|
| |
|
|
A091285
|
|
Numbers n such that sigma[3,n] is divisible by square of phi[n].
|
|
7
| | |
|
|
|
OFFSET
| 1,2
|
|
|
COMMENTS
| First 8 terms are solutions to relations for all j as follows: {sigma[6j+3,x]/phi[x]^2 is integer} for j=1,...300. Full proof is possible in knowledge of divisors corresponding sigma[k,x] and phi[x.]
No more terms through 10^9. - Ryan Propper (rpropper(AT)cs.stanford.edu), Jan 18 2008
|
|
|
EXAMPLE
| n=14: phi[n]^2=36,sigma[3,n]=3096=36.86
|
|
|
MATHEMATICA
| Empirical test for very high powers of divisors is: t = {1, 2, 3, 6, 14, 42, 3810, 13243560} Table[{6*j+3, Union[Table[IntegerQ[DivisorSigma[6*j + 3, Part[t, k]]/EulerPhi[Part[t, k]]^2], {k, 1, 8}]]}, {j, 1, 300}]; output={exponent, True}.
|
|
|
PROG
| (PARI) for(n = 1, 10^9, if(sigma(n, 3) % (eulerphi(n)^2) == 0, print1(n, ", "))) - Ryan Propper (rpropper(AT)cs.stanford.edu), Jan 18 2008
|
|
|
CROSSREFS
| Cf. A000010, A001158, A015773, A015774.
Sequence in context: A188775 A056569 A094468 * A109459 A118986 A091138
Adjacent sequences: A091282 A091283 A091284 * A091286 A091287 A091288
|
|
|
KEYWORD
| more,nonn
|
|
|
AUTHOR
| Labos E. (labos(AT)ana.sote.hu), Feb 03 2004
|
| |
|
|
