A260375 Numbers k such that A260374(k) is a perfect square.

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

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

There are no further terms up to 10^5, so probably the list is complete. - Charles R Greathouse IV, Jul 23 2015

Links

Table of n, a(n) for n=1..13.

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

(PARI) is(n)=my(N=n!, s=sqrtint(N)); issquare(min(N-s^2, (s+1)^2-N)) \\ Charles R Greathouse IV, Jul 23 2015
(Python)
from gmpy2 import isqrt, is_square
A260375_list, g = [0], 1
for i in range(1, 1001):
    g *= i
    s = isqrt(g)
    t = g-s**2
    if is_square(t if t-s <= 0 else 2*s+1-t):
        A260375_list.append(i) # Chai Wah Wu, Jul 23 2015

Crossrefs

Cf. A260373, A260374.

Sequence in context: A086743 A285432 A039079 * A361144 A188160 A047571

Adjacent sequences: A260372 A260373 A260374 * A260376 A260377 A260378

Keyword

nonn,more

Author

Otis Tweneboah, Pratik Koirala, Eugene Fiorini, Nathan Fox, Jul 23 2015

Status

approved