

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.


5



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
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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_{dn^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_{dn^k} 1/d (n = 1,2,3,...) are pairwise distinct. See also A001157 for a similar conjecture.


LINKS

ZhiWei 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 A023850 A175649
Adjacent sequences: A263314 A263315 A263316 * A263318 A263319 A263320


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 14 2015


EXTENSIONS

Definition corrected by Omar E. Pol, Oct 24 2015


STATUS

approved



