The first, 40, terms starting at n=1 of the sequence enumerating 6 by 2*n bal\
    anced 0-1 matrices, i.e. where every row and every column has the same n\
    umber 0s and 1s are

                     In other words OEIS sequence  A2896 :



[6, 90, 1860, 44730, 1172556, 32496156, 936369720, 27770358330, 842090474940, 
25989269017140, 813689707488840, 25780447171287900, 825043888527957000, 
26630804377937061000, 865978374333905289360, 28342398385058078078010, 
932905175625150142902300, 30862498453931119524941700, 
1025612461904076314913090600, 34221233837555288931672244980, 
1146036523273637310151384256280, 38507502835917076808710010791800, 
1297804711481467883526891692041200, 43861095233349763172644603553579100, 
1486134894511368488511298785872590056, 50473042959116456206695526548113874936,
1717935171579831227898110571243045115440, 
58591130504414522992818338034043897226760, 
2002044498542496713444559505063252933261680, 
68529473603957996416207510223654247514473456, 
2349596714984894432382110228494495236586834656, 
80681997360632682235248226291172020032499469370, 
2774511796435675001461553546051852814078630633180, 
95540165693636170851827063699670877946754920344740, 
3294131649118176174850910406317256575226136984268520, 
113715685963242228605749065449983286191670924631338820, 
3930028637507047671218500394850794924034502049231057400, 
135969247090776809839238946061942634518417076025869103000, 
4709025755667844340496023085216048877206005830713755934000, 
163246703984118797490344652201478353553692393611900875694580]


 The sequence, let's call it a(n), enumerating 4 by 2*n balanced 0-1 natrice\
    s probably satisfies the empirically derived linear recurrence



                                                       2
36 (2 n + 3) (2 n + 1) (n + 1) a(n)   2 (2 n + 3) (10 n  + 30 n + 23) a(n + 1)
----------------------------------- - ----------------------------------------
                    3                                        3
             (n + 2)                                  (n + 2)

     + a(n + 2) = 0



                             and in Maple notation



36*(2*n+3)*(2*n+1)*(n+1)/(n+2)^3*a(n)-2*(2*n+3)*(10*n^2+30*n+23)/(n+2)^3*a(n+1)
+a(n+2) = 0


                           This took, 3.022, seconds





                          Just for fun here is a(1000)



1479908968741154176119167487862065281292454548073529147397623010647375301393042\
6548098407836177275382973009950311392311229629609659191416545894883327839631297\
4381516085427189383753564366614624567655767469410404694614376659388553148418759\
3934092091039653161756772324355752374740692655688391698463434304805856153897891\
5034694548308709615793766037997670956464688598980922480533375249882120426094191\
0664345063231818168198810577387955011447501854894555069920845639090890203774525\
6458741923590431743683397208912500132882855612538112565639467585335765392654866\
4342435530017361378483995120877385939023546783533687788954581968442739699436438\
5089866089175780587504142270276342612316989063847724938116766400374609205268529\
5010301143782399721917803434860563607116515209764491701523164966866672171458759\
5890983629554151368518032304825945072859785757977041821116295085612399050892965\
0761220387033326321089976992077147250826345278498209861230836388917069316304486\
9070764047160878950119365779768833135057986675267807834503764639850183052955418\
6325996827747706722106252811827902501488301400640713621670386926577101017150160\
9262814969527586697192886271186373078641265938423944234625349359835625493497370\
7703442619671577815640341684867932013896666406850423275763536049317555735231289\
3370250618019723896032280556943718594188008652716854393662782132106870780034492\
4670510114977880602866176973270200309006559302307400536724764999579992849653481\
6405904320611928774007395361161913554444003606802289450196183523386098870530490\
533207130999675662717321046270279393392001407382080


                     This took, 0.087,  additional seconds



                              --------------------



Now let's do it fully rigorously, using the multivariable Almkvist-Zeilberge\
    r algorithm

                                  trying L=, 0

                                  trying L=, 1

                                  trying L=, 2

                   There is hope for a recurrence of order, 2

`The empirically derived operator annihilating the sequence is`, 36*(2*n+3)*(2*
n+1)*(n+1)/(n+2)^3-2*(2*n+3)*(10*n^2+30*n+23)/(n+2)^3*N+N^2


`The rigorously derived operator annihilating the sequence is`, 36*(2*n+3)*(2*n
+1)*(n+1)/(n+2)^3-2*(2*n+3)*(10*n^2+30*n+23)/(n+2)^3*N+N^2


                           They are indeed the same!

                          This took, 12.837, seconds.