A067080 -id:A067080 - cp's OEIS frontend

A005187 a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.

0, 1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 26, 31, 32, 34, 35, 38, 39, 41, 42, 46, 47, 49, 50, 53, 54, 56, 57, 63, 64, 66, 67, 70, 71, 73, 74, 78, 79, 81, 82, 85, 86, 88, 89, 94, 95, 97, 98, 101, 102, 104, 105, 109, 110, 112, 113, 116, 117, 119, 120, 127, 128

Offset: 0

N. J. A. Sloane, May 20 1991; Allan Wilks, Dec 11 1999

Also exponent of the largest power of 2 dividing (2n)! (A010050) and (2n)!! (A000165).
Write n in binary: 1ab..yz, then a(n) = 1ab..yz + ... + 1ab + 1a + 1. - Ralf Stephan, Aug 27 2003
Also numbers having a partition into distinct Mersenne numbers > 0; A079559(a(n))=1; complement of A055938. - Reinhard Zumkeller, Mar 18 2009
Wikipedia's article on the "Whitney Immersion theorem" mentions that the a(n)-dimensional sphere arises in the Immersion Conjecture proved by Ralph Cohen in 1985. - Jonathan Vos Post, Jan 25 2010
For n > 0, denominators for consecutive pairs of integral numerator polynomials L(n+1,x) for the Legendre polynomials with o.g.f. 1 / sqrt(1-tx+x^2). - Tom Copeland, Feb 04 2016
a(n) is the total number of pointers in the first n elements of a perfect skip list. - Alois P. Heinz, Dec 14 2017
a(n) is the position of the n-th a (indexing from 0) in the fixed point of the morphism a -> aab, b -> b. - Jeffrey Shallit, Dec 24 2020
Numbers that can be expressed as the sum of distinct numbers of the form 2^k - 1 (lenient Mersenne numbers, A000225). This follows from the 2N - Hamming weight definition. A corollary is that these are the numbers with no 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 25 2025

			G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 1000 terms from T. D. Noe)
Jean-Paul Allouche, Jean Bétréma, and Jeffrey Shallit, Sur des points fixes de morphismes d'un monoïde libre, RAIRO-Theor. Inf. Appl. 23 (1989), 235-249.
Laurent Alonso, Edward M. Reingold, and René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
Laurent Alonso, Edward M. Reingold, and René Schott, The average-case complexity of determining the majority, SIAM J. Comput. 26 (1997), no. 1, 1-14.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
Sung-Hyuk Cha, On Integer Sequences Derived from Balanced k-ary Trees, Applied Mathematics in Electrical and Computer Engineering, 2012.
Sung-Hyuk Cha, On Complete and Size Balanced k-ary Tree Integer Sequences, International Journal of Applied Mathematics and Informatics, Issue 2, Volume 6, 2012, pp. 67-75. - From _N. J. A. Sloane_, Dec 24 2012
Ralph L. Cohen, The Immersion Conjecture for Differentiable Manifolds, The Annals of Mathematics, 1985: 237-328.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, preprint, 2016.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Keith Johnson and Kira Scheibelhut, Rational Polynomials That Take Integer Values at the Fibonacci Numbers, American Mathematical Monthly 123.4 (2016): 338-346. See p. 340.
Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.
Anunay Kulshrestha, On Hamming Distance between base-n representations of whole numbers, arXiv:1203.4547 [cs.DM], 2012.
Michael E. Saks and Michael Werman, On computing majority by comparisons, Combinatorica 11 (1991), no. 4, 383-387.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Wikipedia, Whitney Immersion Theorem.
Allan Wilks, Email to N. J. A. Sloane, Jul 7 1988.

Cf. A001790, A011371, A067080, A098844, A132027, A004128, A054899, A046161.
Cf. A001511 (first differences), A122247 (partial sums), A055938 (complement).
Cf. A004134, A010050, A000165.
Cf. A000120, A079559, A085354, A030308, A000225.

a005187 n = a005187_list !! n
a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
-- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011

[n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019

A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014

a[0] = 0; a[n_] := a[n] = a[Floor[n/2]] + n; Table[ a[n], {n, 0, 50}] (* or *)
Table[IntegerExponent[(2n)!, 2], {n, 0, 65}] (* Robert G. Wilson v, Apr 19 2006 *)
Table[2n-DigitCount[2n,2,1],{n,0,70}] (* Harvey P. Dale, May 24 2014 *)

