A374434 Triangle read by rows: T(n, k) = Product_{p in PF(n) symmetric difference PF(k)} p, where PF(a) is the set of the prime factors of a.
1, 1, 1, 2, 2, 1, 3, 3, 6, 1, 2, 2, 1, 6, 1, 5, 5, 10, 15, 10, 1, 6, 6, 3, 2, 3, 30, 1, 7, 7, 14, 21, 14, 35, 42, 1, 2, 2, 1, 6, 1, 10, 3, 14, 1, 3, 3, 6, 1, 6, 15, 2, 21, 6, 1, 10, 10, 5, 30, 5, 2, 15, 70, 5, 30, 1, 11, 11, 22, 33, 22, 55, 66, 77, 22, 33, 110, 1
Offset: 0
Examples
[ 0] 1; [ 1] 1, 1; [ 2] 2, 2, 1; [ 3] 3, 3, 6, 1; [ 4] 2, 2, 1, 6, 1; [ 5] 5, 5, 10, 15, 10, 1; [ 6] 6, 6, 3, 2, 3, 30, 1; [ 7] 7, 7, 14, 21, 14, 35, 42, 1; [ 8] 2, 2, 1, 6, 1, 10, 3, 14, 1; [ 9] 3, 3, 6, 1, 6, 15, 2, 21, 6, 1; [10] 10, 10, 5, 30, 5, 2, 15, 70, 5, 30, 1; [11] 11, 11, 22, 33, 22, 55, 66, 77, 22, 33, 110, 1;
Crossrefs
Programs
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Maple
PF := n -> ifelse(n = 0, {}, NumberTheory:-PrimeFactors(n)): A374434 := (n, k) -> mul(symmdiff(PF(n), PF(k))): seq(print(seq(A374434(n, k), k = 0..n)), n = 0..11);
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Mathematica
nn = 12; Do[Set[s[i], FactorInteger[i][[All, 1]]], {i, 0, nn}]; s[0] = {1}; Table[Times @@ SymmetricDifference[s[k], s[n]], {n, 0, nn}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 11 2024 *)
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Python
# Function A374434 defined in A374433. for n in range(11): print([A374434(n, k) for k in range(n + 1)])