1,1

All terms are divisible by 4.

a(n)=(5n-1)!/(n!(4n-1)!).

a(1)=4 because there are 4 terms in A163142 with 1 zero in binary representation {23,27,29,30}_10 ={10111,11011,11101,11110}_2

a(2)=36 because there are 36 terms in A163142 with 2 zeros in binary representation: {639,703,735,751,759,763,765,766,831,863,879,887,891,893,894,927,943,951,955,957,958,975,983,987,989,990,999,1003,1005,1006,1011,1013,1014,1017,1018,1020}_10={1001111111,...,1111111100}_2

a(3)=364 terms in A163142 from 18431 to 32760 with 3 zeros in binary representation 18431_10=100011111111111_2 and 32760_10=111111111111000_2

a(4)=3876 terms in A163142 from 557055 to 1048560 with 4 zeros in binary representation, etc.

Table[(5*n-1)!/ n!/(4*n-1)!, {n, 20}]

Sequence in context: A093186 A000765 A026334 this_sequence A138736 A019999 A002894

Adjacent sequences: A163452 A163453 A163454 this_sequence A163456 A163457 A163458

nonn

Zak Seidov (zakseidov(AT)yahoo.com), Jul 28 2009