A239441 Number of invertible octonions over Z/nZ.

1, 128, 4320, 32768, 312000, 552960, 4939200, 8388608, 28343520, 39936000, 194858400, 141557760, 752955840, 632217600, 1347840000, 2147483648, 6565340160, 3627970560, 16089567840, 10223616000, 21337344000, 24941875200, 74905892160, 36238786560, 121875000000, 96378347520

Offset: 1

Author

José María Grau Ribas, Mar 18 2014

Keywords

• nonn
• easy
• mult

Comments

Number of octonions over Z/nZ with invertible norm; i.e., number of solutions of the equation gcd(x_1^2 + ... + x_8^2, n)=1 with 0 < x_i <= n.

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000
• 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.

Crossrefs

Sequences giving the number of solutions to the equation gcd(x_1^2+...+x_k^2, n) = 1 with 0 < x_i <= n: A000010 (k=1), A079458 (k=2), A053191 (k=3), A227499 (k=4), A238533 (k=5), A238534 (k=6), A239442 (k=7), A239441 (k=8), A239443 (k=9).

Programs

Mathematica

fa=FactorInteger;lon[n_]:=Length[fa[n]];Phi[k_, n_] := Which[Mod[k, 2] == 1, n^(k - 1)*EulerPhi[n], Mod[k, 4] ==0, n^(k - 1)*EulerPhi[n]*Product[1 - 1/fa[2n][[i, 1]]^(k/2), {i, 2, lon[2 n]}],True, n^(k - 1)*EulerPhi[n]*Product[Which[ Mod[fa[ n][[i, 1]], 4] == 3 , 1 + 1/fa[ n][[i, 1]]^(k/2), Mod[fa[ n][[i, 1]], 4] == 1, 1 - 1/fa[ n][[i, 1]]^(k/2), True, 1], {i, 1, lon[ n]}]]; Table[Phi[8,n],{n,1,100}]
f[p_, e_] := (p-1)*p^(8*e-1) * If[p == 2, 1, 1 - 1/p^4]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 30] (* Amiram Eldar, Feb 13 2024 *)

PARI

a(n)={my(p=lift(Mod(sum(i=0, n-1, x^(i^2%n)), x^n-1)^8)); sum(i=0, n-1, if(gcd(i,n)==1, polcoeff(p,i)))} \\ Andrew Howroyd, Aug 06 2018

PARI

a(n)={my(f=factor(n)); prod(i=1, #f~, my([p,e]=f[i,]); if(p==2, 2^(8*e-1), (p - 1)*p^(8*e - 5)*(p^4 - 1)))} \\ Andrew Howroyd, Aug 06 2018

Formula

Multiplicative with a(2^e) = 2^(8*e-1), a(p^e) = (p - 1)*p^(8*e - 5)*(p^4 - 1) for odd prime p. - Andrew Howroyd, Aug 06 2018

Sum_{k=1..n} a(k) ~ c * n^9, where c = (16/141) * Product_{p prime} (1 - 1/p^2 - 1/p^5 + 1/p^6) = 0.06731687367... . - Amiram Eldar, Nov 30 2022

From Amiram Eldar, Feb 13 2024: (Start)

Dirichlet g.f.: zeta(s-8) * (1 - 1/2^(s-7)) * Product_{p prime > 2} (1 - 1/p^(s-7) - (p-1)/p^(s-3)).

Sum_{n>=1} 1/a(n) = (257*Pi^14/1312151400) * Product_{p prime} (1 - 1/p^2 - 1/p^4 + 1/p^6 + 1/p^9 + 1/p^10 + 1/p^12 - 1/p^14) = 1.00807991170717322545... . (End)