{a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */

a(n)=my(s=n);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Apr 07 2012

a(n)=2*n-hammingweight(n) \\ Charles R Greathouse IV, Jan 07 2013

def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021

@CachedFunction
def A005187(n): return A005187(n//2) + n if n > 0 else 0
[A005187(n) for n in range(66)]  # Peter Luschny, Dec 13 2012

a(n) = A011371(2n+1) = A011371(n) + n, n >= 0.
A046161(n) = 2^a(n).
For m>0, let q = floor(log_2(m)); a(2m+1) = 2^q + 3m + Sum_{k>=1} floor((m-2^q)/2^k); a(2m) = a(2m+1) - 1. - Len Smiley
a(n) = Sum_{k >= 0} floor(n/2^k) = n + A011371(n). - Henry Bottomley, Jul 03 2001
G.f.: A(x) = Sum_{k>=0} x^(2^k)/((1-x)*(1-x^(2^k))). - Ralf Stephan, Dec 24 2002
a(n) = Sum_{k=1..n} A001511(k), sum of binary Hamming distances between consecutive integers up to n. - Gary W. Adamson, Jun 15 2003
Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003 [true as a(n) = 2*n - hamming_weight(2*n). Joerg Arndt, Jun 10 2019]
Sum_{n=2^k..2^(k+1)-1} a(n) = 3*4^k - (k+4)*2^(k-1) = A085354(k). - Philippe Deléham, Feb 19 2004
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence: a(n) = n + a(floor(n/2)); a(2n) = 2n + a(n); a(n*2^m) = 2*n*(2^m-1) + a(n).
  a(2^m) = 2^(m+1) - 1, m >= 0.
Asymptotic behavior: a(n) = 2n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= 2n-1; equality holds for powers of 2.
a(n) >= 2n-1-floor(log_2(n)); equality holds for n = 2^m-1, m > 0.
lim inf (2n - a(n)) = 1, for n-->oo.
lim sup (2n - log_2(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 1, for n-->oo. (End)
a(n) = 2n - A000120(n). - Paul Barry, Oct 26 2007
PURRS demo results: Bounds for a(n) = n + a(n/2) with initial conditions a(1) = 1: a(n) >= -2 + 2*n - log(n)*log(2)^(-1), a(n) <= 1 + 2*n for each n >= 1. - Alexander R. Povolotsky, Apr 06 2008
If n = 2^a_1 + 2^a_2 + ... + 2^a_k, then a(n) = n-k. This can be used to prove that 2^n maximally divides (2n!)/n!. - Jon Perry, Jul 16 2009
a(n) = Sum_{k>=0} A030308(n,k)*A000225(k+1). - Philippe Deléham, Oct 16 2011
a(n) = log_2(denominator(binomial(-1/2,n))). - Peter Luschny, Nov 25 2011
a(2n+1) = a(2n) + 1. - M. F. Hasler, Jan 24 2015
a(n) = A004134(n) - n. - Cyril Damamme, Aug 04 2015
G.f.: (1/(1 - x))*Sum_{k>=0} (2^(k+1) - 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

A011371 a(n) = n minus (number of 1's in binary expansion of n). Also highest power of 2 dividing n!.

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 22, 22, 23, 23, 25, 25, 26, 26, 31, 31, 32, 32, 34, 34, 35, 35, 38, 38, 39, 39, 41, 41, 42, 42, 46, 46, 47, 47, 49, 49, 50, 50, 53, 53, 54, 54, 56, 56, 57, 57, 63, 63, 64, 64, 66, 66, 67, 67, 70

