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%I #16 Feb 20 2023 07:52:20
%S 788,210999,466255,4669455,25916396,51122994,204732428,204732429,
%T 549769529,2309049600,883426096,5108177043,2258007227,15750496273,
%U 22958443910,11162458684,41157474821,32790221027,130700807239
%N a(n) = smallest number k such that six consecutive prime numbers prime(n), prime(n+1),...,prime(n+5) are divisors of k, k+1,..., k+5 respectively.
%H Charles R Greathouse IV, <a href="/A180159/b180159.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4)= 4669455 because prime(4) = 7 =>
%e 4669455 = 7*667065 ;
%e 4669456 = 11*424496 ;
%e 4669457 = 13*359189 ;
%e 4669458 = 17*274674 ;
%e 4669459 = 19*245761 ;
%e 4669460 = 23*203020.
%p with(numtheory):for p from 1 to 15 do: p1:=ithprime(p):p2:=ithprime(p+1):p3:=ithprime(p+2):p4:=ithprime(p+3):p5:=ithprime(p+4):p6:=ithprime(p+5):it:=0:for
%p n from 1 to 50000000 while(it=0) do:if irem(n, p1)=0 and irem(n+1, p2)=0 and
%p irem(n+2, p3)=0 and irem(n+3, p4)=0 and irem(n+4, p5)=0 and irem(n+5, p6)=0
%p then it:=1:printf(`%d, `, n):else fi:od:od:
%o (Sage) def A180159(n): return crt([-5..0][::-1], [nth_prime(i) for i in [n..n+5]]) # _D. S. McNeil_, Jan 16 2011
%o (PARI) a(n)=my(p=prime(n),r=Mod(0,p));for(i=1,5,p=nextprime(p+1);r=chinese(r,Mod(-i,p)));lift(r) \\ _Charles R Greathouse IV_, Jan 16 2011
%Y Cf. A077338, A180095, A180096, A180100.
%K nonn
%O 1,1
%A _Michel Lagneau_, Jan 16 2011
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