A359070
Smallest k > 1 such that k^n - 1 is the product of n distinct primes.
Original entry on oeis.org
3, 4, 15, 12, 39, 54, 79, 86, 144, 318, 1591, 144, 20131, 2014, 1764, 1308, 46656, 1296
Offset: 1
a(3) = 15 since 15^3 - 1 = 3374 = 2*7*241 is the product of 3 distinct primes and 15 is the smallest number with this property.
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isok(k, n) = my(f=factor(k^n - 1)); issquarefree(f) && (omega(f) == n);
a(n) = my(k=2); while (!isok(k, n), k++); k; \\ Michel Marcus, Dec 15 2022
A219018
Smallest number k > 0 such that k^n + 1 has exactly n distinct prime factors.
Original entry on oeis.org
1, 3, 5, 43, 17, 47, 51, 1697, 59, 512, 521, 3255, 8189, 18951, 656
Offset: 1
a(3) = 5 is the smallest number of the set {k(i)} = {5, 9, 10, 11, 12, 13, 14, 19,….} where k(i)^3 + 1 has exactly 3 distinct prime factors.
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with(numtheory) :for n from 1 to 10 do:ii:=0:for k from 1 to 10^10 while(ii=0) do:x:=k^n+1:y:=factorset(x):n1:=nops(y):if n1=n then ii:=1: printf ( "%d %d \n",n,k):
else fi:od:od:
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L = {}; Do[n = 1; While[Length[FactorInteger[n^k + 1]] != k, n++]; Print@AppendTo[L, n], {k, 15}] (* Giovanni Resta, Nov 09 2012 *)
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a(n) = my(k=1); while (omega(k^n+1) != n, k++); k; \\ Daniel Suteu, Feb 06 2023
A368162
a(n) is the smallest number k > 0 such that bigomega(k^n + 1) = n.
Original entry on oeis.org
1, 3, 3, 43, 7, 32, 23, 643, 17, 207, 251, 3255, 255, 1568, 107
Offset: 1
a(5) = 7 is the smallest number of the set {k(i)} = {7, 14, 24, 26, 46, 51, ...} where k(i)^5 + 1 has exactly 5 prime factors counted with multiplicity.
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