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a(n) = product of total number of 0's and total number of 1's in binary expansions of 0, ..., n.
2

%I #10 Mar 01 2023 19:51:36

%S 0,1,4,8,20,35,54,72,117,165,221,280,352,425,504,576,726,875,1036,

%T 1200,1386,1575,1776,1976,2214,2451,2700,2944,3216,3479,3750,4000,

%U 4455,4897,5355,5808,6300,6789,7296,7800,8364,8925,9504,10080,10695,11305,11931,12544,13260,13965,14688

%N a(n) = product of total number of 0's and total number of 1's in binary expansions of 0, ..., n.

%H Alois P. Heinz, <a href="/A301896/b301896.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F a(n) = A059015(n)*A000788(n).

%F a(2^k-1) = 2^(k-2)*(2^k*(k - 2) + 4)*k.

%e +---+-----+---+---+---+---+----------+

%e | n | bin.|0's|sum|1's|sum| a(n) |

%e +---+-----+---+---+---+---+----------+

%e | 0 | 0 | 1 | 1 | 0 | 0 | 1*0 = 0 |

%e | 1 | 1 | 0 | 1 | 1 | 1 | 1*1 = 1 |

%e | 2 | 10 | 1 | 2 | 1 | 2 | 2*2 = 4 |

%e | 3 | 11 | 0 | 2 | 2 | 4 | 2*4 = 8 |

%e | 4 | 100 | 2 | 4 | 1 | 5 | 4*5 = 20 |

%e | 5 | 101 | 1 | 5 | 2 | 7 | 5*7 = 35 |

%e | 6 | 110 | 1 | 6 | 2 | 9 | 6*9 = 54 |

%e +---+-----+---+---+---+---+----------+

%e bin. - n written in base 2;

%e 0's - number of 0's in binary expansion of n;

%e 1's - number of 1's in binary expansion of n;

%e sum - total number of 0's (or 1's) in binary expansions of 0, ..., n.

%p b:= proc(n) option remember; `if`(n=0, [1, 0], b(n-1)+

%p (l-> [add(1-i, i=l), add(i, i=l)])(Bits[Split](n)))

%p end:

%p a:= n-> (l-> l[1]*l[2])(b(n)):

%p seq(a(n), n=0..50); # _Alois P. Heinz_, Mar 01 2023

%t Accumulate[DigitCount[Range[0, 50], 2, 0]] Accumulate[DigitCount[Range[0, 50], 2, 1]]

%o (Python)

%o def A301896(n): return (2+(n+1)*(m:=(n+1).bit_length())-(1<<m)-(k:=sum(i.bit_count() for i in range(1,n+1))))*k # _Chai Wah Wu_, Mar 01 2023

%Y Cf. A000120, A000788, A023416, A059015, A071295, A301336.

%K nonn,base

%O 0,3

%A _Ilya Gutkovskiy_, Mar 28 2018