A322070 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, 0, 1, 8, 22, 98, 844, 3831, 20922, 88902, 358253

Offset: 1

Author

Zhi-Wei Sun, Nov 25 2018

Keywords

• nonn
• more

Comments

Conjecture 1: a(n) > 0 for all n > 6. In other words, for each n = 7,8,... 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 > 7, there is a permutation g of {1,...,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.

Conjecture 3. For any integer n > 5, there is a permutation p of {1,...,n} such that Sum_{k=1..n-1} 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(n-1)-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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A322069.

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