Offset: 0

N. J. A. Sloane

Terms of A005187 repeated. - Lekraj Beedassy, Jul 06 2004
This sequence shows why in binary 0 and 1 are the only two numbers n such that n equals the sum of its digits raised to the consecutive powers (equivalent to the base-10 sequence A032799). 1 raised to any consecutive power is still 1 and thus any sum of digits raised to consecutive powers for any n > 1 falls short of equaling the value of n by the n-th term of this sequence. - Alonso del Arte, Jul 27 2004
Also the number of trailing zeros in the base-2 representation of n!. - Hieronymus Fischer, Jun 18 2007
Partial sums of A007814. - Philippe Deléham, Jun 21 2012
If n is in A089633 and n > 0, then a(n) = n - floor(log_2(n+1)). - Douglas Latimer, Jul 25 2012
For n > 1, denominators of integral numerator polynomials L(n,x) for the Legendre polynomials with o.g.f. 1/sqrt(1 - t*x + x^2). - Tom Copeland, Feb 04 2016
The definition of this sequence explains why, for n > 1, the highest power of 2 dividing n! added to the number of 1's in the binary expansion of n is equal to n. This result is due to the French mathematician Adrien Legendre (1752-1833) [see the Honsberger reference]. - Bernard Schott, Apr 07 2017
a(n) is the total number of 2's in the prime factorizations over the first n positive integers. The expected number of 2's in the factorization of an integer n is 1 (as n->infinity). Generally, the expected number of p's (for a prime p) is 1/(p-1). - Geoffrey Critzer, Jun 05 2017

			a(3) = 1 because 3 in binary is 11 (two 1's) and 3 - 2 = 1.
a(4) = 3 because 4 in binary is 100 (one 1 and two 0's) and 4 - 1 = 3.
a(5) = 3 because 5 in binary is 101 (a zero between two 1's) and 5 - 2 = 3.
a(100) = 97.
a(10^3) = 994.
a(10^4) = 9995.
a(10^5) = 99994.
a(10^6) = 999993.
a(10^7) = 9999992.
a(10^8) = 99999988.
a(10^9) = 999999987.
G.f. = x^2 + x^3 + 3*x^4 + 3*x^5 + 4*x^6 + 4*x^7 + 7*x^8 + 7*x^9 + 8*x^10 + ...

K. Atanassov, On Some of Smarandache's Problems, section 7, on the 61st problem, page 42, American Research Press, 1999, 16-21.
G. Bachman, Introduction to p-Adic Numbers and Valuation Theory, Academic Press, 1964; see Lemma 3.1.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 305.
H. Davenport, The Higher Arithmetic, 7th ed. 1999, Cambridge University Press, p. 216, exercise 1.07.
R. Honsberger, Mathematical Gems II, Dolciani Mathematical Expositions, 1976, pp. 1-6.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
Laurent Alonso, Edward M. Reingold, and René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
Laurent Alonso, Edward M. Reingold, and René Schott, The average-case complexity of determining the majority, SIAM J. Comput. 26 (1997), no. 1, 1-14.
Sung-Hyuk Cha, On Integer Sequences Derived from Balanced k-ary Trees, Applied Mathematics in Electrical and Computer Engineering, 2012.
Sung-Hyuk Cha, On Complete and Size Balanced k-ary Tree Integer Sequences, International Journal of Applied Mathematics and Informatics, Issue 2, Volume 6, 2012, pp. 67-75. - From _N. J. A. Sloane_, Dec 24 2012
R. Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) 463-535, doi, Section 4.4.
Keith Johnson and Kira Scheibelhut, Rational polynomials that take integer values at the Fibonacci numbers, American Mathematical Monthly 123.4 (2016): 338-346. See omega_2.
S-C Liu and J. C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, eq. (5).
K. Matthews, Computing the prime-power factorization of n!.
A. Mir, F. Rossello and L. Rotger, A new balance index for phylogenetic trees, arXiv preprint arXiv:1202.1223 [q-bio.PE], 2012.
A. M. Oller-Marcen and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8.
Michael E. Saks and Michael Werman, On computing majority by comparisons, Combinatorica 11 (1991), no. 4, 383-387.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions.
Zhujun Zhang, A Note on Counting Binomial Heaps, ResearchGate (2019).

Cf. A000120, A005187, A054861, A032799, A067080, A098844, A132027.

a(n) = Sum_{k=1..n} A007814(k), n >= 1, a(0) = 0.

a011371 n = n - a000120 n  -- Reinhard Zumkeller, Jan 24 2014

[Valuation(Factorial(n), 2): n in [0..80]]; // Bruno Berselli, Aug 05 2013

A011371(n) = RETURN(((2^(l))-1)+sum('(j*floor((n-(2^l)+2^j)/(2^(j+1))))','j'=1..l)); # after K. Atanassov. Here l is [ log2(n) ].
A011371 := n -> n - add(i,i=convert(n,base,2)): # Peter Luschny, May 02 2009
read("transforms") : A011371 := proc(n) n-wt(n) ; end proc: # R. J. Mathar, May 15 2013

