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A307719
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Number of partitions of n into 3 mutually coprime parts.
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33
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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
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OFFSET
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0,7
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COMMENTS
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The Heinz numbers of these partitions are the intersection of A014612 (triples) and A302696 (pairwise coprime). - Gus Wiseman, Oct 16 2020
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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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 *)
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CROSSREFS
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A337599 is the pairwise non-coprime instead of pairwise coprime version.
A337601 only requires the distinct parts to be pairwise coprime.
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KEYWORD
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AUTHOR
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STATUS
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approved
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