A067564 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, 2, 9, 217, 12021

Offset

1,2

Comments

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.

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

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

a(6) > 10^9. - Sean A. Irvine, Dec 19 2023

Links

Table of n, a(n) for n=1..5.

Example

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.

Mathematica

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}]
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 *)

Prog

(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

Crossrefs

Sequence in context: A114563 A217017 A112311 * A267408 A359608 A165797

Adjacent sequences: A067561 A067562 A067563 * A067565 A067566 A067567

Keyword

nonn,more

Author

Joseph L. Pe, Jan 29 2002

Status

approved