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%I #22 Apr 16 2024 02:39:37
%S 1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,
%T 1,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,
%U 1,1,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0
%N a(n) = 1 if A053669(6*n) [the smallest prime not dividing 6*n] is of the form 6m-1, otherwise a(n) = 0.
%H Antti Karttunen, <a href="/A358847/b358847.txt">Table of n, a(n) for n = 1..100000</a>
%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>.
%F a(n) = A358755(6*n).
%F a(n) = A358846(n-1) XOR A358846(n), where XOR is bitwise-XOR, A003987. See comments in A358755.
%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 6 * Sum_{p prime, p == 5 (mod 6)} ((p-1)/(Product_{q prime, q <= p} q)) = 0.8261626908... . - _Amiram Eldar_, Apr 16 2024
%e For n = 85085 = 5*7*11*13*17, 6*n = A002110(7) = 510510, whose smallest nondividing prime is 19, which is of the form 6m+1, therefore a(85085) = 0.
%e For n = 1616615 = 5*7*11*13*17*19, 6*n = A002110(8), whose smallest nondividing prime is 23, which is of the form 6m-1, therefore a(1616615) = 1.
%e For n = 37182145 = 5*7*11*13*17*19*23, 6*n = A002110(9), whose smallest nondividing prime is 29, which is of the form 6m-1, therefore a(37182145) = 1. This is the first case where the alternating pattern following A276084 breaks.
%o (PARI)
%o A053669(n) = forprime(p=2, , if(n%p, return(p)));
%o A358847(n) = (5 == (A053669(6*n)%6));
%o (PARI) a(n)=forprime(p=5, , if(n%p, return(p%6==5))) \\ _Charles R Greathouse IV_, Dec 03 2022, corrected by _Antti Karttunen_, Apr 14 2024
%Y Characteristic function of A358849, whose complement A358848 gives the positions of zeros.
%Y Cf. A002110, A003987, A276084, A358755, A358846.
%Y Cf. also asymptotics of A353528 and A353529.
%K nonn,easy
%O 1
%A _Antti Karttunen_, Dec 03 2022
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