A000195 -id:A000195 - cp's OEIS frontend

A336018 a(n) = floor(frac(log_2(n))*n), where frac denotes the fractional part.

0, 0, 1, 0, 1, 3, 5, 0, 1, 3, 5, 7, 9, 11, 13, 0, 1, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 27, 29, 0, 1, 2, 4, 6, 7, 9, 11, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 45, 47, 49, 52, 54, 56, 59, 61, 0, 1, 2, 4, 5, 7, 9, 10, 12, 13

Offset: 1

Andres Cicuttin, Jul 04 2020

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000079, A000523, A000195, A336017, A326299, A340301.

a:= n-> floor(n*log[2](n))-n*ilog2(n):
seq(a(n), n=1..80);  # Alois P. Heinz, Jan 04 2021

a[n_]:=Floor[FractionalPart[Log[2, n]]*n];
Table[a[n], {n, 1, 100}]

a(n) = floor(n*frac(log(n)/log(2))); \\ Michel Marcus, Jul 07 2020

def A336018(n):
    return len(bin(n**n//(2**((len(bin(n))-3)*n))))-3 # Chai Wah Wu, Jul 09 2020

a(n) = floor((log_2(n) - floor(log_2(n)))*n).
From Alois P. Heinz, Jan 04 2021: (Start)
a(n) = A326299(n) - A340301(n).
a(n) = 0 <=> n in { A000079 }. (End)

A003619 Not of form [ e^m ], m >= 1.

Original entry on oeis.org

1, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53

Offset: 1

N. J. A. Sloane, Mira Bernstein

nonn

If 1 is excluded (of form [e^0]) then complement of A000149. - Michel Marcus, Jun 16 2013

J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 11.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J. Lambek and L. Moser, Inverse and complementary sequences of natural numbers, Amer. Math. Monthly, 61 (1954), 454-458.

Cf. A000195.

a003619 n = n + floor (log (x + fromIntegral (floor $ log x)))
            where x = fromIntegral n + 1
-- Reinhard Zumkeller, Mar 17 2015

Table[n + Floor@ Log[n + 1 + Floor@ Log[n + 1]], {n, 50}] (* Michael De Vlieger, Oct 06 2017 *)

a(n) = n + floor( log (n + 1 + floor( log (n + 1) )) ) \\ Michel Marcus, Jun 16 2013

a(n) = n + [ log (n + 1 + [ log (n + 1) ]) ].

A092919 Partial sums of A000193 (round(log(n))).

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 190, 194, 198, 202, 206, 210

Offset: 1

Jorge Coveiro, Apr 13 2004

nonn

			a(1) = round(log(1)).
a(2) = a(1) + round(log(2)).
a(3) = a(2) + round(log(3)).
...

Cf. A000195, A000193, A092755.

A292621 a(n) = a(n-1) + a(floor(log(n))) with a(1) = 1, a(2) = 2.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 139, 143, 147, 151, 155, 159, 163, 167

Offset: 1

Yi Yang, Sep 20 2017

a(n) > c*n*log(n)*log(log(n))*log(log(log(n)))*...*log(log...(log(n))...) (k layers) for any sufficient large n, any constant c and any positive integer k.

The sum of 1/a(i) for i = 1, 2, 3, ... diverges extremely slowly.

Robert Israel, Table of n, a(n) for n = 1..10000
KeyTo9_Fans, A Chinese post discussing the sum of 1/a(i)
Fedor Petrov, The proof of the divergence of the sum of 1/a(i), Mathoverflow, Sep 2017.

Cf. A000195, A031876, A292620.

f:= proc(n) option remember;
procname(n-1)+procname(floor(log(n)))
end proc:
f(1):= 1: f(2):= 2:
map(f, [$1..100]); # Robert Israel, Sep 28 2017

a[n_] := a[n] = If[n <= 2, n, a[n - 1] + a[Floor@ Log@ n]]; Array[a, 62] (* Michael De Vlieger, Sep 21 2017 *)

a(n) = if (n<=2, n, a(n-1) + a(floor(log(n)))); \\ Michel Marcus, Sep 21 2017

A004237 a(n) = floor(100*log(n)).

