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A082024
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Number of partitions of n into 3 parts which have common divisors.
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15
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0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 4, 0, 4, 3, 5, 0, 9, 0, 9, 5, 10, 0, 16, 2, 14, 7, 17, 0, 27, 0, 21, 11, 24, 6, 36, 0, 30, 15, 37, 0, 51, 0, 41, 25, 44, 0, 64, 4, 58, 25, 57, 0, 81, 12, 69, 31, 70, 0, 108, 0, 80, 43, 85, 16, 123, 0, 97, 45, 120, 0, 144, 0, 114, 69, 121, 14, 171, 0
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OFFSET
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0,11
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COMMENTS
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a(p) = 0 if p is a prime. Can anyone suggest a formula?
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LINKS
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EXAMPLE
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a(14) = 4 and the partitions are (10,2,2), (8,4,2),(6,6,2) and (6,4,4).
a(13) = 0 as for all r + s + t = 13,r > 0, s > 0,t> 0 gcd(r,s,t) = 1.
a(100) = 233. The squarefree part of 100 is 10. The divisors of 10 are 1, 2, 5 and 10.
These are the possible squarefree divisors of parts. As parts must not be coprime, we exclude 1, leaving 2, 5 and 10. We then compute 100/k for each of these numbers.
This gives 50, 20 and 10 respectively. Now a(100) is found by adding -(round(50^2/12)*(-1)^omega(2) + round(20^2/12)*(-1)^omega(5) + round(10^2/12)*(-1)^omega(10)) = -(-208 - 33 + 8) = 233 where omega(m) is the number of distinct divisors of m (Cf. A001221) and round(m^2/12) is the number of partitions of m into 3 parts (Cf. A069905) (End)
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MATHEMATICA
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a[n_] := Length[Select[Flatten[Table[{a, b, n-a-b}, {a, 1, Floor[n/3]}, {b, a, Floor[(n-a)/2]}], 1], GCD@@#1>1&]]
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PROG
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(PARI) a(n) = if(n==0, return(0)); cn = factorback(factor(n)[, 1]); d = divisors(cn); -sum(i = 2, #d, round((n/d[i])^2/12) * (-1)^omega(d[i])) \\ David A. Corneth, Aug 24 2020
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 20 2003 and Dean Hickerson, Apr 22 2003
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STATUS
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approved
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