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%I #20 Aug 22 2022 04:50:18
%S 0,0,0,0,0,0,0,1,4,4,55,78,552,3484,12720
%N 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.
%C 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.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1811.10503">On restricted permutations of {1,...,n}</a>, arXiv:1811.10503 [math.CO], 2018.
%H Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/316208">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?</a>, Question 316208 on Mathoverflow, Nov. 25, 2018.
%e 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.
%e 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.
%e 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.
%t V[n_]:=V[n]=Permutations[Table[i,{i,1,n}]];
%t 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}]
%o (PARI) a(n)={my(s=0); forperm(n, f, if(f[1]<f[n] && sum(k=1, #f-1, 1/(f[k]^2-f[k+1]^2))==0, s++)); s} \\ _Andrew Howroyd_, Nov 27 2018
%Y Cf. A322069, A322070, A356446.
%K nonn,more
%O 1,9
%A _Zhi-Wei Sun_, Nov 26 2018
%E a(12) from _Andrew Howroyd_, Nov 27 2018
%E a(13)-a(15) from _Hugo Pfoertner_, Aug 22 2022
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