

A260375


Numbers k such that A260374(k) is a perfect square.


1



0, 1, 2, 4, 5, 6, 7, 8, 10, 11, 14, 15, 16
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OFFSET

1,3


COMMENTS

There are a surprising number of small terms in this sequence.
Heuristic: The square root of x has an average distance of 1/4 to an integer, so x  round(sqrt(x))^2 is around x  (sqrt(x)  1/4)^2 or about sqrt(x)/2, hence A260374(n) is around sqrt(n!)/2. By Stirling's approximation this is around (n/e)^(n/2) which is a square with probability (n/e)^(n/4). The integral of this function converges, so this sequence should be finite. This heuristic is crude, though, because it does not model the extreme values of A260374.  Charles R Greathouse IV, Jul 23 2015


LINKS



EXAMPLE

6! = 720. The nearest perfect square is 729. The difference is 9, which is itself a perfect square. So, 6 is in this sequence.


PROG

(Python)
from gmpy2 import isqrt, is_square
for i in range(1, 1001):
g *= i
s = isqrt(g)
t = gs**2
if is_square(t if ts <= 0 else 2*s+1t):


CROSSREFS



KEYWORD

nonn,more


AUTHOR



STATUS

approved



