the set of residues mod M[m] for numbers n with A187824(n)>=m

%p # A[m] is the set of residues mod M[m] for numbers n with A187824(n)=m-1

%p for m from 4 to N+1 do

%p M[m]:= ilcm(M[m-1], m); p:= M[m]/M[m-1];

%p if p = 1 then T:= S[m-1]

%p else T:= {seq(seq(a+b*M[m-1], a=S[m-1]), b=0..p-1)}

%p end if;

%p S[m],A[m]:= selectremove(t -> member(mods(t, m), {1, 0, -1}), T);

%p if A[m] = {} then V[m-1]:= -1

%p else V[m-1]:= min(A[m])

%p end if;

%p end do:

%p seq(V[j], j=0..N);

%p # _Robert Israel_, Dec 31 2012

%Y Cf. A187824, A056697, A220891.

%K sign

%O 0,4

%A _N. J. A. Sloane_, Dec 30 2012

%E a(26) = 39412801. Double-checked all lower given values. - _M. F. Hasler_, Dec 30 2012