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A322069
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Number of permutations f of {1,...,n} with f(1) < f(n) such that Sum_{k=1..n-1} 1/(f(k)*f(k+1)) = 1.
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3
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0, 0, 0, 0, 0, 3, 1, 7, 13, 83, 84, 540, 480, 5488, 48922
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OFFSET
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1,6
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COMMENTS
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Conjecture 1: a(n) > 0 for all n > 5. In other words, for each n = 6,7,... we have Sum_{k=1..n-1} 1/(f(k)*f(k+1)) = 1 for some permutation f in the symmetric group S_n.
Conjecture 2: For any integer n > 6, there is an undirected circular permutation g different from the circular permutation (1,2,...,n) such that 1/(g(1)*g(2)) + 1/(g(2)*g(3)) + ... + 1/(g(n-1)*g(n)) + 1/(g(n)*g(1)) = 1.
We have verified both conjectures for n up to 11. For Conjecture 2 with n = 7, we may take (g(1),...,g(7)) = (3,2,1,6,5,4,7) since 1/(3*2) + 1/(2*1) + 1/(1*6) + 1/(6*5) + 1/(5*4) + 1/(4*7) + 1/(7*3) = 1.
See also A322070 for a similar conjecture.
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LINKS
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EXAMPLE
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a(7) = 1, and for the permutation (2,1,3,7,4,5,6) of {1,...,7} we have 1/(2*1) + 1/(1*3) + 1/(3*7) + 1/(7*4) + 1/(4*5) + 1/(5*6) = 1.
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MATHEMATICA
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V[n_]:=V[n]=Permutations[Table[i, {i, 1, n}]];
Do[r=0; Do[If[Part[V[n], k][[1]]>=Part[V[n], k][[n]]||Sum[1/(Part[V[n], k][[i]]*Part[V[n], k][[i+1]]), {i, 1, n-1}]!=1, Goto[aa]]; r=r+1; Label[aa], {k, 1, n!}]; Print[n, " ", r], {n, 1, 11}]
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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