A321651 Number of even permutations f of {1,...,n} such that k^3 + f(k)^3 is a practical number for every k = 1,...,n.

1, 1, 1, 2, 1, 3, 6, 24, 36, 180, 840

Offset: 1

Author

Zhi-Wei Sun, Nov 15 2018

Keywords

• nonn
• more

Comments

Conjecture 1: a(n) > 0 for all n > 0.

Conjecture 2: For any positive integer n, there is a permutation f of {1,...,n} such that k*f(k) is practical for every k = 1,...,n.

P. Bradley proved in arXiv:1809.01012 that for any positive integer n there is a permutation f of {1,...,n} such that all the numbers k + f(k) (k = 1,...,n) are prime. Modifying his proof slightly we see that for each n = 1,2,3,... there is a permutation f of {1,...,n} such that k + f(k) is practical for every k = 1,...,n.

Examples

			a(5) = 1, and (5,4,3,2,1) is an even permutation of {1,2,3,4,5} with 1^3 + 5^3 = 126, 2^3 + 4^3 = 72, 3^3 + 3^3 = 54, 4^3 + 2^3 = 72 and 5^3 + 1^3 = 126 all practical.
		

Links

• Paul Bradley, Prime number sums, arXiv:1809.01012 [math.GR], 2018.
• Zhi-Wei Sun, Primes arising from permutations, Question 315259 on Mathoverflow, Nov. 14, 2018.
• Zhi-Wei Sun, Primes arising from permutations (II), Question 315341 on Mathoverflow, Nov. 14, 2018.
• Zhi-Wei Sun, A mysterious connection between primes and squares, Question 315351 on Mathoverflow, Nov. 15, 2018.

Crossrefs

Cf. A000578, A005153, A073364, A321597, A321610, A321611.

Programs

Mathematica

f[n_]:=f[n]=FactorInteger[n];
Pow[n_, i_]:=Pow[n,i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2]);
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}];
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0);
V[n_]:=V[n]=Permutations[Table[i,{i,1,n}]];
Do[r=0;Do[If[Signature[Part[V[n],k]]==-1,Goto[aa]];Do[If[pr[i^3+Part[V[n],k][[i]]^3]==False,Goto[aa]],{i,1,n}];r=r+1;Label[aa],{k,1,n!}];Print[n," ",r],{n,1,11}]