A082024 Number of partitions of n into 3 parts which have common divisors.
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
Offset: 0
Examples
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. From _David A. Corneth_, Aug 24 2020: (Start) 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)
Links
- David A. Corneth, Table of n, a(n) for n = 0..9999 (first 5001 terms from Alois P. Heinz)
Programs
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Mathematica
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&]] -
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
Extensions
More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 20 2003 and Dean Hickerson, Apr 22 2003
Comments