Original entry on oeis.org

0, 69, 109, 138, 160, 179, 194, 207, 219, 230, 239, 248, 256, 263, 270, 277, 283, 289, 294, 299, 304, 309, 313, 317, 321, 325, 329, 333, 336, 340, 343, 346, 349, 352, 355, 358, 361, 363, 366, 368, 371, 373, 376

Offset: 1

N. J. A. Sloane

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000195, A004234.

seq(floor(100*log(n)),n=1..100); # Robert Israel, Aug 04 2019

A067283 Numbers n such that the number of divisors of n is floor(log(n)).

Original entry on oeis.org

11, 13, 17, 19, 25, 49, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 115, 118, 119, 122, 123, 125, 129, 133, 134, 141, 142, 143, 145, 146, 404, 412, 423, 425, 428, 436, 452, 475, 477, 507, 508, 524, 531, 539, 548, 549, 556, 575, 578

Offset: 1

Benoit Cloitre, Feb 24 2002

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000005, A000195.

Select[Range[600],DivisorSigma[0,#]==Floor[Log[#]]&] (* Harvey P. Dale, May 07 2018 *)

n such that A000005(n) = A000195(n).

A108173 Let beta = A058265. Sequence gives a(n) = 1 + ceiling((n-1)*beta^2).

Original entry on oeis.org

1, 5, 8, 12, 15, 18, 22, 25, 29, 32, 35, 39, 42, 45, 49, 52, 56, 59, 62, 66, 69, 73, 76, 79, 83, 86, 89, 93, 96, 100, 103, 106, 110, 113, 117, 120, 123, 127, 130, 133, 137, 140, 144, 147, 150, 154, 157, 160, 164, 167, 171, 174, 177, 181, 184, 188, 191, 194, 198, 201

Offset: 1

Roger L. Bagula, Jun 13 2005

nonn

Tribonacci version of A007066 using positive real Pisot root of x^3-x^2-x-1.

Cf. A007066, A076662, A000195.

NSolve[x^3 - x^2 - x - 1 == 0, x] beta = 1.8392867552141612 a[n_] = 1 + Ceiling[(n - 1)*beta^2] (* A007066 like*) aa = Table[a[n], {n, 1, 100}] (*A076662 like*) b = Table[a[n] - a[n - 1], {n, 2, Length[aa]}] F[1] = 2; F[n_] := F[n] = F[n - 1] + b[[n]] (* A000195 like *) c = Table[F[n], {n, 1, Length[b] - 1}]

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, May 31 2007

A171728 Numbers k which establish records for floor(log(log(log(2^k)))).

Original entry on oeis.org

2, 3, 4, 22, 2335, 762451795, 742762245454927736743542, 41133018324375596439235122590123953570787986963829981156569123587

Offset: 1

Jonathan Vos Post, Dec 16 2009

Morris writes: E. Thorp introduced the following card shuffling model. Suppose the number of cards n is even. Cut the deck into two equal piles. Drop the first card from the left pile or from the right pile according to the outcome of a fair coin flip. Then drop from the other pile. Continue this way until both piles are empty. We show that if n is a power of 2 then the mixing time of the Thorp shuffle is O(log^3 n). Previously, the best known bound was O(log^4 n).

This sequence seems to be unrelated to the Thorp shuffle in which the bound is log^3 x = (log x)^3 rather than log log log x. - Charles R Greathouse IV, Sep 04 2015

			a(1) = 2 because log(log(log(2^2))) ~ -1.1189142050548055457 whose floor is -2.
a(2) = 3 because log(log(log(2^3))) ~ -0.31183902548187902095 whose floor is -1.

Robert G. Wilson v, Table of n, a(n) for n = 1..11
Ben Morris, Improved mixing time bounds for the Thorp shuffle, arXiv:0912.2759 [math.PR], Dec 14, 2009.

