A054900 a(n) = Sum_{j >= 1} floor(n/16^j).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset

0,33

Links

G. C. Greubel, Table of n, a(n) for n = 0..10000

Formula

a(n) = (n - A053836(n))/15.

From Hieronymus Fischer, Aug 14 2007: (Start)

Recurrence:

a(n) = a(floor(n/16)) + floor(n/16).

a(16*n) = a(n) + n.

a(n*16^m) = a(n) + n*(16^m - 1)/15.

a(k*16^m) = k*(16^m - 1)/15, for 0 <= k < 16, m>=0.

Asymptotic behavior:

a(n) = n/15 + O(log(n)).

a(n+1) - a(n) = O(log(n)) (this follows from the inequalities below).

a(n) <= (n-1)/15; equality holds for powers of 16.

a(n) >= (n-15)/15 - floor(log_16(n)); equality holds for n = 16^m - 1, m > 0.

Limits:

lim inf (n/15 - a(n)) = 1/15, for n --> oo.

lim sup (n/15 - log_16(n) - a(n)) = 0, for n --> oo.

lim sup (a(n+1) - a(n) - log_16(n)) = 0, for n --> oo.

Series:

G.f.: (1/(1-x))*Sum_{k > 0} x^(16^k)/(1-x^(16^k)). (End)

Mathematica

a[n_, m_]:= If[n==0, 0, a[Floor[n/m], m] +Floor[n/m]];
Table[a[n, 16], {n, 0, 127}] (* G. C. Greubel, Apr 28 2023 *)

Prog

(Magma)
m:=16;
function a(n) // a = A054900, m = 16
  if n eq 0 then return 0;
  else return a(Floor(n/m)) + Floor(n/m);
  end if; end function;
[a(n): n in [0..127]]; // G. C. Greubel, Apr 28 2023
(SageMath)
m=16 # a = A054900
def a(n): return 0 if (n==0) else a(n//m) + (n//m)
[a(n) for n in range(128)] # G. C. Greubel, Apr 28 2023

Crossrefs

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.

Cf. A053836, A054897, A054899, A067080, A098844, A132032.

Sequence in context: A373095 A056811 A097430 * A046042 A287866 A071841

Adjacent sequences: A054897 A054898 A054899 * A054901 A054902 A054903

Keyword

nonn

Author

Henry Bottomley, May 23 2000

Status

approved