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A092147
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Number of even-length palindromes among the k-tuples of partial quotients of the continued fraction expansions of n/r, r=1,...,n.
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0
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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
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OFFSET
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1,5
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COMMENTS
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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
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LINKS
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FORMULA
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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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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