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A061549
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Denominator of probability that there is no error when average of n numbers is computed, assuming errors of +1, -1 are possible and they each occur with p=1/4.
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14
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1, 8, 128, 1024, 32768, 262144, 4194304, 33554432, 2147483648, 17179869184, 274877906944, 2199023255552, 70368744177664, 562949953421312, 9007199254740992, 72057594037927936, 9223372036854775808, 73786976294838206464, 1180591620717411303424, 9444732965739290427392
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OFFSET
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0,2
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COMMENTS
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Using WolframAlpha, it appears that 2*a(n) gives the coefficients of Pi in the denominators of the residues of f(z) = 2z choose z at odd negative half integers. E.g., the residues of f(z) at z = -1/2, -3/2, -5/2 are 1/(2*Pi), 1/(16*Pi), and 3/(256*Pi) respectively. - Nicholas Juricic, Mar 31 2022
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LINKS
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FORMULA
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a(n) = denominator of binomial(2*n-1/2, -1/2).
a(n) are denominators of coefficients of 1/(sqrt(1+x)-sqrt(1-x)) power series. - Benoit Cloitre, Mar 12 2002
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EXAMPLE
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For n=1, the binomial(2*n-1/2, -1/2) yields the term 3/8. The denominator of this term is 8, which is the second term of the sequence.
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MAPLE
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seq(denom(binomial(2*n-1/2, -1/2)), n=0..20);
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MATHEMATICA
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Table[Denominator[(4*n)!/(2^(4*n)*(2*n)!^2) ], {n, 0, 19}] (* Indranil Ghosh, Mar 11 2017 *)
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PROG
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def a(n): return 1 << (4*n - A000120(n))
(PARI) for(n=0, 19, print1(denominator((4*n)!/(2^(4*n)*(2*n)!^2)), ", ")) \\ Indranil Ghosh, Mar 11 2017
(Python)
import math
from fractions import gcd
f = math.factorial
def A061549(n): return (2**(4*n)*f(2*n)**2)/ gcd(f(4*n), (2**(4*n)*f(2*n)**2)) # Indranil Ghosh, Mar 11 2017
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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Leah Schmelzer (leah2002(AT)mit.edu), May 16 2001
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EXTENSIONS
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STATUS
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approved
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