

A322070


Number of permutations f of {1,...,n} with f(1) < f(n) such that Sum_{k=1..n1} 1/(f(k)+f(k+1)) = 1.


5



0, 0, 0, 0, 0, 0, 1, 8, 22, 98, 844, 3831, 20922, 88902, 358253
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OFFSET

1,8


COMMENTS

Conjecture 1: a(n) > 0 for all n > 6. In other words, for each n = 7,8,... we have Sum_{k=1..n1} 1/(f(k)+f(k+1)) = 1 for some permutation f in the symmetric group S_n.
Conjecture 2: For any integer n > 7, there is a permutation g of {1,...,n} such that 1/(g(1)+g(2)) + 1/(g(2)+g(3)) + ... + 1/(g(n1)+g(n)) + 1/(g(n)+g(1)) = 1.
Conjecture 3. For any integer n > 5, there is a permutation p of {1,...,n} such that Sum_{k=1..n1} 1/(p(k)p(k+1)) = 0.
Conjecture 4. For each integer n > 7, there is a permutation q of {1,...,n} such that 1/(q(1)q(2)) + 1/(q(2)q(3)) + ... + 1/(q(n1)q(n)) + 1/(q(n)q(1)) = 0.
We have verified all the four conjectures for n up to 11. For Conjecture 2 with n = 8, we may take (g(1),...,g(8)) = (1,3,5,4,6,2,7,8) since 1/(1+3) + 1/(3+5) + 1/(5+4) + 1/(4+6) + 1/(6+2) + 1/(2+7) + 1/(7+8) + 1/(8+1) = 1.
See also A322069 for a similar conjecture.


LINKS



EXAMPLE

a(7) = 1, and for the permutation (4,5,7,2,1,3,6) of {1,...,7} we have 1/(4+5) + 1/(5+7) + 1/(7+2) + 1/(2+1) + 1/(1+3) + 1/(3+6) = 1.


MATHEMATICA

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, n1}]!=1, Goto[aa]]; r=r+1; Label[aa], {k, 1, n!}]; Print[n, " ", r], {n, 1, 11}]


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



