OFFSET
1,2
COMMENTS
This sequence is the product of n-regular numbers.
A number m is said to be "regular" to n or "n-regular" if all the prime factors p of m also divide n.
The divisor is a special case of a regular m such that m also divides n in addition to all of its prime factors p | n.
Analogous to A007955 (Product of divisors of n).
If n is 1 or prime, a(n) = n.
If n is a prime power, a(n) = A007955(n).
Note: b-file ends at n = 4619, because a(4620) has more than 1000 decimal digits.
Product of the numbers 1 <= k <= n such that (floor(n^k/k) - floor((n^k - 1)/k)) = 1. - Michael De Vlieger, May 26 2016
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..4619
Encyclopedia Britannica, Regular Number (base-neutral definition)
Eric W. Weisstein, Regular Number (decimal definition)
Wikipedia, Regular Number (sexagesimal / Hamming number definition)
FORMULA
a(n) = product of terms of n-th row of irregular triangle A162306(n,k).
a(n) = Product_{k=1..n} k^( floor(n^k/k)-floor((n^k -1)/k) ). - Anthony Browne, Jul 06 2016
From Antti Karttunen, Mar 22 2017: (Start)
a(n) = Product_{k=2..n, A123275(n,k)=1} k.
(End)
EXAMPLE
a(12) = 124416 since 1 * 2 * 3 * 4 * 6 * 8 * 9 * 12 = 124416. These numbers are products of prime factors that are the distinct prime divisors of 12 = {2, 3}.
From David A. Corneth, Feb 09 2015: (Start)
Let p# be the product of primes up to p, A002110. Then
a(13#) ~= 8.3069582 * 10 ^ 4133
a(17#) ~= 1.3953000 * 10 ^ 22689
a(19#) ~= 3.8258936 * 10 ^ 117373
a(23#) ~= 6.7960327 * 10 ^ 594048
a(29#) ~= 1.3276817 * 10 ^ 2983168
a(31#) ~= 2.8152792 * 10 ^ 14493041
a(37#) ~= 1.9753840 * 10 ^ 69927040
Up to n = 11# already in the table.
(End)
MAPLE
A:= proc(n) local F, S, s, j, p;
F:= numtheory:-factorset(n);
S:= {1};
for p in F do
S:= {seq(seq(s*p^j, j=0..floor(log[p](n/s))), s=S)}
od;
convert(S, `*`)
end proc:
seq(A(n), n=1..100); # Robert Israel, Feb 09 2015
MATHEMATICA
regularQ[m_Integer, n_Integer] := Module[{omega = First /@ FactorInteger @ m }, If[Length[Select[omega, Divisible[n, #] &]] == Length[omega], True, False]]; a20140819[n_Integer] := Times @@ Flatten[Position[Thread[regularQ[Range[1, n], n]], True]]; a20140819 /@ Range[41]
regulars[n_] := Block[{f, a}, f[x_] := First /@ FactorInteger@ x; a = f[n]; {1}~Join~Select[Range@ n, SubsetQ[a, f@ #] &]]; Array[Times @@ regulars@ # &, 12] (* Michael De Vlieger, Feb 09 2015 *)
Table[Times @@ Select[Range@ n, (Floor[n^#/#] - Floor[(n^# - 1)/#]) == 1 &], {n, 41}] (* Michael De Vlieger, May 26 2016 *)
PROG
(PARI) lista(nn) = {vf = vector(nn, n, Set(factor(n)[, 1])); vector(nn, n, prod(i=1, n, if (setintersect(vf[i], vf[n]) == vf[i], i, 1))); } \\ Michel Marcus, Aug 23 2014
(PARI) for(n=1, 100, print1(prod(k=1, n, k^(floor(n^k/k) - floor((n^k - 1)/k))), ", ")) \\ Indranil Ghosh, Mar 22 2017
(Python)
from sympy import primefactors
def A243103(n):
y, pf = 1, set(primefactors(n))
for m in range(2, n+1):
if set(primefactors(m)) <= pf:
y *= m
return y # Chai Wah Wu, Aug 28 2014
(Scheme)
;; A naive implementation, code for A123275bi given under A123275:
(define (A243103 n) (let loop ((k n) (m 1)) (cond ((= 1 k) m) ((= 1 (A123275bi n k)) (loop (- k 1) (* m k))) (else (loop (- k 1) m)))))
;; Antti Karttunen, Mar 22 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael De Vlieger, Aug 19 2014
STATUS
approved