A357033 a(n) is the smallest number that has exactly n divisors that are cyclops numbers (A134808).

1, 101, 202, 404, 606, 1212, 2424, 7272, 21816, 41208, 84048, 123624, 144144, 336336, 288288, 504504, 432432, 865368, 864864, 1009008, 2378376, 1729728, 3459456, 3027024, 4756752, 6054048, 9081072, 11099088, 12108096, 16648632, 23207184, 29405376, 36324288

Offset: 0

Author

Marius A. Burtea, Sep 20 2022

Keywords

• nonn
• base

Examples

			The divisors of 101 are 1 and 101. Of those, only 101 is a cyclops number; it is the smallest cyclops number, so a(1) = 101.
The divisors of 202 are 1, 2, 101, and 202, the cyclops numbers being 101 and 202, so a(2) = 202.
The divisors of 404 are 1, 2, 4, 101, 202, and 404, the cyclops numbers being 101, 202 and 404, so a(3) = 404.
		

Crossrefs

Cf. A134808.

Programs

Magma

ints:=func; cyc:=func; a:=[]; for n in [0..32]  do k:=1; while #[s:s in Divisors(k)| cyc(s)] ne n do k:=k+1; end while; Append(~a,k); end for; a;

Maple

L:= Vector(10^8):
C:= [0]:
for d from 3 to 7 by 2 do
  C:= [seq(seq(seq(a*10^(d-1)+10*b+c,c=1..9),b=C),a=1..9)];
  for x in C do
    Mx:= [seq(i,i=x..10^8,x)];
    L[Mx]:= map(`+`,L[Mx],1)
  od;
od:
V:= Array(0..max(L)):
for n from 1 to 10^8 do
  if V[L[n]] = 0 then V[L[n]]:= n; fi
od:
if member(0,V,'k') then convert(V[0..k-1],list)
else convert(V,list)
fi; # Robert Israel, Sep 20 2022

Mathematica

cyclopQ[n_] := Module[{d = IntegerDigits[n], len}, OddQ[len = Length[d]] && Position[d, 0] == {{(len + 1)/2}}]; f[n_] := DivisorSum[n, 1 &, cyclopQ[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[10, 10^5] (* Amiram Eldar, Sep 26 2022 *)