-1 + Length[ Last[ Split[ IntegerDigits[ 2(n!), 2 ] ] ] ], FoldList[ Plus, 0, Fold[ Flatten[ {#1, #2, #1} ]&, 0, Range[ 6 ] ] ]
Table[IntegerExponent[n!, 2], {n, 0, 127}]
Table[n - DigitCount[n, 2, 1], {n, 0, 127}]
Table[t = 0; p = 2; While[s = Floor[n/p]; t = t + s; s > 0, p *= 2]; t, {n, 0, 100} ]

{a(n) = if( n<0, 0, valuation(n!, 2))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, sum(k=1, n, n\2^k))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, n - subst( Pol( binary( n ) ), x, 1))}; /* Michael Somos, Aug 28 2007 */

a(n)=sum(k=1,log(n+1)\log(2),n>>k) \\ Charles R Greathouse IV, Oct 03 2012

a(n)=my(s);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Aug 09 2013

a(n) = n - hammingweight(n); \\ Michel Marcus, Jun 05 2014

[n - bin(n)[2:].count("1") for n in range(101)] # Indranil Ghosh, Apr 09 2017

# 3.10+
def A011371(n): return n-n.bit_count() # Chai Wah Wu, Jul 09 2022

a(n) = a(floor(n/2)) + floor(n/2) = floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ... - Henry Bottomley, Apr 24 2001
G.f.: A(x) = (1/(1 - x))*Sum_{k>=1} x^(2^k)/(1 - x^(2^k)). - Ralf Stephan, Apr 11 2002
a(n) = n - A000120(n). - Lekraj Beedassy, Sep 01 2003
a(n) = A005187(n) - n, n >= 0.
a(n) = A007814(A000142(n)). - Reinhard Zumkeller, Apr 09 2004
From Hieronymus Fischer, Jun 25 and Aug 13 2007: (Start)
a(n) = Sum_{k=2..n} Sum_{j|k, j >= 2} (floor(log_2(j)) - floor(log_2(j - 1))).
The g.f. can be expressed in terms of a Lambert series, in that g(x) = L[b(k)](x)/(1 - x), where
L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1 - x^k) is a Lambert series with b(k) = 1, if k is a power of 2, otherwise b(k) = 0.
G.f.: g(x) = (1/(1-x))*Sum_{k>0} c(k)*x^k, where c(k) = Sum_{j>1, j|k} (floor(log_2(j)) - floor(log_2(j-1))).
Recurrence:
a(n) = floor(n/2) + a(floor(n/2));
a(2*n) = n + a(n);
a(n*2^m) = n*(2^m - 1) + a(n).
a(2^m) = 2^m - 1, m >= 0.
Asymptotic behavior:
a(n) = n + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= n - 1; equality holds for powers of 2.
a(n) >= n - 1 - floor(log_2(n)); equality holds for n = 2^m - 1, m > 0.
lim inf (n - a(n)) = 1, for n->oo.
lim sup (n - log_2(n) - a(n)) = 0, for n->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 0, for n->oo. (End)
a(n) = Sum_{k >= 0} A030308(n, k)*A000225(k). - Philippe Deléham, Oct 16 2011
a(n) = Sum_{k=0..floor(log_2(n+1))} f^(k+1)(n), where f(n) = (n - (n mod 2))/2 and f^(k+1) is the (k+1)-th composition of f. - Joseph Wheat, Mar 01 2018
a(n) = Sum_{k=1..floor(n/2)} floor(log_2(n/k)). - Ammar Khatab, Feb 01 2025

Examples added by Hieronymus Fischer, Jun 06 2012

A048651 Decimal expansion of Product_{k >= 1} (1 - 1/2^k).

