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A222423 Sum of (n AND k) for k = 0, 1, 2, ..., n, where AND is the bitwise AND operator.

%I A222423 #39 Jun 08 2020 07:21:30
%S A222423 0,1,2,6,4,11,18,28,8,21,34,50,60,79,98,120,16,41,66,94,116,147,178,
%T A222423 212,216,253,290,330,364,407,450,496,32,81,130,182,228,283,338,396,
%U A222423 424,485,546,610,668,735,802,872,816,889,962,1038,1108,1187,1266,1348,1400
%N A222423 Sum of (n AND k) for k = 0, 1, 2, ..., n, where AND is the bitwise AND operator.
%C A222423 If n = 2^x, (n AND k) = 0 for k < n, therefore a(n) = n if and only if n = 0 or n = 2^x.
%C A222423 Row sums of A080099. - _R. J. Mathar_, Apr 26 2013
%C A222423 The associated incomplete sum_{0<=k<n} (n AND k) = a(n)-n starts 0, 0, 0, 3, 0, 6, 12, 21, 0, 12, 24, 39, 48, 66, 84, 105, .... and appears to be 3*A213673(n). - _R. J. Mathar_, Aug 22 2013
%H A222423 Ivan Neretin, <a href="/A222423/b222423.txt">Table of n, a(n) for n = 0..8192</a>
%F A222423 a(2^n-1) = A006516(n) for all n, since k AND 2^n-1 = k for all k<2^n. - _M. F. Hasler_, Feb 28 2013
%e A222423 a(3) = 6 because 1 AND 3 = 1; 2 AND 3 = 2; 3 AND 3 = 3; and 1 + 2 + 3 = 6.
%e A222423 a(4) = 4 because 1 AND 4 = 0; 2 AND 4 = 0; 3 AND 4 = 0; 4 AND 4 = 4; and 0 + 0 + 0 + 4 = 4.
%e A222423 a(5) = 11 because 1 AND 5 = 1; 2 AND 5 = 0; 3 AND 5 = 1; 4 AND 5 = 4; 5 AND 5 = 5; and 1 + 0 + 1 + 4 + 5 = 11.
%t A222423 Table[Sum[BitAnd[n, k], {k, 0, n}], {n, 0, 63}] (* _Alonso del Arte_, Feb 24 2013 *)
%o A222423 (Python)
%o A222423 for n in range(99):
%o A222423     s = 0
%o A222423     for k in range(n+1):
%o A222423         s += n & k
%o A222423     print(s, end=",")
%o A222423 (PARI) a(n) = sum(k=0, n, bitand(n, k)); \\ _Michel Marcus_, May 17 2015
%Y A222423 Cf. A004125.
%K A222423 nonn,base
%O A222423 0,3
%A A222423 _Alex Ratushnyak_, Feb 23 2013