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 A322099 Number of permutations f of {1,...,n} with f(1) < f(n) and Sum_{k=1..n-1} 1/(f(k)^2 - f(k+1)^2) = 0. 2
 0, 0, 0, 0, 0, 0, 0, 1, 4, 4, 55, 78, 552, 3484, 12720 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,9 COMMENTS Conjecture: a(n) > 0 for all n > 7. In other words, for each n = 8,9,... we have Sum_{k=1..n-1} 1/(f(k)^2 - f(k+1)^2) = 0 for some permutation f in the symmetric group S_n. LINKS Table of n, a(n) for n=1..15. Zhi-Wei Sun, On restricted permutations of {1,...,n}, arXiv:1811.10503 [math.CO], 2018. Zhi-Wei Sun, Is there a permutation pi in S_n with Sum_{k=1..n-1} 1/(pi(k)^2-pi(k+1)^2) = 0 for each n > 7?, Question 316208 on Mathoverflow, Nov. 25, 2018. EXAMPLE a(8) = 1, and for the permutation (4,5,2,7,3,1,6,8) of {1,...,8} we have 1/(4^2-5^2) + 1/(5^2-2^2) + 1/(2^2-7^2) + 1/(7^2-3^2) + 1/(3^2-1^2) + 1/(1^2-6^2) + 1/(6^2-8^2) = 0. a(12) > 0 since for the permutation (1,3,7,5,4,8,6,2,10,11,9,12) of {1,...,12} we have 1/(1^2-3^2) + 1/(3^2-7^2) + 1/(7^2-5^2) + 1/(5^2-4^2) + 1/(4^2-8^2) + 1/(8^2-6^2) + 1/(6^2-2^2) + 1/(2^2-10^2) + 1/(10^2-11^2) + 1/(11^2-9^2) + 1/(9^2-12^2) = 0. a(13) > 0 since for the permutation (1,6,2,9,11,5,3,7,13,8,4,10,12) of {1,...,13} we have 1/(1^2-6^2) + 1/(6^2-2^2) + 1/(2^2-9^2) + 1/(9^2-11^2) + 1/(11^2-5^2) + 1/(5^2-3^2) + 1/(3^2-7^2) + 1/(7^2-13^2) + 1/(13^2-8^2) + 1/(8^2-4^2) + 1/(4^2-10^2) + 1/(10^2-12^2) = 0. 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]]^2-Part[V[n], k][[i+1]]^2), {i, 1, n-1}]!=0, Goto[aa]]; r=r+1; Label[aa], {k, 1, n!}]; Print[n, " ", r], {n, 1, 11}] PROG (PARI) a(n)={my(s=0); forperm(n, f, if(f[1]

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