Original entry on oeis.org

2, 8, 8, 7, 8, 8, 0, 9, 5, 0, 8, 6, 6, 0, 2, 4, 2, 1, 2, 7, 8, 8, 9, 9, 7, 2, 1, 9, 2, 9, 2, 3, 0, 7, 8, 0, 0, 8, 8, 9, 1, 1, 9, 0, 4, 8, 4, 0, 6, 8, 5, 7, 8, 4, 1, 1, 4, 7, 4, 1, 0, 6, 6, 1, 8, 4, 9, 0, 2, 2, 4, 0, 9, 0, 6, 8, 4, 7, 0, 1, 2, 5, 7, 0, 2, 4, 2, 8, 4, 3, 1, 9, 3, 3, 4, 8, 0, 7, 8, 2

Offset: 0

N. J. A. Sloane

This is the limiting probability that a large random binary matrix is nonsingular (cf. A002884).
This constant is very close to 2^(13/24) * sqrt(Pi/log(2)) / exp(Pi^2/(6*log(2))) = 0.288788095086602421278899775042039398383022429351580356839... - Vaclav Kotesovec, Aug 21 2018
This constant is irrational (see Penn link). - Paolo Xausa, Dec 09 2024

			(1/2)*(3/4)*(7/8)*(15/16)*... = 0.288788095086602421278899721929230780088911904840685784114741...

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 318, 354-361.

Harry J. Smith, Table of n, a(n) for n = 0..20000
Steven R. Finch, Digital Search Tree Constants. [Broken link]
Steven R. Finch, Digital Search Tree Constants. [From the Wayback machine]
Marvin Geiselhart, Ahmed Elkelesh, Moustafa Ebada, Sebastian Cammerer, and Stephan ten Brink, On the Automorphism Group of Polar Codes, arXiv:2101.09679 [cs.IT], 2021.
T. S. Jayram, Hellinger strikes back: a note on the multi-party information complexity of AND, LNCS 5687 (2009) 562-573.
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.
Victor S. Miller, Counting Matrices that are Squares, arXiv:1606.09299 [math.GR], 2016.
Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.
Michael Penn, A non-standard irrationality proof, YouTube video, 2024.
V. Arvind Rameshwar, Shreyas Jain, and Navin Kashyap, Sampling-Based Estimates of the Sizes of Constrained Subcodes of Reed-Muller Codes, arXiv:2309.08907 [cs.IT], 2023.
László Tóth, Alternating sums concerning multiplicative arithmetic functions, arXiv preprint arXiv:1608.00795 [math.NT], 2016.
Eric Weisstein's World of Mathematics, Infinite Product.
Eric Weisstein's World of Mathematics, Tree Searching.
Wikipedia, Pentagonal number theorem.
Wanchen Zhang, Yu Ning, Fei Shi, and Xiande Zhang, Extremal Maximal Entanglement, arXiv:2411.12208 [quant-ph], 2024. See p. 15.

Cf. A002884, A001318, A005327, A005329, A048652, A065446, A079555, A098844, A067080, A100220, A132019, A132020, A132026, A132038, A070933, A261584, A335764.

RealDigits[ Product[1 - 1/2^i, {i, 100}], 10, 111][[1]] (* Robert G. Wilson v, May 25 2011 *)
RealDigits[QPochhammer[1/2], 10, 100][[1]] (* Jean-François Alcover, Nov 18 2015 *)

default(realprecision, 20080); x=prodinf(k=1, -1/2^k, 1); x*=10; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b048651.txt", n, " ", d)); \\ Harry J. Smith, May 07 2009

