1,2

Comment from Ralf Stephan, May 17 2004: Add terms of an increasingly bigger diamond-shaped part of Pascal's triangle:

.......................... 1

............ 1 .......... 1 1

.. 1 ...... 1 1 ........ 1 2 1

. 1 1 =5 . 1 2 1 =19 .. 1 3 3 1 =69

.. 2 ...... 3 3 ........ 4 6 4

............ 6 ......... 10 10

.......................... 20

p divides a((p-1)/2) for prime p=5,13,17,29,37,41,53,61,73,89,97..=A002144[n] Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 04 2006

Also, number of square submatrices of a square matrix. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008

2*(2*n-1)!/(n!*(n-1)!)-1.

a(n)=sum(k=1, n, C(n, k)^2) - Benoit Cloitre (benoit7848c(AT)orange.fr), Aug 20 2002

Sum(sum(binomial(i, j), i=j..n+j), j=0..n) - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jul 23 2003

Also for n>1: a(n)=(2*n)!/(n!)^2-1 - Hugo Pfoertner (hugo(AT)pfoertner.org), Feb 10 2004

a(n) = Sum[Sum[(2n-i-j)!/(n-i)!/(n-j)!,{i,1,n}],{j,1,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 04 2006

Equals A115112 + 1. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008

seq(sum((binomial(n, m))^2, m=1..n), n=1..23); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 19 2008

Table[Sum[Sum[(2n-i-j)!/(n-i)!/(n-j)!, {i, 1, n}], {j, 1, n}], {n, 1, 20}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 04 2006

2*A001700 - 1. Cf. A047909, A091908.

Equals A000984 - 1. Central column of triangle A014473.

Right-hand column 2 of triangle A102541.

Cf. A002144.

Sequence in context: A070857 A047145 A055991 this_sequence A026590 A095073 A128349

Adjacent sequences: A030659 A030660 A030661 this_sequence A030663 A030664 A030665

nonn

Donald Mintz (djmintz(AT)home.com)