A374435 Triangle read by rows: T(n, k) = Product_{p in PF(n) difference PF(k)} p, where PF(a) is the set of the prime factors of a.
1, 1, 1, 2, 2, 1, 3, 3, 3, 1, 2, 2, 1, 2, 1, 5, 5, 5, 5, 5, 1, 6, 6, 3, 2, 3, 6, 1, 7, 7, 7, 7, 7, 7, 7, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 3, 3, 1, 3, 3, 1, 3, 3, 1, 10, 10, 5, 10, 5, 2, 5, 10, 5, 10, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1
Offset: 0
Examples
[ 0] 1; [ 1] 1, 1; [ 2] 2, 2, 1; [ 3] 3, 3, 3, 1; [ 4] 2, 2, 1, 2, 1; [ 5] 5, 5, 5, 5, 5, 1; [ 6] 6, 6, 3, 2, 3, 6, 1; [ 7] 7, 7, 7, 7, 7, 7, 7, 1; [ 8] 2, 2, 1, 2, 1, 2, 1, 2, 1; [ 9] 3, 3, 3, 1, 3, 3, 1, 3, 3, 1; [10] 10, 10, 5, 10, 5, 2, 5, 10, 5, 10, 1; [11] 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1;
Crossrefs
Programs
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Maple
PF := n -> ifelse(n = 0, {}, NumberTheory:-PrimeFactors(n)): A374435 := (n, k) -> mul(PF(n) minus PF(k)): seq(print(seq(A374435(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[Apply[Times, Complement[s[n], s[k]]], {n, 0, nn}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 11 2024 *)
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Python
# Function A374435 defined in A374433. for n in range(12): print([A374435(n, k) for k in range(n + 1)])