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A243103
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Product of numbers m with 2 <= m <= n whose prime divisors all divide n.
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8
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1, 2, 3, 8, 5, 144, 7, 64, 27, 3200, 11, 124416, 13, 6272, 2025, 1024, 17, 35831808, 19, 1024000, 3969, 247808, 23, 859963392, 125, 346112, 729, 2809856, 29, 261213880320000000, 31, 32768, 264627, 18939904, 30625, 26748301344768, 37, 23658496, 369603, 32768000000, 41
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
a(n) = Product_{k=2..n, A123275(n,k)=1} k.
(End)
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EXAMPLE
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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}.
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)
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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(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
y, pf = 1, set(primefactors(n))
for m in range(2, n+1):
if set(primefactors(m)) <= pf:
y *= m
(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)))))
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CROSSREFS
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Cf. A162306 (irregular triangle of regular numbers of n), A010846 (number of regular numbers of n), A244974 (sum of regular numbers of n), A007955, A244052 (record transform of regular numbers of n).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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