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A319516
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Number of integers x such that 1 <= x <= n and gcd(x,n) = gcd(x+2,n) = gcd(x+6,n) = gcd(x+8,n) = 1.
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5
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1, 1, 1, 2, 1, 1, 3, 4, 3, 1, 7, 2, 9, 3, 1, 8, 13, 3, 15, 2, 3, 7, 19, 4, 5, 9, 9, 6, 25, 1, 27, 16, 7, 13, 3, 6, 33, 15, 9, 4, 37, 3, 39, 14, 3, 19, 43, 8, 21, 5, 13, 18, 49, 9, 7, 12, 15, 25, 55, 2, 57, 27, 9, 32, 9, 7, 63, 26, 19, 3, 67, 12, 69, 33, 5, 30, 21, 9, 75, 8
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OFFSET
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1,4
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COMMENTS
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Equivalently, a(n) is the number of "admissible" residue classes modulo n which are allowed (by divisibility considerations) to contain infinitely many initial primes in prime quadruples (p, p+2, p+6, p+8). This is a generalization of Euler's totient function: the number of residue classes modulo n containing infinitely many primes.
If n is prime, a(n) = max(1,n-4).
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REFERENCES
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V. A. Golubev, Sur certaines fonctions multiplicatives et le problème des jumeaux. Mathesis 67 (1958), 11-20.
József Sándor and Borislav Crstici, Handbook of Number Theory II, Kluwer, 2004, p. 289.
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LINKS
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FORMULA
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Multiplicative with a(p^e) = p^(e-1) if p <= 5; (p-4)*p^(e-1) if p > 5.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (49/200) * Product_{p prime >= 7} (1 - 4/p^2) = 0.1987646881... . - Amiram Eldar, Nov 01 2022
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EXAMPLE
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Some prime quadruples start with a prime congruent to 1 mod 4; others start with a prime congruent to 3 mod 4; that is, there are 2 "admissible" residue classes mod 4; therefore a(4)=2. All initial primes in prime quadruples are 5 mod 6; that is, there is only one "admissible" residue class mod 6; therefore a(6) = 1.
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MATHEMATICA
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a[n_] := Sum[Boole[CoprimeQ[n, x] && CoprimeQ[n, x+2] && CoprimeQ[n, x+6] && CoprimeQ[n, x+8]], {x, 1, n}]; Array[a, 80] (* Jean-François Alcover, Jan 29 2019 *)
f[p_, e_] := If[p < 7, p^(e-1), (p-4)*p^(e-1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jan 22 2020 *)
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PROG
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(PARI) phi4(n) = sum(x=1, n, (gcd(n, x)==1) && (gcd(n, x+2)==1) && (gcd(n, x+6)==1) && (gcd(n, x+8)==1));
for(n=1, 80, print1(phi4(n)", "))
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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STATUS
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approved
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