exp(-Sum_{k>0} sigma_1(k)/k*2^(-k)) = exp(-Sum_{k>0} A000203(k)/k*2^(-k)). - Hieronymus Fischer, Jul 28 2007
From Hieronymus Fischer, Aug 13 2007: (Start)
Equals lim inf Product_{k=0..floor(log_2(n))} floor(n/2^k)*2^k/n for n->oo.
Equals lim inf A098844(n)/n^(1+floor(log_2(n)))*2^(1/2*(1+floor(log_2(n)))*floor(log_2(n))) for n->oo.
Equals lim inf A098844(n)/n^(1+floor(log_2(n)))*2^A000217(floor(log_2(n))) for n->oo.
Equals lim inf A098844(n)/(n+1)^((1+log_2(n+1))/2) for n->oo.
Equals (1/2)*exp(-Sum_{n>0} 2^(-n)*Sum_{k|n} 1/(k*2^k)). (End)
Limit of A177510(n)/A000079(n-1) as n->infinity (conjecture). - Mats Granvik, Mar 27 2011
Product_{k >= 1} (1-1/2^k) = (1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 27 2015
exp(Sum_{n>=1}(1/n/(1 - 2^n))) (according to Mathematica). - Mats Granvik, Sep 07 2016
(Sum_{k>0} (4^k-1)/(Product_{i=1..k} ((4^i-1)*(2*4^i-1))))*2 = 2/7 + 2/(3*7*31) + 2/(3*7*15*31*127)+2/(3*7*15*31*63*127*511) + ... (conjecture). - Werner Schulte, Dec 22 2016
Equals Sum_{k=-oo..oo} (-1)^k/2^((3*k+1)*k/2) (by Euler's pentagonal number theorem). - Amiram Eldar, Aug 13 2020
From Peter Bala, Dec 15 2020: (Start)
Constant C = Sum_{n >= 0} (-1)^n/( Product_{k = 1..n} (2^k - 1) ). The above conjectural result by Schulte follows by adding terms of this series in pairs.
C = (1/2)*Sum_{n >= 0} (-1/2)^n/( Product_{k = 1..n} (2^k - 1) ).
C = (3/8)*Sum_{n >= 0} (-1/4)^n/( Product_{k = 1..n} (2^k - 1) ).
1/C = Sum_{n >= 0} 2^(n*(n-1)/2)/( Product_{k = 1..n} (2^k - 1) ).
C = 1 - Sum_{n >= 0} (1/2)^(n+1)*Product_{k = 1..n} (1 - 1/2^k).
This latter identity generalizes as:
C = Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
3*C = 1 - Sum_{n >= 0} (1/8)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
3*7*C = 6 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
3*7*15*C = 91 - Sum_{n >= 0} (1/32)^(n+1)*Product_{k = 1..n} (1 - 1/2^k),
and so on, where the sequence [1, 0, 1, 6, 91, ...] is A005327.
(End)
From Amiram Eldar, Feb 19 2022: (Start)
Equals sqrt(2*Pi/log(2)) * exp(log(2)/24 - Pi^2/(6*log(2))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(2))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A005329(n).
Equals exp(-A335764). (End)
Equals 1/A065446. - Hugo Pfoertner, Nov 23 2024

Corrected by Hieronymus Fischer, Jul 28 2007

A027868 Number of trailing zeros in n!; highest power of 5 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 19

Offset: 0

Warut Roonguthai

Also the highest power of 10 dividing n! (different from A054899). - Hieronymus Fischer, Jun 18 2007
Alternatively, a(n) equals the expansion of the base-5 representation A007091(n) of n (i.e., where successive positions from right to left stand for 5^n or A000351(n)) under a scale of notation whose successive positions from right to left stand for (5^n - 1)/4 or A003463(n); for instance, n = 7392 has base-5 expression 2*5^5 + 1*5^4 + 4*5^3 + 0*5^2 + 3*5^1 + 2*5^0, so that a(7392) = 2*781 + 1*156 + 4*31 + 0*6 + 3*1 + 2*0 = 1845. - Lekraj Beedassy, Nov 03 2010
Partial sums of A112765. - Hieronymus Fischer, Jun 06 2012
Also the number of trailing zeros in A000165(n) = (2*n)!!. - Stefano Spezia, Aug 18 2024

			a(100)  = 24.
a(10^3) = 249.
a(10^4) = 2499.
a(10^5) = 24999.
a(10^6) = 249998.
a(10^7) = 2499999.
a(10^8) = 24999999.
a(10^9) = 249999998.
a(10^n) = 10^n/4 - 3 for 10 <= n <= 15 except for a(10^14) = 10^14/4 - 2. - _M. F. Hasler_, Dec 27 2019

M. Gardner, "Factorial Oddities." Ch. 4 in Mathematical Magic Show: More Puzzles, Games, Diversions, Illusions and Other Mathematical Sleight-of-Mind from Scientific American. New York: Vintage, 1978, pp. 50-65.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
David S. Hart, James E. Marengo, Darren A. Narayan, and David S. Ross, On the number of trailing zeros in n!, College Math. J., 39(2):139-145, 2008.
Enrique Pérez Herrero, Trailing Zeros in n!, Psychedelic Geometry Blogspot.
Soichi Ikeda and Kaneaki Matsuoka, On transcendental numbers generated by certain integer sequences, Siauliai Math. Semin., 8 (16) 2013, 63-69.
Shyam Sunder Gupta, Fascinating Factorials, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 16, 411-442.
Shu-Chung Liu and Jean C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, eq (5).
Antonio M. Oller-Marcén, A new look at the trailing zeros of n!, arXiv:0906.4868v1 [math.NT], 2009.
Antonio M. Oller-Marcén and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8
Eric Weisstein's World of Mathematics, Factorial.
Index entries for sequences related to factorial numbers

