来自在线整数百科全书的问候语!http://oeis.org/ Search: id:a087694 Showing 1-1 of 1 %I A087694 %S A087694 1,1,3,4,1,3,13,4,9,1,1,12,25,13,3,16,1,9,37,4,39,1,1,12,25,25,27,52, %T A087694 1,3,61,16,3,1,13,36,73,37,75,4,1,39,85,4,9,1,1,48,133,25,3,100,1,27, %U A087694 1,52,111,1,1,12,121,61,117,64,25,3,133,4,3,13 %N A087694 Number of solutions to x^2 + xy + y^2 == 0 (mod n). %H A087694 Andrew Howroyd,n,a(n)n=1…10000的表%f A08699-乘以A(3 ^ E)=3 ^ E,A(p^ e)=((p-1)*e+p)*p^(E-1),如果p mod 3=1,则(p ^ e)=p^(2 *楼层(E/2)),如果p mod 3=2。- _Vladeta Jovovic_, Sep 27 2003 %p A087694 A087694 := proc(n) option remember; local pf,p,f,e ; if n = 1 then 1; else pf := ifactors(n)[2] ; if nops(pf) = 1 then f := op(1,pf) ; p := op(1,f) ; e := op(2,f) ; if p = 3 then n ; elif p mod 3 =1 then ((p-1)*e+p)*p^(e-1) ; else p^(2*floor(e/2)) ; end if; else mul(procname(op(1,p)^op(2,p)),p=pf) ; end if; end if; end proc: %p A087694 seq(A087694(n),n=1..70) ; # _R. J. Mathar_, Jan 07 2011 %t A087694 a[n_] := If[n==1, 1, Product[{p, e} = pe; Which[p==3, 3^e, Mod[p, 3] == 2, (p^2)^Quotient[e, 2], True, ((p-1) e + p) p^(e-1)], {pe, FactorInteger[n] }]]; %t A087694 a /@ Range[1, 100] (* _Jean-François Alcover_, Sep 20 2019, from PARI *) %o A087694 (PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); if(p==3, 3^e, if(p%3==2, (p^2)^(e\2), ((p-1)*e+p)*p^(e-1))))} \\ _Andrew Howroyd_, Jul 09 2018 %Y A087694 Cf. A000086. %K A087694 mult,nonn %O A087694 1,3 %A A087694 Yuval Dekel (dekelyuval(AT)hotmail.com), 9月27日在OEIS终端用户许可协议下可用的2003μl内容:HTTP:/OEIS.Org/许可证