

A126848


Arises in lower bound of the spectral norm of n X n symmetric random matrices.


0



2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

Abstract: "We show that the spectral radius of an N times N random symmetric matrix with i.i.d. bounded centered but nonsymmetrically distributed entries is bounded from below by 2 *sigma  o(N^{6/11+\epsilon}), where sigma^2 is the variance of the matrix entries and epsilon is an arbitrary small positive number. Combining with our previous result from [6], this proves that for any epsilon > 0, one has A_N =2*sigma + o(N^{6/11+epsilon}) with probability going to 1 as N goes to infinity." In this sequence, we computer a lower bound under the artificial assumption that sigma = n. Records for a(n) are for n = 1, 3, 5, 8, 12, 16, 22, 28, 35, 43, 52, 62, 73.


LINKS

Table of n, a(n) for n=1..75.
Sandrine Peche and Alexander Soshnikov, On the lower bound of the spectral norm of symmetric random matrices with independent entries


FORMULA

a(n) = floor(2*n*(n^(6/11))).


EXAMPLE

a(10) = 5 because 5 = floor(2 * 10 * (10^((6/11))) = floor(5.69607174).
a(100) = 16 = floor(2 * 100 * (100^((6/11)) = floor(16.2226166).
a(1000) = 46 = floor(2 * 1000 * (1000^((6/11)) = floor(46.202594).
a(10000) = 131 = floor(2 * 10000 * (10000^((6/11)) = floor(131.586645).
a(100000) = 374 = floor(2 * 100000 * (1000000^((6/11)) = floor(374.763485).
a(1000000) = 1067 = floor(2 * 1000000 * (1000000^((6/11)) = floor(1067.33985).


CROSSREFS

Sequence in context: A048686 A278959 A090501 * A232753 A067085 A321578
Adjacent sequences: A126845 A126846 A126847 * A126849 A126850 A126851


KEYWORD

easy,nonn


AUTHOR

Jonathan Vos Post, Jun 07 2007


STATUS

approved



