A092147 Number of even-length palindromes among the k-tuples of partial quotients of the continued fraction expansions of n/r, r=1,...,n.

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, 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, 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

Offset: 1

Author

John W. Layman, Mar 31 2004

Keywords

• nonn

Comments

Suggested by R. K. Guy, Mar 26 2004.

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

Crossrefs

Cf. A092089.

Formula

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.