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A214916
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a(0) = a(1) = 1, a(n) = n! / a(n-2).
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3
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1, 1, 2, 6, 12, 20, 60, 252, 672, 1440, 5400, 27720, 88704, 224640, 982800, 5821200, 21288960, 61102080, 300736800, 1990850400, 8089804800, 25662873600, 138940401600, 1007370302400, 4465572249600, 15397724160000, 90311261040000, 707173952284800, 3375972620697600
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OFFSET
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0,3
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COMMENTS
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a(n) is least k > a(n-1) such that k*a(n-2) is a factorial.
Two periodic subsets of these numbers appear in the coefficients of a series involved in a solution of a Riccati-type differential equation addressed by the Bernoulli brothers: z = 1 - x^4/12 + x^8/672 - x^12/88704 + ... = 1 - 2 * x^4/4! + 60 * x^8/8! - 5400 * x^12/12! + ... . See the MathOverflow question. - Tom Copeland, Jan 24 2017
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LINKS
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FORMULA
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a(0) = a(1) = 1, for n>=2, a(n) = n! / a(n-2).
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PROG
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(Python)
import math
prpr = prev = 1
for n in range(2, 33):
print prpr,
cur = math.factorial(n) / prpr
prpr = prev
prev = cur
(Magma) [1] cat [n le 2 select n else Factorial(n) div Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 25 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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