A259397 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.

6, 12, 14, 28, 30, 48, 62, 124, 126, 222, 224, 254, 448, 476, 496, 510, 768, 876, 1022, 1792, 1806, 2032, 2034, 2046, 2625, 2850, 2898, 3204, 3246, 3560, 3705, 3850, 4064, 4094, 7722, 7744, 7920, 7980, 7992, 8060, 8094, 8136, 8148, 8150, 8164, 8190, 11880, 13365

Offset: 1

Author

Paolo P. Lava, Jun 26 2015

Keywords

• nonn
• base

Comments

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.

Examples

			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).
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);
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.
		

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..150

Crossrefs

Cf. A000010, A001274, A258813, A258843, A258844.

Programs

Maple

with(numtheory): P:=proc(q) local a,b,c,k,n;
for n from 1 to q do c:=convert(n,binary,decimal);
for k from 1 to ilog10(c) do
a:=convert(trunc(c/10^k),decimal,binary);
b:=convert((c mod 10^k),decimal,binary);
if a*b>0 then if phi(a+b)=phi(n) then print(n); break;
fi; fi; od; od; end: P(10^8);