A220890 a(n) = smallest m such that A187824(m) = n, or -1 if A187824 never takes the value n.

-1, -1, -1, 2, 3, 4, 5, 29, 41, 55, 71, 881, 791, 9360, 10009, 1079, 30239, -1, 246960, -1, 636481, 1360800, 3160079, -1, 2162161

Offset: 0

Author

N. J. A. Sloane, Dec 30 2012

Keywords

• sign

Comments

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

From M. F. Hasler, Dec 30 2012 and Dec 31 2012: (Start)

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).

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.

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)

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

Links

• Robert Israel, Table of n, a(n) for n = 0..70
• Don Reble, Division gets rough: OEIS A187824 and A220890.

Crossrefs

Cf. A187824, A056697, A220891.

Programs

Maple

  N:= 70: # maximum m
V[0]:= -1: V[1]:= -1: V[2]:= -1:
S[3]:= {$0..5}: M[3]:= 6:
# M[m] is the lcm of 1..m
# S[m] is the set of residues mod M[m] for numbers n with A187824(n)>=m
# A[m] is the set of residues mod M[m] for numbers n with A187824(n)=m-1
for m from 4 to N+1 do
   M[m]:= ilcm(M[m-1], m); p:= M[m]/M[m-1];
   if p = 1 then T:= S[m-1]
   else T:= {seq(seq(a+b*M[m-1], a=S[m-1]), b=0..p-1)}
   end if;
   S[m],A[m]:= selectremove(t -> member(mods(t, m), {1, 0, -1}), T);
   if A[m] = {} then V[m-1]:= -1
   else V[m-1]:= min(A[m])
   end if;
end do:
seq(V[j], j=0..N);
# Robert Israel, Dec 31 2012

Extensions

a(26) = 39412801. Double-checked all lower given values. - M. F. Hasler, Dec 30 2012