A214561 Number of 1's in binary expansion of n^n.

1, 1, 1, 4, 1, 6, 6, 11, 1, 14, 11, 20, 10, 28, 23, 33, 1, 31, 27, 41, 26, 49, 36, 59, 16, 58, 41, 68, 37, 62, 51, 83, 1, 79, 61, 88, 58, 97, 85, 98, 53, 115, 96, 116, 63, 123, 96, 128, 41, 138, 105, 144, 90, 163, 128, 164, 81, 172, 148, 181, 124, 167, 134, 201, 1

Offset

0,4

Comments

a(n) + A214562(n) = 1+floor(log_2(n^n)) = 1, 1, 3, 5, 9, 12, 16, 20, 25, 29, 34, 39, 44, 49... is the number of binary digits in n^n. - R. J. Mathar, Jul 22 2012

Links

Alois P. Heinz, Table of n, a(n) for n = 0..16384

Formula

a(n) = A000120(A000312(n)).

a(2^k)=1.

Maple

a:= proc(n) option remember; local m, r;
      m, r:= n^n, 0;
      while m>0 do r:= r +irem(m, 2, 'm') od; r
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 21 2012

Mathematica

Table[Count[IntegerDigits[n^n, 2], 1], {n, 1, 64}] (* Geoffrey Critzer, Sep 30 2013 *)
Join[{1}, Table[DigitCount[n^n, 2, 1], {n, 100}]] (* Harvey P. Dale, Oct 13 2022 *)

Prog

(Python)
for n in range(300):
    c = 0
    b = n**n
    while b>0:
        c += b&1
        b//=2
    print(c, end=', ')
(Python)
def a(n): return bin(n**n)[2:].count('1')
print([a(n) for n in range(65)]) # Michael S. Branicky, May 22 2021
(PARI) vector(66, n, b=binary((n-1)^(n-1)); sum(j=1, #b, b[j])) /* Joerg Arndt, Jul 21 2012 */

Crossrefs

Cf. A000120, A159918, A079584.

Sequence in context: A021100 A021028 A193529 * A066575 A070251 A134234

Adjacent sequences: A214558 A214559 A214560 * A214562 A214563 A214564

Keyword

nonn

Author

Alex Ratushnyak, Jul 21 2012

Status

approved