%N Numbers with 3 prime factors a < b < c such that 2c = a^4 + b^2 and a = round(sqrt(b)).
%C This sequence is interesting because the sequence of any member's divisors increases approximately exponentially, without members having any repeated prime factors.
%C Removing the limitation a=round(sqrt(b)) results in A261657.
%C This sequence would be particularly interesting if the b and c of one member were the a and b of another, as it could give us a number with 4 prime factors and similar properties as here. The resulting compound would be a member of A261658. No such pairs exist in this sequence with any lowest factor < 10000.
%H Charles R Greathouse IV, <a href="/A261656/b261656.txt">Table of n, a(n) for n = 1..10000</a>
%e The prime factors of 3333 are 3, 11, and 101. 3^4=81, 11^2=121, and the average of 81 and 121 gives 101. Thus, 3333 is a member.
%e The divisors of 3333 are 1, 3, 11, 33, 101, 303, 1111, 3333. This sequence is approximately exponential.
%p n := 100; for a from 3 to n do if isprime(a) then for b from ceil((a-.5)^2) to floor((a+.5)^2) do if isprime(b) then c := (a^4+b^2)*(1/2); if isprime(c) then print(a*b*c) end if end if end do end if end do
%o (PARI) list(lim)=my(v=List(),t,a,c); forprime(b=2,, a=round(sqrt(b)); c=(a^4+b^2)/2; if(isprime(a) && denominator(c)==1 && isprime(c), t=a*b*c; if(t>lim, break); listput(v,a*b*c))); Set(v) \\ _Charles R Greathouse IV_, Aug 29 2015
%Y Cf. A261657, A261658.
%A _David Ferris_, Aug 28 2015