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A351166
a(1)=1; for n > 1, a(n) is the smallest number that has n divisors and is coprime to a(n-1).
1
1, 2, 9, 8, 81, 20, 729, 40, 441, 80, 59049, 140, 531441, 320, 3969, 440, 43046721, 700, 387420489, 560, 88209, 5120, 31381059609, 1120, 1185921, 20480, 53361, 4160, 22876792454961, 2800, 205891132094649, 3080, 9979281, 327680, 1750329, 8800, 150094635296999121
OFFSET
1,2
COMMENTS
For every odd prime p, a(p) = 3^(p-1).
For n > 1, the smallest prime factor of a(n) is 2 + (n mod 2); see Examples.
For n > 4, a(n) is a multiple of 5 iff n is even.
Is 4 the largest composite m such that a(m) is a prime power?
There are no other such composite numbers below 10^5. - Amiram Eldar, Jul 29 2024
LINKS
EXAMPLE
Given that a(27) = 53361 = 3^2 * 7^2 * 11^2, the prime factors of a(28) cannot include 3, 7, or 11, so the smallest primes available as prime factors of a(28) are 2, 5, and 13. The only prime signatures yielding 28 divisors are p^27, p^13 * q, p^6 * q^3, and p^6 * q * r, and the smallest candidate value with each of these signatures is 2^27 = 134217728, 2^13 * 5 = 40960, 2^6 * 5^3 = 8000, and 2^6 * 5 * 13 = 4160. The smallest of these is 4160, so a(28) = 4160. (This is the smallest term that is a multiple of 13.)
The table below lists the first several terms and their prime factorizations.
n a(n)
-- ----
1 1 = 1
2 2 = 2^1
3 9 = 3^2
4 8 = 2^3
5 81 = 3^4
6 20 = 2^2 * 5^1
7 729 = 3^6
8 40 = 2^3 * 5^1
9 441 = 3^2 * 7^2
10 80 = 2^4 * 5^1
MATHEMATICA
b[n_, primes_] := Module[{mulpar = mp[n] - 1, np, primes2}, np = Max[Length /@ mulpar]; primes2 = Complement[Prime[Range[np + Length[primes]]], primes]; Min[Table[Product[primes2[[s]]^mulpar[[j, s]], {s, 1, Length[mulpar[[j]]]}], {j, 1, Length[mulpar]}]]]; a[1] = 1; a[n_] := a[n] = b[n, FactorInteger[a[n-1]][[;; , 1]]]; Array[a, 50] (* Amiram Eldar, Jul 29 2024, after Vaclav Kotesovec at A005179 and using his function mp[n_, m_] *)
CROSSREFS
Sequence in context: A214300 A092397 A272348 * A092270 A249225 A191351
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Mar 27 2022
STATUS
approved