A092147 - OEIS

%I #8 Nov 28 2013 15:20:31

%S 0,1,0,1,2,1,0,1,0,5,0,1,2,1,2,1,2,1,0,5,0,1,0,1,4,5,0,1,2,5,0,1,0,5,

%T 2,1,2,1,2,5,2,1,0,1,2,1,0,1,0,9,2,5,2,1,2,1,0,5,0,5,2,1,0,1,8,1,0,5,

%U 0,5,0,1,2,5,4,1,0,5,0,5,0,5,0,1,8,1,2,1,2,5,2,1,0,1,2,1,2,1,0,9,2,5,0,5,2

%N Number of even-length palindromes among the k-tuples of partial quotients of the continued fraction expansions of n/r, r=1,...,n.

%C Suggested by _R. K. Guy_, Mar 26 2004.

%C Theorem. Let n=2^a*p1^a1*p2^a2*...*pk^ak*q, where the pi are distinct primes of the form 4m+1 and q contains only primes of the form 4m+3. Then a(n) is given by (1) 0, if a=0 and k=0, (2) 1, if a>0 and k=0, (3) Sum[2^t*s_t(a1,a2,...,ak), if a=0 and k>0 and (4) 1+Sum[2^(t+1)*s_t(a1,a2,...,ak), if a>0 and k>0, where the s_i are the symmetric polynomials s1(a1,a2a,...,ak)=a1+a2+...+ak, s2(a1,a2,...ak)=a1a2+a1a3+...+a2a3+a2a4+...+...+a(k-1)ak, etc. - James E. Shockley (shockley(AT)math.vt.edu), Jul 13 2004

%F Conjecture. Let n=(2^r)(p^s) where p is an odd prime and s>0. Then if p=4k+1, we have a(n)=2s if r=0, a(n)=4s+1 if r>0. On the other hand, if p=4k+3, we get a(n)=0 if r=0, a(n)=1 if r>0. Finally, if n=2^r we get a(n)=1.

%Y Cf. A092089.

%K nonn

%O 1,5

%A _John W. Layman_, Mar 31 2004