See A000966 for the missing numbers. See A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A054899, A007953, A112765, A067080, A098844, A132027, A067080, A098844, A132029, A054999, A112765, A191610, A000351.
Cf. also A000142, A004154.
Cf. A000165, A008904.

a027868 n = sum $ takeWhile (> 0) $ map (n `div`) $ tail a000351_list
-- Reinhard Zumkeller, Oct 31 2012

[Valuation(Factorial(n), 5): n in [0..80]]; // Bruno Berselli, Oct 11 2021

0, seq(add(floor(n/5^i),i=1..floor(log[5](n))), n=1..100); # Robert Israel, Nov 13 2014

Table[t = 0; p = 5; While[s = Floor[n/p]; t = t + s; s > 0, p *= 5]; t, {n, 0, 100} ]
Table[ IntegerExponent[n!], {n, 0, 80}] (* Robert G. Wilson v *)
zOF[n_Integer?Positive]:=Module[{maxpow=0},While[5^maxpow<=n,maxpow++];Plus@@Table[Quotient[n,5^i],{i,maxpow-1}]]; Attributes[zOF]={Listable}; Join[{0},zOF[ Range[100]]] (* Harvey P. Dale, Apr 11 2022 *)

a(n)={my(s);while(n\=5,s+=n);s} \\ Charles R Greathouse IV, Nov 08 2012, edited by M. F. Hasler, Dec 27 2019

a(n)=valuation(n!,5) \\ Charles R Greathouse IV, Nov 08 2012

apply( A027868(n)=(n-sumdigits(n,5))\4, [0..99]) \\ M. F. Hasler, Dec 27 2019

from sympy import multiplicity
A027868, p5 = [0,0,0,0,0], 0
for n in range(5,10**3,5):
    p5 += multiplicity(5,n)
    A027868.extend([p5]*5) # Chai Wah Wu, Sep 05 2014

def A027868(n): return 0 if n<5 else n//5 + A027868(n//5) # David Radcliffe, Jun 26 2016

from sympy.ntheory.factor_ import digits
def A027868(n): return n-sum(digits(n,5)[1:])>>2 # Chai Wah Wu, Oct 18 2024

a(n) = Sum_{i>=1} floor(n/5^i).
a(n) = (n - A053824(n))/4.
From Hieronymus Fischer, Jun 25 2007 and Aug 13 2007, edited by M. F. Hasler, Dec 27 2019: (Start)
G.f.: g(x) = Sum_{k>0} x^(5^k)/(1-x^(5^k))/(1-x).
a(n) = Sum_{k=5..n} Sum_{j|k, j>=5} (floor(log_5(j)) - floor(log_5(j-1))).
G.f.: g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k) = 1, if k>1 is a power of 5, else b(k) = 0.
G.f.: g(x) = Sum_{k>0} c(k)*x^k/(1-x), where c(k) = Sum_{j>1, j|k} floor(log_5(j)) - floor(log_5(j - 1)).
Recurrence:
a(n) = floor(n/5) + a(floor(n/5));
a(5*n) = n + a(n);
a(n*5^m) = n*(5^m-1)/4 + a(n).
a(k*5^m) = k*(5^m-1)/4, for 0 <= k < 5, m >= 0.
Asymptotic behavior:
a(n) = n/4 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n-1)/4; equality holds for powers of 5.
a(n) >= n/4 - 1 - floor(log_5(n)); equality holds for n = 5^m-1, m > 0.
lim inf (n/4 - a(n)) = 1/4, for n -> oo.
lim sup (n/4 - log_5(n) - a(n)) = 0, for n -> oo.
lim sup (a(n+1) - a(n) - log_5(n)) = 0, for n -> oo.
(End)
a(n) <= A027869(n). - Reinhard Zumkeller, Jan 27 2008
10^a(n) = A000142(n) / A004154(n). - Reinhard Zumkeller, Nov 24 2012
a(n) = Sum_{k=1..floor(n/2)} floor(log_5(n/k)). - Ammar Khatab, Feb 01 2025

Examples added by Hieronymus Fischer, Jun 06 2012

A054899 a(n) = Sum_{k>0} floor(n/10^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11, 11, 11

