A322069 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.

0, 0, 0, 0, 0, 3, 1, 7, 13, 83, 84, 540, 480, 5488, 48922

Offset: 1

Author

Zhi-Wei Sun, Nov 25 2018

Keywords

• nonn
• more

Comments

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.

Examples

			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.
		

Links

• Zhi-Wei Sun, On permutations of {1, ..., n} and related topics, arXiv:1811.10503 [math.CO], 2018.

Crossrefs

Cf. A322070.

Programs

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

Extensions

a(12)-a(15) from Hugo Pfoertner, Aug 20 2022