|
%I #3 Mar 31 2012 13:21:55
%S 21,31,21,31,21,11,31,21,11,31,22,11,32,41,42,21,32,41,22,12,11,31,41,
%T 21,32,21,32,41,22,11,11,32,41,22,31,21,11,12,31,42,42,22,11,12,12,31,
%U 42,21,32,21,32,21,31,41,22,31,21,11,12,31,42,41,41,21,11,11,32,42,42
%N Compare t(n) and t(n+1) where t = Ramanujan's tau function (A000594); a(n) = pq where p and q are given below.
%C p = 1 if t(n) < 0, t(n+1) < 0; p = 2 if t(n) >= 0, t(n+1) < 0; p = 3 if t(n) < 0, t(n+1) >= 0; p = 4 if t(n) >= 0, t(n+1) >= 0.
%C q = 1 if |t(n)| <= t(n+1); q = 2 if |t(n)| > t(n+1.
%e A000594(1)=1, A000594(2)=-24. Hence the sign changes from + to - and the modulus increasing. Consulting the key, this gives a(1)=21.
%o (PARI) T(n)=n*(n+1)/2 rtau3(n)=local(y,j); y=0; j=1; while (T(j-1)<n,j++); j--; for (i=1,j,y=y-(-1)^i*(2*i-1)*x^(T(i-1))); y=y^8; polcoeff(y,n-1) for (n=1,100, r3=rtau3(n); r31=rtau3(n+1); rn=(sign(r3)+1)/2; rn1=sign(r31)+1; print1((rn+rn1+1)""(abs(r3)>abs(r31))+1","))
%Y Cf. A000594.
%K nonn
%O 0,1
%A _Jon Perry_, Jan 06 2003
|
