A374433
Triangle read by rows: T(n, k) = Product_{p in PF(n) intersect PF(k)} p, where PF(a) is the set of the prime factors of a.
Original entry on oeis.org
1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 5, 1, 1, 2, 3, 2, 1, 6, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 2, 3, 2, 1, 6, 1, 2, 3, 2, 1, 6
Offset: 0
[ 0] 1;
[ 1] 1, 1;
[ 2] 1, 1, 2;
[ 3] 1, 1, 1, 3;
[ 4] 1, 1, 2, 1, 2;
[ 5] 1, 1, 1, 1, 1, 5;
[ 6] 1, 1, 2, 3, 2, 1, 6;
[ 7] 1, 1, 1, 1, 1, 1, 1, 7;
[ 8] 1, 1, 2, 1, 2, 1, 2, 1, 2;
[ 9] 1, 1, 1, 3, 1, 1, 3, 1, 1, 3;
[10] 1, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10;
[11] 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11;
Family: this sequence (intersection),
A374434 (symmetric difference),
A374435 (difference),
A374436 (union).
-
PF := n -> ifelse(n = 0, {}, NumberTheory:-PrimeFactors(n)):
A374433 := (n, k) -> mul(PF(n) intersect PF(k)):
seq(seq(A374433(n, k), k = 0..n), n = 0..12);
-
nn = 12; Do[Set[s[i], FactorInteger[i][[All, 1]]], {i, 0, nn}]; s[0] = {1};
Table[Times @@ Intersection[s[k], s[n]], {n, 0, nn}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 11 2024 *)
-
from math import prod
from sympy import primefactors
def PF(n): return set(primefactors(n)) if n > 0 else set({})
def PrimeIntersect(n, k): return prod(PF(n).intersection(PF(k)))
def PrimeSymDiff(n, k): return prod(PF(n).symmetric_difference(PF(k)))
def PrimeUnion(n, k): return prod(PF(n).union(PF(k)))
def PrimeDiff(n, k): return prod(PF(n).difference(PF(k)))
A374433 = PrimeIntersect; A374434 = PrimeSymDiff
A374435 = PrimeDiff; A374436 = PrimeUnion
for n in range(11): print([A374433(n, k) for k in range(n + 1)])
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.
Original entry on oeis.org
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
[ 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;
Family:
A374433 (intersection), this sequence (symmetric difference),
A374435 (difference),
A374436 (union).
-
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);
-
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 *)
-
# Function A374434 defined in A374433.
for n in range(11): print([A374434(n, k) for k in range(n + 1)])
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.
Original entry on oeis.org
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
[ 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;
Family:
A374433 (intersection),
A374434 (symmetric difference), this sequence (difference),
A374436 (union).
-
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);
-
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 *)
-
# Function A374435 defined in A374433.
for n in range(12): print([A374435(n, k) for k in range(n + 1)])
Showing 1-3 of 3 results.