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A003827 'Core' alternating sign n X n matrices, i.e., those that are not 'blown up' from a smaller matrix by inserting row i, column j with a_ij = 1 and all other entries in that row and column equal to 0. 5
1, 2, 59, 1292, 53862, 3615208, 392961340, 68986099580, 19595297946515, 9048133666290540, 6832278662513786160, 8489106538840284343800, 17456177529017536829265000, 59700294731704834466701403040, 340945552945616104095546549396336, 3261527521637774696821080128931389072 (list; graph; refs; listen; history; text; internal format)
OFFSET

3,2

REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, D1.

LINKS

Alois P. Heinz, Table of n, a(n) for n = 3..100

FORMULA

Let b(n) = Product_{i=0..n-1} (3*i+1)!/(n+i)! be the number of alternating sign n X n matrices (i.e., sequence A005130), and a(n) the number of core alternating sign n X n matrices considered here, with the sequence [1,2,59,...] starting at offset n=3. Then it is not hard to show that for n>3: a(n) = b(n) - n! - Sum_{k=1..n-3} binomial(n,k)^2 * k! *a(n-k). - Christine Bessenrodt, Oct 02 2015

a(n) ~ exp(1/36) * Pi^(1/3) * 2^(5/12 - 2*n^2) * 3^(-7/36 + 3*n^2/2) / (A^(1/3) * Gamma(1/3)^(2/3) * n^(5/36)), where A = A074962 is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Apr 25 2016

PROG

(PARI) \\ rather inefficient, should use memoization

b(n) = prod(i=0, n-1, (3*i+1)!/(n+i)! );

a(n) = b(n) - n! - sum(k=1, n-3, binomial(n, k)^2 * k! *a(n-k) );

vector(20, n, a(n)) \\ Joerg Arndt, Oct 03 2015

CROSSREFS

Cf. A051055, A005130.

Sequence in context: A142666 A283489 A051055 * A241324 A139190 A050283

Adjacent sequences:  A003824 A003825 A003826 * A003828 A003829 A003830

KEYWORD

nonn

AUTHOR

Christine Bessenrodt

EXTENSIONS

Corrected and extended by Christine Bessenrodt, Oct 02 2015

STATUS

approved

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Last modified November 20 06:06 EST 2018. Contains 317385 sequences. (Running on oeis4.)