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A227499 Number of the Lipschitz quaternions in a reduced system modulo n. 16
1, 8, 48, 128, 480, 384, 2016, 2048, 3888, 3840, 13200, 6144, 26208, 16128, 23040, 32768, 78336, 31104, 123120, 61440, 96768, 105600, 267168, 98304, 300000, 209664, 314928, 258048, 682080, 184320, 892800, 524288, 633600, 626688, 967680, 497664, 1822176 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
Catalina Calderón, Jose Maria Grau, A. Oller-Marcén, and László Tóth, Counting invertible sums of squares modulo n and a new generalization of Euler's totient function, Publicationes Mathematicae-Debrecen, Vol. 87 (1-2) (2015), pp. 133-145; arXiv preprint, arXiv:1403.7878 [math.NT], 2014.
FORMULA
Multiplicative: a(2^s) = 2^(4s-1); a(3^s) = 16*3^(4s-3); a(5^s) = 32*3*5^(4s-3).
From Amiram Eldar, Feb 13 2024: (Start)
Multiplicative with a(2^e) = 2^(4*e-1), and a(p^e) = p^(4*e-3) * (p-1)^2 * (p+1) for an odd prime p.
Dirichlet g.f.: zeta(s-4) * (1 - 1/2^(s-3)) * Product_{p prime > 2} (1 - 1/p^(s-3) - (p-1)/p^(s-1)).
Sum_{k=1..n} a(k) = (12/55) * c * n^5 + O(n^4 * log(n)), where c = Product_{p prime} (1 - 1/p^2 - 1/p^3 + 1/p^4) = 0.53589615382833799980... (Calderón et al., 2015).
Sum_{n>=1} 1/a(n) = (17*Pi^8/57240) * Product_{p prime} (1 - 2/p^2 + 1/p^4 + 1/p^5 + 2/p^6 - 1/p^8) = 1.16039588611967540703... . (End)
MATHEMATICA
cuater[n_] := Flatten[Table[{a, b, c, d}, {a, n}, {b, n}, {c, n}, {d, n}], 3]; a[n_] := Length@Select[cuater[n], GCD[#.#, n] == 1 &]; Array[a, 20]
f[p_, e_] := (p-1)*p^(4*e-1) * If[p == 2, 1, 1 - 1/p^2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Feb 13 2024 *)
PROG
(PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i, 1]; e = f[i, 2]; (p-1)*p^(4*e-1) * if(p == 2, 1, 1 - 1/p^2)); } \\ Amiram Eldar, Feb 13 2024
CROSSREFS
Sequence in context: A067239 A152750 A121355 * A168012 A222816 A280056
KEYWORD
nonn,easy,mult
AUTHOR
STATUS
approved

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Last modified April 28 06:27 EDT 2024. Contains 372020 sequences. (Running on oeis4.)