OFFSET
1,1
COMMENTS
The sequence contains the smaller member of every pair of twin primes (A001359) and all squarefree semiprimes m such that m+4 is also a squarefree semiprime (A255746). Can one prove that this is an infinite sequence? - Vladimir Shevelev, Jul 11 2015
The sequence does not contain perfect squares. Indeed, let a(m)=k^2. Then d(k^2)+d(k^2))=d(k^2). Note that d(k^2) is odd. On the other hand, it is known (A046522) that d(k^2)<2*k. Hence, (k+1)^2 - k^2 > d(k^2). Thus k^2<k^2+d(k^2)<(k+1)^2 and k^2+d(k^2) cannot be perfect squares. So, k^2+d(k^2) is even and we have a contradiction. - Vladimir Shevelev, Feb 10 2017
If p is prime and t+1 is odd prime, then p^t is not in the sequence. Indeed, if d(p^t+t+1)=t+1, then p^t+t+1=q^t, where q is prime > p (if p^t+t+1= say q^l*r^m, then (l+1)*(m+1)=t+1 which is impossible by the condition). But q>=p+2 and p^t+t+1>=p^t+2*t*p^(t-1) or t+1>=2*t*p^(t-1) which trivially has only solution t=1; however, by the condition t>=2. - Vladimir Shevelev, Feb 18 2017
If an odd integer k is in this sequence, so is 2k. - Charlie Neder, Jan 14 2019
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000
EXAMPLE
10 is in the sequence because d(10)=4 and d(10+d(10))=d(14)=4. - Emeric Deutsch, Apr 08 2010
MAPLE
with(numtheory): a := proc (n) if tau(n+tau(n)) = tau(n) then n else end if end proc: seq(a(n), n = 1 .. 230); # Emeric Deutsch, Apr 08 2010
MATHEMATICA
Select[Range@ 224, Function[n, DivisorSigma[0, n + #] == # &@ DivisorSigma[0, n]]](* Michael De Vlieger, Sep 27 2015 *)
Position[#, 0][[All, 1]] &@ Table[DivisorSigma[0, n + DivisorSigma[0, n]] - DivisorSigma[0, n], {n, 222}] (* Michael De Vlieger, May 21 2017 *)
PROG
(PARI) is(n)=numdiv(n+n=numdiv(n))==n \\ M. F. Hasler, Sep 27 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, Mar 24 2010
EXTENSIONS
More terms from Emeric Deutsch, Apr 08 2010
STATUS
approved