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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.
0
1, 1, 1, 2, 1, 3, 6, 24, 36, 180, 840
OFFSET
1,4
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.
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Zhi-Wei Sun, Nov 15 2018
STATUS
approved