A389880 Number of permutations f of {1,...,n} with f(1) = 1 such that k*f(k) (k = 1..n) are distinct practical numbers.

1, 1, 0, 2, 0, 4, 0, 44, 24, 106, 4, 1096, 48, 224, 128, 155756, 288, 850540, 2128, 12605232, 5038304, 127394036, 189264, 3250295568, 42035712, 1040509776, 1623406080

Offset

1,4

Comments

If f is a permutation of {1,...,n} with f(1) = 1, and k*f(k) (k = 1..n) are distinct practical numbers, then f^{-1} (the inverse of f) is also a permutation of {1,...,n} with f^{-1}(1) = 1, and k*f^{-1}(k) (k = 1..n) are distinct practical numbers.

Conjecture: a(n) > 0 except for n = 3, 5, 7.

This is stronger than Conjecture 2 in A321651.

Links

Table of n, a(n) for n=1..27.

Example

a(4) = 2 since we can only take (f(1),f(2),f(3),f(4)) = (1,3,4,2), (1,4,2,3).
a(6) = 4 since we can only take (f(1),...,f(6)) = (1,2,4,5,6,3), (1,2,6,3,4,5), (1,3,6,2,4,5), (1,4,2,5,6,3).
a(11) > 0 since we may take (f(1),...,f(11)) = (1,3,10,5,8,7,4,11,2,9,6). Note that 1*1 = 1, 2*3 = 6, 3*10 = 30, 4*5 = 20, 5*8 = 40, 6*7 = 42, 7*4 = 28, 8*11 = 88, 9*2 = 18, 10*9 = 90 and 11*6 = 66 are 11 distinct practical numbers.

Mathematica

(* A program for computing a(8): *)
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[i_]:=V[i]=Part[Permutations[{2, 3, 4, 5, 6, 7, 8}], i];
m=0; Do[U={}; Do[q=j*V[i][[j-1]]; If[pr[q], U=Append[U, q]], {j, 2, 8}]; If[Length[Union[U]]==7, m=m+1], {i, 1, 7!}]; Print[m]

Crossrefs

Cf. A005153, A321651.

Sequence in context: A285774 A262589 A019215 * A326799 A112081 A265158

Adjacent sequences: A389877 A389878 A389879 * A389881 A389882 A389883

Keyword

nonn,more

Author

Zhi-Wei Sun, Nov 15 2025

Extensions

a(11)-a(27) from Bert Dobbelaere, Nov 18 2025

Status

approved