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A364925
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If the Sylow 2-subgroup of the multiplicative group modulo n is isomorphic to the direct product of (C_(2^i))^(e_i) for 1 <= i <= k, where C_m denotes the cyclic group of order m, then a(n) = Product_{i=1..k} prime(i)^(e_i).
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1
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1, 1, 2, 2, 3, 2, 2, 4, 2, 3, 2, 4, 3, 2, 6, 6, 7, 2, 2, 6, 4, 2, 2, 8, 3, 3, 2, 4, 3, 6, 2, 10, 4, 7, 6, 4, 3, 2, 6, 12, 5, 4, 2, 4, 6, 2, 2, 12, 2, 3, 14, 6, 3, 2, 6, 8, 4, 3, 2, 12, 3, 2, 4, 14, 9, 4, 2, 14, 4, 6, 2, 8, 5, 3, 6, 4, 4, 6, 2, 18, 2, 5, 2, 8, 21, 2, 6, 8, 5, 6
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OFFSET
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1,3
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COMMENTS
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Every positive integer occurs. Proof: it suffices to show that every prime number occurs infinitely many times in {a(prime(m)):m>=1}. Given prime p = prime(i), by Dirichlet's theorem on arithmetic progressions, there exist infinitely many primes congruent to 1 modulo 2^i but not 2^(i+1), hence the result.
For example, take n = 2*3^2*5^3. Note that a(3) = 2, a(5) = a(13) = 3, a(41) = a(73) = a(89) = 5, so a(3*5*13*41*73*89) = n.
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LINKS
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FORMULA
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Multiplicative with a(2) = 1, a(4) = 2, a(2^e) = 2*prime(e-2) for e >= 3; a(p^e) = prime(v(p-1,2)), where v(,2) is the 2-adic valuation.
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EXAMPLE
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(Z/80Z)* = C_2 X C_4 X C_4, so a(80) = prime(1)*prime(2)*prime(2) = 18.
(Z/85Z)* = C_4 X C_16, so a(85) = prime(2)*prime(4) = 21.
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MATHEMATICA
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f[p_, e_] := If[p == 2, Switch[e, 1, 1, 2, 2, _, 2*Prime[e-2]], Prime[IntegerExponent[p-1, 2]]]; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Aug 30 2023 *)
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PROG
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(PARI) a(n) = my(L=znstar(n)[2]); factorback(vector(#L, i, prime(valuation(L[i], 2))))
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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STATUS
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approved
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