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%I #30 Feb 15 2019 18:24:48
%S 25,49,50,64,81,75,121,100,169,128,127,147,157,163,181,193,199,289,
%T 229,243,239,257,361,283,293,313,343,337,349,353,373,379,397,409,421,
%U 529,439,457,463,467,487,499,509,523,541,547,557,577,625,601,613,619,631,643,661,673,677,691,841,667,733
%N a(n) is the largest level N such that genus of modular curve X_0(N) is n (or -1 if no such curve exists).
%C a(10^7) = 120000007 is the largest value in the first 1+10^7 terms of the sequence.
%C The exception occurs first at a(150) = -1. - _Georg Fischer_, Feb 15 2019
%H Gheorghe Coserea, <a href="/A273510/b273510.txt">Table of n, a(n) for n = 0..200010</a>
%H J. A. Csirik, M. Zieve, and J. Wetherell, <a href="http://arXiv.org/abs/math/0006096">On the genera of X0(N)</a>, arXiv:math/0006096 [math.NT], 2000.
%F Let S(n) = {k, n = A001617(k)}; if the level set S(n) is not empty then a(n) = max S(n) and A054728(n) = min S(n) and A273445(n) = card S(n), otherwise a(n) = A054728(n) = -1 and A273445(n) = 0.
%e For n = 0 we have 0 = A001617(k) when k is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18, 25 (A091401); the largest of this values is 25 therefore a(0) = 25.
%e For n = 1 we have 1 = A001617(k) when k is 11, 14, 15, 17, 19, 20, 21, 24, 27, 32, 36, 49 (A091403); the largest of this values is 49 therefore a(1) = 49.
%e For n = 2 we have 2 = A001617(k) when k is 22, 23, 26, 28, 29, 31, 37, 50 (A091404); the largest of this values is 50 therefore a(2) = 50.
%e For n = 150 (= A054729(1)) we have 150 <> A001617(k) for all k therefore a(150) = -1.
%t a1617[n_] := If[n < 1, 0, 1 + Sum[MoebiusMu[d]^2 n/d/12 - EulerPhi[GCD[d, n/d]]/2, {d, Divisors[n]}] - Count[(#^2 - # + 1)/n& /@ Range[n], _?IntegerQ]/3 - Count[(#^2 + 1)/n& /@ Range[n], _?IntegerQ]/4];
%t seq[n_] := Module[{a, bnd}, a = Table[-1, {n+1}]; bnd = 12n + 18 Floor[Sqrt[n] ] + 100; For[k = 1, k <= bnd, k++, g = a1617[k]; If[g <= n, a[[g+1]] = k]]; a];
%t seq[60] (* _Jean-François Alcover_, Nov 20 2018, after _Gheorghe Coserea_ and _Michael Somos_ in A001617 *)
%o (PARI)
%o A000089(n) = {
%o if (n%4 == 0 || n%4 == 3, return(0));
%o if (n%2 == 0, n \= 2);
%o my(f = factor(n), fsz = matsize(f)[1]);
%o prod(k = 1, fsz, if (f[k, 1] % 4 == 3, 0, 2));
%o };
%o A000086(n) = {
%o if (n%9 == 0 || n%3 == 2, return(0));
%o if (n%3 == 0, n \= 3);
%o my(f = factor(n), fsz = matsize(f)[1]);
%o prod(k = 1, fsz, if (f[k, 1] % 3 == 2, 0, 2));
%o };
%o A001615(n) = {
%o my(f = factor(n), fsz = matsize(f)[1],
%o g = prod(k=1, fsz, (f[k, 1]+1)),
%o h = prod(k=1, fsz, f[k, 1]));
%o return((n*g)\h);
%o };
%o A001616(n) = {
%o my(f = factor(n), fsz = matsize(f)[1]);
%o prod(k = 1, fsz, f[k, 1]^(f[k, 2]\2) + f[k, 1]^((f[k, 2]-1)\2));
%o };
%o A001617(n) = 1 + A001615(n)/12 - A000089(n)/4 - A000086(n)/3 - A001616(n)/2;
%o seq(n) = {
%o my(a = vector(n+1, g, -1), bnd = 12*n + 18*sqrtint(n) + 100, g);
%o for (k = 1, bnd, g = A001617(k); if (g <= n, a[g+1] = k));
%o return(a);
%o };
%o seq(60)
%Y Cf. A001617, A054728, A054729, A091401, A091403, A091404, A273445.
%K sign
%O 0,1
%A _Gheorghe Coserea_, May 23 2016
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