A263317 Least prime p > n such that the numbers sigma(k^2)/k^2 (k = 1,...,n) are pairwise incongruent modulo p, where sigma(m) is the sum of the divisors of m.

2, 5, 5, 7, 7, 29, 37, 37, 37, 37, 37, 43, 43, 43, 53, 79, 101, 101, 101, 101, 101, 101, 101, 101, 131, 131, 131, 131, 131, 131, 131, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 173, 283, 317, 389, 389, 389, 389, 389, 389, 389, 389, 389

Offset

1,1

Comments

Conjecture: a(n) exists for any n > 0, and a(n) < n^2 for all n > 2.

This implies that all the rational numbers sigma(n^2)/n^2 = Sum_{d|n^2} 1/d (n = 1,2,3,...) are pairwise distinct. We have verified that the numbers sigma(n^2)/n^2 (n = 1..10^5) are indeed pairwise distinct, and noted that sigma(26334^2)/26334^2 - sigma(6^2)/6^2 = 127/36 - 91/36 = 1.

We guess that for each k = 2,3,... all the numbers sigma(n^k)/n^k = Sum_{d|n^k} 1/d (n = 1,2,3,...) are pairwise distinct. See also A001157 for a similar conjecture.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

Example

a(1) = 2 since 2 is the least prime greater than sigma(1^2)/1^2 = 1.
a(2) = 5 since sigma(1^2)/1^2 = 1 and sigma(2^2)/2^2 = 7/4 are incongruent modulo the prime 5 > 2, but 1 is congruent to 7/4 modulo the prime 3.

Mathematica

rMod[m_, n_]:=rMod[m, n]=Mod[Numerator[m]*PowerMod[Denominator[m], -1, n], n, -n/2]
f[n_]:=f[n]=DivisorSigma[1, n^2]/n^2
Le[n_, m_]:=Le[m, n]=Length[Union[Table[rMod[f[k], Prime[m]], {k, 1, n}]]]
Do[n=1; m=1; Label[aa]; If[m>PrimePi[n]&&Le[n, m]==n, Goto[bb], m=m+1; Goto[aa]]; Label[bb]; Print[n, " ", Prime[m]]; If[n<60, n=n+1; Goto[aa]]]

Crossrefs

Cf. A000203, A000290, A001157, A065764.

Sequence in context: A233565 A121359 A082087 * A256300 A342852 A023850

Adjacent sequences: A263314 A263315 A263316 * A263318 A263319 A263320

Keyword

nonn

Author

Zhi-Wei Sun, Oct 14 2015

Extensions

Definition corrected by Omar E. Pol, Oct 24 2015

Status

approved