A307719 Number of partitions of n into 3 mutually coprime parts.
0, 0, 0, 1, 1, 1, 2, 1, 3, 2, 4, 2, 7, 2, 8, 4, 8, 4, 15, 4, 16, 7, 15, 7, 26, 7, 23, 11, 26, 10, 43, 9, 35, 16, 38, 16, 54, 14, 49, 23, 54, 18, 79, 18, 66, 31, 64, 25, 100, 25, 89, 36, 85, 31, 127, 35, 104, 46, 104, 39, 167, 36, 125, 58, 129, 52, 185, 45
Offset: 0
Examples
There are 2 partitions of 9 into 3 mutually coprime parts: 7+1+1 = 5+3+1, so a(9) = 2. There are 4 partitions of 10 into 3 mutually coprime parts: 8+1+1 = 7+2+1 = 5+4+1 = 5+3+2, so a(10) = 4. There are 2 partitions of 11 into 3 mutually coprime parts: 9+1+1 = 7+3+1, so a(11) = 2. There are 7 partitions of 12 into 3 mutually coprime parts: 10+1+1 = 9+2+1 = 8+3+1 = 7+4+1 = 6+5+1 = 7+3+2 = 5+4+3, so a(12) = 7.
Links
- Fausto A. C. Cariboni, Table of n, a(n) for n = 0..10000 (terms 0..2000 from Robert Israel)
- Index entries for sequences related to partitions
Crossrefs
A023022 is the version for pairs.
A327516 counts these partitions of any length, with strict version A305713 and Heinz numbers A302696.
A337461 is the ordered version.
A337563 is the case with no 1's.
A337599 is the pairwise non-coprime instead of pairwise coprime version.
A337601 only requires the distinct parts to be pairwise coprime.
Programs
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Maple
N:= 200: # to get a(0)..a(N) A:= Array(0..N): for a from 1 to N/3 do for b from a to (N-a)/2 do if igcd(a,b) > 1 then next fi; ab:= a*b; for c from b to N-a-b do if igcd(ab,c)=1 then A[a+b+c]:= A[a+b+c]+1 fi od od od: convert(A,list); # Robert Israel, May 09 2019
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Mathematica
Table[Sum[Sum[Floor[1/(GCD[i, j] GCD[j, n - i - j] GCD[i, n - i - j])], {i, j, Floor[(n - j)/2]}], {j, Floor[n/3]}], {n, 0, 100}] Table[Length[Select[IntegerPartitions[n,{3}],CoprimeQ@@#&]],{n,0,100}] (* Gus Wiseman, Oct 15 2020 *)
Formula
a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} [gcd(i,j) * gcd(j,n-i-j) * gcd(i,n-i-j) = 1], where [] is the Iverson bracket.
a(n > 2) = A220377(n) + 1. - Gus Wiseman, Oct 15 2020
Comments