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%I #27 Jul 24 2020 07:12:45
%S 15,18,21,26,28,30,31,32,44,45,50,52,56,57,62,64,68,75,76,85,86,91,92,
%T 93,98,99,110,111,116,117,129,133,146,147,148,153,164,175,183,188,207,
%U 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
%N Brazilian numbers which have exactly two Brazilian representations.
%C These numbers could be called 2-Brazilian numbers.
%C 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.
%C Following the Goormaghtigh conjecture, only two primes, 31 and 8191, which are both Mersenne numbers, are Brazilian in two different bases (A119598).
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Goormaghtigh_conjecture">Goormaghtigh conjecture</a>
%e 18 = 2 * 9 = 22_8 = 3 * 6 = 33_5.
%e 26 = 2 * 13 = 2 * 111_3 = 222_3 = 22_12.
%e 31 = 11111_2 = 111_5;
%e 8191 = 1111111111111_2 = 111_90.
%p bresilienbaseb:=proc(n,b)
%p local r,q,coupleq:
%p if n<b then
%p return [1,n]
%p else
%p r:=(n mod b):
%p q:=(n-r)/b
%p coupleq:=bresilienbase(q,b):
%p if r=coupleq[2] and r>0 then
%p return [couple[1]+1,r]
%p else
%p return [0,0]
%p end if
%p end if
%p end proc;
%p bresil:=proc(n)
%p local b,L,k,t:
%p k:=0:
%p for b from 2 to (n-2) do
%p t:=bresilienbase(n,b):
%p if t[1]>0 then
%p k:=k+1
%p L[k]:=[b,t[1],t[2]]:
%p end if:
%p end do:
%p seq(L[i],i=1..k);
%p end proc;
%p nbbresil:=n->nops([bresil(n)]);
%p #Numbers 2 times Brazilian
%p for n from 1 to 100 do if nbbresil(n)=2 then print(n,bresil(n)) else fi; od:
%t 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 *)
%Y Cf. A085104, A125134, A220570, A220571, A257521, A284758, A288783, A290016, A290017, A290018, A329383.
%K nonn,base
%O 1,1
%A _Bernard Schott_, Jul 17 2017
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