A290015 Brazilian numbers which have exactly two Brazilian representations.

15, 18, 21, 26, 28, 30, 31, 32, 44, 45, 50, 52, 56, 57, 62, 64, 68, 75, 76, 85, 86, 91, 92, 93, 98, 99, 110, 111, 116, 117, 129, 133, 146, 147, 148, 153, 164, 175, 183, 188, 207, 212, 215, 219, 236, 243, 244, 245, 259, 261, 268, 275, 279, 284, 314, 316, 325, 332, 338, 341, 343, 356, 363, 365, 369, 381, 387, 388

Offset: 1

Author

Bernard Schott, Jul 17 2017

Keywords

• nonn
• base

Comments

These numbers could be called 2-Brazilian numbers.

The smallest number of this sequence is 15 which is also the smallest odd composite Brazilian in A257521 with 15 = 11111_2 = 33_4. The number 15 is highly Brazilian in A329383.

Following the Goormaghtigh conjecture, only two primes, 31 and 8191, which are both Mersenne numbers, are Brazilian in two different bases (A119598).

Examples

			18 = 2 * 9 = 22_8 = 3 * 6 = 33_5.
26 = 2 * 13 = 2 * 111_3 = 222_3 = 22_12.
31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90.
		

Links

• Wikipedia, Goormaghtigh conjecture

Crossrefs

Cf. A085104, A125134, A220570, A220571, A257521, A284758, A288783, A290016, A290017, A290018, A329383.

Programs

Maple

bresilienbaseb:=proc(n,b)
local r,q,coupleq:
if n0 then
return [couple[1]+1,r]
else
return [0,0]
end if
end if
end proc;
bresil:=proc(n)
local b,L,k,t:
k:=0:
for b from 2 to (n-2) do
t:=bresilienbase(n,b):
if t[1]>0 then
k:=k+1
L[k]:=[b,t[1],t[2]]:
end if:
end do:
seq(L[i],i=1..k);
end proc;
nbbresil:=n->nops([bresil(n)]);
#Numbers 2 times Brazilian
for n from 1 to 100 do if nbbresil(n)=2 then print(n,bresil(n)) else fi; od:

Mathematica

Flatten@ Position[#, 2] &@ Table[Count[Range[2, n - 2], ?(And[Length@ # != 1, Length@ Union@ # == 1] &@ IntegerDigits[n, #] &)], {n, 400}] (* _Michael De Vlieger, Jul 18 2017 *)