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A015650
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Number of ordered 5-tuples of integers from [ 1..n ] with no global factor.
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5
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1, 5, 19, 49, 118, 225, 434, 729, 1209, 1850, 2850, 4059, 5878, 8044, 11020, 14566, 19410, 24789, 32103, 40213, 50615, 62260, 77209, 93099, 113504, 135431, 162341, 191396, 227355, 264463, 310838, 359322, 417212, 478408, 551944, 626971
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OFFSET
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1,2
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LINKS
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FORMULA
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G.f.: (1/(1 - x)) * Sum_{k>=1} mu(k) * x^k / (1 - x^k)^5. - Ilya Gutkovskiy, Feb 14 2020
a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)/120 - Sum_{j=2..n} a(floor(n/j)) = A000389(n+4) - Sum_{j=2..n} a(floor(n/j)). - Chai Wah Wu, Apr 18 2021
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PROG
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(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
if n == 0:
return 0
c, j = n+1, 2
k1 = n//j
while k1 > 1:
j2 = n//k1 + 1
j, k1 = j2, n//j2
return n*(n+1)*(n+2)*(n+3)*(n+4)//120-c+j # Chai Wah Wu, Apr 18 2021
(PARI) a(n) = sum(k=1, n, sumdiv(k, d, moebius(k/d)*binomial(d+3, 4))); \\ Seiichi Manyama, Jun 12 2021
(PARI) a(n) = binomial(n+4, 5)-sum(k=2, n, a(n\k)); \\ Seiichi Manyama, Jun 12 2021
(PARI) my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*x^k/(1-x^k)^5)/(1-x)) \\ Seiichi Manyama, Jun 12 2021
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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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