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A222423
Sum of (n AND k) for k = 0, 1, 2, ..., n, where AND is the bitwise AND operator.
7
0, 1, 2, 6, 4, 11, 18, 28, 8, 21, 34, 50, 60, 79, 98, 120, 16, 41, 66, 94, 116, 147, 178, 212, 216, 253, 290, 330, 364, 407, 450, 496, 32, 81, 130, 182, 228, 283, 338, 396, 424, 485, 546, 610, 668, 735, 802, 872, 816, 889, 962, 1038, 1108, 1187, 1266, 1348, 1400
OFFSET
0,3
COMMENTS
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.
Row sums of A080099. - R. J. Mathar, Apr 26 2013
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
LINKS
FORMULA
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
EXAMPLE
a(3) = 6 because 1 AND 3 = 1; 2 AND 3 = 2; 3 AND 3 = 3; and 1 + 2 + 3 = 6.
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.
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.
MATHEMATICA
Table[Sum[BitAnd[n, k], {k, 0, n}], {n, 0, 63}] (* Alonso del Arte, Feb 24 2013 *)
PROG
(Python)
for n in range(99):
s = 0
for k in range(n+1):
s += n & k
print(s, end=", ")
(PARI) a(n) = sum(k=0, n, bitand(n, k)); \\ Michel Marcus, May 17 2015
CROSSREFS
Cf. A004125.
Sequence in context: A242901 A353731 A266013 * A059909 A145177 A007517
KEYWORD
nonn,base
AUTHOR
Alex Ratushnyak, Feb 23 2013
STATUS
approved