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%I #20 Jan 03 2022 07:37:48
%S 2,12,30,18,90,132,24,240,306,60,462,300,162,756,870,180,360,1260,264,
%T 1560,1722,216,2070,1470,480,2652,1080,612,3306,3540,540,1584,4290,
%U 924,4830,5112,600,2700,6006,1458,6642,2880,1512,7656,3960,1740,3672,9120
%N Number of solutions to x + y + z = 0 mod (2n+1) such that x,y,z are units modulo 2n+1, i.e., gcd(x, 2n+1) = gcd(y, 2n+1) = gcd(z, 2n+1) = 1.
%C This sequence is not multiplicative. What is multiplicative is the sequence b = 1,0,2,0,12,0,30, ... such that a(n) = b(2n+1) and b(2n)=0. - _Robert Israel_, Jan 29 2017
%H Robert Israel, <a href="/A061780/b061780.txt">Table of n, a(n) for n = 1..10000</a>
%H Masao Arai and Jiyu Gakuen, <a href="http://www.jstor.org/stable/2311477">Problem E 1460</a>, American Mathematical Monthly, Vol. 68, No. 3 (1961), p. 295 and <a href="http://www.jstor.org/stable/2311716">A Number-Theoretic Function</a>, solution by Leonard Carlitz, American Mathematical Monthly, Vol. 68, No. 9 (1961), pp. 932-933.
%H László Tóth, <a href="https://doi.org/10.1007/s11139-020-00353-z">Another generalization of Euler's arithmetic function and Menon's identity</a>, The Ramanujan Journal (2021).
%F If 2n+1 = p^k is a prime power with p an odd prime then a(n) = p^(2k-2) * (p^2 - 3p + 2).
%F a(n) = (2n+1)^2 * Product_{primes p | 2n+1} ((1 - 3/p + 2/p^2). - _Robert Israel_, Jan 29 2017
%F Sum_{k=1..n} a(k) ~ c * (2*n)^3/3 + O(n^2*log(n)^3), where c = A065473 (Tóth, 2021). - _Amiram Eldar_, Jan 03 2022
%e The only solutions modulo 3 in units are 1+1+1 = 0 mod 3, 2+2+2 = 0 mod 3 so the first element of the sequence is 2.
%p f:= n -> n^2*mul((1-1/p)*(1-2/p),p=numtheory:-factorset(n)):
%p seq(f(2*n+1),n=1..100); # _Robert Israel_, Jan 29 2017
%t a[n_] := (2*n+1)^2 * Product[(1-1/p)*(1-2/p), {p, FactorInteger[2*n+1][[;;,1]]}]; Array[a, 50] (* _Amiram Eldar_, Jan 03 2022 *)
%Y Cf. A065473.
%K nonn,look
%O 1,1
%A Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 22 2001
%E More terms from _Vladeta Jovovic_, Jun 23 2001
%E Corrected by _Robert Israel_, Jan 29 2017
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