

A282391


Numbers n such that d(n) = d(n+3*d(n)), where d(n) is the number of divisors of n.


1



5, 7, 10, 11, 13, 14, 15, 17, 21, 22, 23, 26, 27, 30, 31, 32, 34, 37, 39, 41, 42, 45, 46, 47, 50, 53, 54, 57, 60, 61, 62, 65, 67, 72, 73, 74, 78, 82, 83, 90, 94, 96, 97, 98, 99, 101, 103, 104, 106, 107, 111, 114, 120, 122, 128, 129, 130, 131, 133, 134, 143
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OFFSET

1,1


COMMENTS

The sequence contains the smaller member of every pair of sexy primes (A023201).
The sequence does not contain perfect squares. Indeed, let for some m a(m)=k^2. Then, by the definition, d(k^2+3d(k^2))=d(k^2). Note that d(k^2) is odd. On the other hand, it is known (cf. A046522) that d(k^2)<2*k. Hence (k+3)^2k^2=6*k+1>3d(k^2). Thus k^2< k^2+3*d(k^2) <(k+3)^2. Note that, evidently, k^2+3*d(k^2) cannot be (k+2)^2. Let us show that also k^2+3*d(k^2) cannot be (k+1)^2, or, the same, 3*d(k^2) cannot be equal 2*k+1. Indeed, let 3*d(k^2)=2*k+1. Let, for some prime p, p^ak, a>0, so 2*k+1==1(mod p). But now we have 3*p^(a+1)  3*d(k^2) and thus 3*p^(a+1)2*k+1, so 2*k+1==0(mod p). Contradiction. Therefore, we conclude that k^2 + 3*d(k^2) cannot be a square. Hence, d(k^2+3*d(k^2)) is even, which is contradiction.


LINKS



MAPLE

with(numtheory): P:=proc(q); if tau(q)=tau(q+3*tau(q)) then q; fi; end:


PROG

(PARI) isok(n) = numdiv(n) == numdiv(n+3*numdiv(n)); \\ Michel Marcus, Feb 14 2017


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