Cf. A000079, A000195.

a[n_] := Ceiling[Exp[Exp[n - 3] - Log@ Log@ 2]]; Array[a, 11] (* Robert G. Wilson v, Feb 05 2013 *)

a(n)=ceil(exp(exp(n-3))/log(2)) \\ Charles R Greathouse IV, Dec 20 2011

a(n) = Min(n such that floor(log(log(log(2^n)))) > floor(log(log(log(2^(n-1)))))).

a(n) = ceiling(exp(exp(n-3)-log(log(2)))). - R. J. Mathar, Mar 31 2010

Two more terms from R. J. Mathar, Mar 31 2010

A355703 a(n) = binomial(n, floor(log(n))).

Original entry on oeis.org

1, 1, 3, 4, 5, 6, 7, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 1330, 1540, 1771, 2024, 2300, 2600, 2925, 3276, 3654, 4060, 4495, 4960, 5456, 5984, 6545, 7140, 7770, 8436, 9139, 9880, 10660, 11480, 12341, 13244, 14190, 15180, 16215, 17296, 18424, 19600, 20825

Offset: 1

Christoph B. Kassir, Jul 14 2022

nonn

Mathematics Stack Exchange, How to compute asymptotic growth of binomial(n, log(n)).

Cf. A000195, A007318.

Cf. A001405, A051033, A051036, A051052, A051053.

a:= n-> binomial(n, ilog(n)):
seq(a(n), n=1..60);  # Alois P. Heinz, Jul 31 2022

a[n_] := Binomial[n, Floor[Log[n]]]; Array[a, 50] (* Amiram Eldar, Jul 31 2022 *)

a(n) = binomial(n, floor(log(n))); \\ Michel Marcus, Jul 31 2022

from numpy import log
from math import comb, floor
for n in range(1, 50):
    x = comb(n, floor(log(n)))
    print("{}, ".format(x), end='')

A108165 a(n)=a(n-1) +A108173(n+1) -A108173(n).

Original entry on oeis.org

2, 5, 9, 12, 15, 19, 22, 26, 29, 32, 36, 39, 42, 46, 49, 53, 56, 59, 63, 66, 70, 73, 76, 80, 83, 86, 90, 93, 97, 100, 103, 107, 110, 114, 117, 120, 124, 127, 130, 134, 137, 141, 144, 147, 151, 154, 157, 161, 164, 168, 171, 174, 178, 181, 185, 188, 191, 195, 198, 201

Offset: 1

Roger L. Bagula, Jun 13 2005

nonn

Tribonacci version of A001950 using beta, the positive real root of x^3-x^2-x-1, as the constant.

Cf. A007066, A076662, A001950.

NSolve[x^3 - x^2 - x - 1 = 0, x] beta = 1.8392867552141612 a[n_] = 1 + Ceiling[(n - 1)*beta^2] (* A007066-like*) aa = Table[a[n], {n, 1, 100}] (*A076662-like*) b = Table[a[n] - a[n - 1], {n, 2, Length[aa]}] F[1] = 2; F[n_] := F[n] = F[n - 1] + b[[n]] (* A000195-like*) c = Table[F[n], {n, 1, Length[b] - 1}]

Partially edited by N. J. A. Sloane, May 04 2007

Correction of definition and offset by R. J. Mathar, Mar 12 2012

cp's OEIS Frontend

A336018 a(n) = floor(frac(log_2(n))*n), where frac denotes the fractional part.

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A003619 Not of form [ e^m ], m >= 1.

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A092919 Partial sums of A000193 (round(log(n))).

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A292621 a(n) = a(n-1) + a(floor(log(n))) with a(1) = 1, a(2) = 2.

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A004237 a(n) = floor(100*log(n)).

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A067283 Numbers n such that the number of divisors of n is floor(log(n)).

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Mathematica

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A108173 Let beta = A058265. Sequence gives a(n) = 1 + ceiling((n-1)*beta^2).

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Mathematica

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A171728 Numbers k which establish records for floor(log(log(log(2^k)))).

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