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Numbers n such that determinant[{{n, phi(n), sigma(n)}, {n+1, phi(n+1), sigma(n+1)}, {n+2, phi(n+2), sigma(n+2)}}] is a nonnegative cube.
1

%I #27 Jan 03 2024 06:31:26

%S 1,2,9,217,12021

%N Numbers n such that determinant[{{n, phi(n), sigma(n)}, {n+1, phi(n+1), sigma(n+1)}, {n+2, phi(n+2), sigma(n+2)}}] is a nonnegative cube.

%C If n is a term of the sequence, then the parallelepiped formed by the vectors {n, phi(n), sigma(n)}, {n+1, phi(n+1), sigma(n+1)}, {n+2, phi(n+2), sigma(n+2)} has the same volume as that of an integral cube.

%C The corresponding sides of the cube are: 0,1,2,6,24,... and a(6) > 10^8. - _Michel Marcus_, May 18 2013

%C If negative values had been allowed, then the sequence would have been : 1, 2, 9, 217, 347, 12021, 37589, 4386257, 9231569, 14031359, 15692057, 36773279, 77018021, ... The negative cube roots for the new values being: -8, -42, -174, -269, -408, -251, -498, ... - _Michel Marcus_, May 19 2013

%C a(6) > 10^9. - _Sean A. Irvine_, Dec 19 2023

%e For 217, the corresponding matrix is {{217,phi(217), sigma(217)},{218,phi(218), sigma(218)},{219,phi(219), sigma(219)}} = {{217,180,256},{218,108,330},{219,144,296}}, whose determinant is 216 = 6^3. Therefore 217 is a term of the sequence.

%t p[n_] := Det[{{n, EulerPhi[n], DivisorSigma[1, n]}, {n + 1, EulerPhi[n + 1], DivisorSigma[1, n + 1]}, {n + 2, EulerPhi[n + 2], DivisorSigma[1, n + 2]}}]; Do[If[IntegerQ[p[i]^(1/3)], Print[i]], {i, 1, 10^5}]

%t Position[Table[Det[Table[{n,EulerPhi[n],DivisorSigma[1,n]},{n,k,k+2}]],{k,13000}],_?(#>=0&&IntegerQ[Surd[#,3]]&)]//Flatten (* _Harvey P. Dale_, Jul 19 2020 *)

%o (PARI) lista(nn) = {for (n=1, nn, m = matrix (3, 3, x, y, if (y==1, x+n-1, if (y==2, eulerphi(x+n-1), if (y==3, sigma(x+n-1))))); md = matdet (m); if (md >= 0 && ispower(md, 3), print1(n, ", ")););} \\ _Michel Marcus_, May 18 2013

%K nonn,more

%O 1,2

%A _Joseph L. Pe_, Jan 29 2002