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A059583
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Each c(i) is "multiply" (*) or "divide" (/); a(n) is number of choices for c(1),..,c(n) so that the reduced fraction 1 c(1) 2 c(2) 3 c(3) 5 ... c(n) prime(n) is equal to (k*m+1)/m for a positive integer m and a nonnegative integer k.
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1
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1, 2, 4, 5, 6, 6, 6, 3, 5, 6, 7, 9, 8, 7, 7, 9, 6, 8, 6, 9, 6, 8, 9, 12, 11, 5, 9, 8, 7, 7, 11, 9, 11, 11, 9, 10
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OFFSET
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1,2
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LINKS
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EXAMPLE
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For n = 4 there are five possibilities: 1/2/3/5/7 = 1/210, 1/2*3/5*7 = 21/10, 1/2*3*5/7 = 15/14, 1/2*3*5*7 = 105/2, and 1*2/3*5*7 = 70/3. So a(4) = 5.
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PROG
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(Python)
from math import prod
from itertools import combinations
from sympy import prime, primerange
def a(n):
pset = set(primerange(2, prime(n)+1))
m, c = prod(pset), 1 # count 1/2/3/.../prime(n)
for r in range(1, len(pset)):
first = 1
for dens in combinations(pset, r):
den = prod(dens)
num = m//den
if dens[0] != first:
if den > num: break
first = dens[0]
if num%den == 1: c += 1
return c
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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