A076978 Product of the distinct primes dividing the product of composite numbers between consecutive primes.
1, 2, 6, 30, 6, 210, 6, 2310, 2730, 30, 39270, 7410, 42, 7590, 46410, 1272810, 30, 930930, 82110, 6, 21111090, 1230, 48969690, 1738215570, 2310, 102, 144690, 6, 85470, 29594505363092670, 16770, 49990710, 138, 7849357706190, 30, 300690390, 20223210, 1122990, 37916970
Offset: 1
Keywords
Examples
a(4) = product of prime divisors of the product of composite numbers between 7 and 11 = 2 * 3 * 5 = 30. a(5)=6 because 12 is the only composite number between the 5th and the 6th primes (11 and 13) and largest squarefree divisor of 12 is 6.
Links
- Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Programs
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Maple
with(numtheory): b:=proc(j) if issqrfree(j) then j else fi end: a:=proc(n) local B,BB: B:=divisors(product(i,i=ithprime(n)+1..ithprime(n+1)-1)): BB:=(seq(b(B[j]),j=1..nops(B))): max(BB); end: seq(a(n),n=1..33); # Emeric Deutsch, Jul 28 2006
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Mathematica
f[n_] := Times @@ (First@# & /@ FactorInteger[Times @@ Range[Prime[n] + 1, Prime[n + 1] - 1]]); Array[f, 50] (* Robert G. Wilson v, Dec 02 2020 *)
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PARI
a(n) = my(p=1); forcomposite(c=prime(n), prime(n+1), p*=c); factorback(factorint(p)[, 1]); \\ Michel Marcus, May 29 2022
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Python
from sympy import sieve as p, primefactors def A076978(n): result = 1 for composites in range(p[n]+1, p[n+1]): for primefactor in primefactors(composites): if result % primefactor != 0: result *= primefactor return result # Karl-Heinz Hofmann, May 30 2022
Formula
From Michel Marcus, May 29 2022: (Start)
Extensions
More terms from Emeric Deutsch, Jul 28 2006
More terms from Robert G. Wilson v, Dec 02 2020
Entry revised by N. J. A. Sloane, Dec 02 2020
Comments