A055227
Nearest integer to sqrt( n! ).
Original entry on oeis.org
1, 1, 1, 2, 5, 11, 27, 71, 201, 602, 1905, 6318, 21886, 78911, 295260, 1143536, 4574144, 18859677, 80014834, 348776577, 1559776269, 7147792818, 33526120082, 160785623545, 787685471323, 3938427356615, 20082117944246, 104349745809074, 552166953567228
Offset: 0
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Round[Sqrt[Range[0,30]!]] (* Harvey P. Dale, Aug 27 2013 *)
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from gmpy2 import isqrt
A055227_list, g = [1], 1
for i in range(1, 101):
g *= i
s = isqrt(g)
A055227_list.append(int(s if g-s*(s+1) <= 0 else s+1)) # Chai Wah Wu, Jul 24 2015
A260374
The distance between n! and the nearest perfect square.
Original entry on oeis.org
0, 0, 1, 2, 1, 1, 9, 1, 81, 476, 225, 324, 4604, 74879, 176400, 215296, 3444736, 11551671, 45680444, 255004929, 1158920361, 2657058876, 24923993276, 130518272975, 97216010329, 2430400258225, 1553580508516, 4666092737476, 538347188396016, 2137864362693921
Offset: 0
6!=720. The nearest perfect square is 729. The difference between these is 9, so a(6)=9.
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a(n)=abs(n!-round(sqrt(n!))^2) \\ Charles R Greathouse IV, Jul 23 2015
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from gmpy2 import isqrt
A260374_list, g = [0], 1
for i in range(1, 1001):
g *= i
s = isqrt(g)
t = g-s**2
A260374_list.append(int(t if t-s <= 0 else 2*s+1-t)) # Chai Wah Wu, Jul 23 2015
A260375
Numbers k such that A260374(k) is a perfect square.
Original entry on oeis.org
0, 1, 2, 4, 5, 6, 7, 8, 10, 11, 14, 15, 16
Offset: 1
6! = 720. The nearest perfect square is 729. The difference is 9, which is itself a perfect square. So, 6 is in this sequence.
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is(n)=my(N=n!,s=sqrtint(N)); issquare(min(N-s^2, (s+1)^2-N)) \\ Charles R Greathouse IV, Jul 23 2015
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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
Showing 1-3 of 3 results.
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