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%I #15 Jun 29 2015 14:07:38
%S 6,12,14,28,30,48,62,124,126,222,224,254,448,476,496,510,768,876,1022,
%T 1792,1806,2032,2034,2046,2625,2850,2898,3204,3246,3560,3705,3850,
%U 4064,4094,7722,7744,7920,7980,7992,8060,8094,8136,8148,8150,8164,8190,11880,13365
%N Numbers n with the property that it is possible to write the base 2 expansion of n as concat(a_2,b_2), with a_2>0 and b_2>0 such that, converting a_2 and b_2 to base 10 as a and b, we have phi(a + b) = phi(n), where phi(n) is the Euler totient function of n.
%C It appears that a or b is equal to 1. In particular, if b=1 we have 2625, 3705, 13365, 25545, 57645, ... that are a subset of A001274.
%H Paolo P. Lava, <a href="/A259397/b259397.txt">Table of n, a(n) for n = 1..150</a>
%e 6 in base 2 is 110. If we take 110 = concat(1,10) then 1 and 10 converted to base 10 are 1 and 2. Finally phi(1 + 2) = 2 = phi(6).
%e 12 in base 2 is 1100. If we take 1100 = concat(1,100) then 1 and 100 converted to base 10 are 1 and 4. Finally phi(1 + 4) = 4 = phi(12);
%e 2625 in base 2 is 101001000001. If we take 101001000001 = concat(10100100000,1) then 10100100000 and 1 converted to base 10 are 1312 and 1. Finally phi(1312 + 1) = 1200 = phi(2625); etc.
%p with(numtheory): P:=proc(q) local a,b,c,k,n;
%p for n from 1 to q do c:=convert(n,binary,decimal);
%p for k from 1 to ilog10(c) do
%p a:=convert(trunc(c/10^k),decimal,binary);
%p b:=convert((c mod 10^k),decimal,binary);
%p if a*b>0 then if phi(a+b)=phi(n) then print(n); break;
%p fi; fi; od; od; end: P(10^8);
%Y Cf. A000010, A001274, A258813, A258843, A258844.
%K nonn,base
%O 1,1
%A _Paolo P. Lava_, Jun 26 2015
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