Offset: 0

Henry Bottomley, May 23 2000

nonn

The old definition of this sequence was "Highest power of 10 dividing n!", but that is wrong (see A027868). For example, the highest power of 10 dividing 5!=120 is 1; however, a(5)=0. - Hieronymus Fischer, Jun 18 2007
Highest power of 10 dividing the quotient of multifactorials Product_{k>=1} M(10^k, 10^k*floor(n/10^k)) /( Product_{k>=1} M(10^(k-1), 10^(k-1) * floor(n/10^k)) ) where M(r,s) is the r-multifactorial (r>=1) of s which is defined by M(r,s) = s*M(r,s-r) for s > 0, M(r,s) = 1 for s <= 0. This is because that quotient of multifactorials evaluates to the product 10^floor(n/10)*10^floor(n/100)*... - Hieronymus Fischer, Jun 14 2007
Partial sums of A122840. - Hieronymus Fischer, Jun 06 2012
Called the "terminating nines function" by Kennedy et al. (1989). a(n) is the number of terminating nines which occur up to n but not including n. - Amiram Eldar, Sep 06 2024

			          a(11) = 1
         a(111) = 12.
        a(1111) = 123.
       a(11111) = 1234.
      a(111111) = 12345.
     a(1111111) = 123456.
    a(11111111) = 1234567.
   a(111111111) = 12345678.
  a(1111111111) = 123456789.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000
Robert K. Kennedy, Curtis N. Cooper, Vladimir Drobot, and Fred Hickling, On the natural density of the range of the terminating nines function, International Journal of Mathematics and Mathematical Sciences, Vol. 12, No. 4 (1989), pp. 805-808.
Eric Weisstein's World of Mathematics, Multifactorial.

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
Different from the highest power of 10 dividing n! (see A027868 for reference).
Cf. A007953, A027868, A067080, A098844, A132027, A122840.

m:=10;
function a(n) // a = A054899, m = 10
  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..103]]; // G. C. Greubel, Apr 28 2023

Table[t=0; p=10; While[s=Floor[n/p]; t=t+s; s>0, p*=10]; t, {n,0,100}]

a(n)=my(s);while(n\=10,s+=n);s \\ Charles R Greathouse IV, Jul 19 2011

m=10 # a = A054899
def a(n): return 0 if (n==0) else a(n//m) + (n//m)
[a(n) for n in range(104)] # G. C. Greubel, Apr 28 2023

a(n) = floor(n/10) + floor(n/100) + floor(n/1000) + ...
a(n) = (n - A007953(n))/9.
From Hieronymus Fischer, Jun 14 2007, Jun 25 2007, and Aug 13 2007: (Start)
a(n) = Sum_{k>0} floor(n/10^k).
a(n) = Sum_{k=10..n} Sum_{j|k, j>=10} ( floor(log_10(j)) -floor(log_10(j-1)) ).
G.f.: g(x) = ( Sum_{k>0} x^(10^k)/(1-x^(10^k)) )/(1-x).
G.f. expressed in terms of Lambert series:
g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k)=1, if k>1 is a power of 10, else b(k)=0.
G.f.: g(x) = ( Sum_{k>0} c(k)*x^k )/(1-x), where c(k) = Sum_{j>1, j|k} (floor(log_10(j)) - floor(log_10(j-1)) ).
a(n) = Sum_{k=0..floor(log_10(n))} ds_10(floor(n/10^k))*10^k - n where ds_10(x) = digital sum of x in base 10.
a(n) = Sum_{k=0..floor(log_10(n))} A007953(floor(n/10^k))*10^k - n.
Recurrence:
a(n) = floor(n/10) + a(floor(n/10)).
a(10*n) = n + a(n).
a(n*10^m) = n*(10^m-1)/9 + a(n).
a(k*10^m) = k*(10^m-1)/9, for 0 <= k < 10, m >= 0.
Asymptotic behavior:
a(n) = n/9 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n - 1)/9; equality holds for powers of 10.
a(n) >= n/9 - 1 - floor(log_10(n)); equality holds for n=10^m-1, m>0.
lim inf (n/9 - a(n)) = 1/9, for n --> oo.
lim sup (n/9 - log_10(n) - a(n)) = 0, for n --> oo.
lim sup (a(n+1) - a(n) - log_10(n)) = 0, for n --> oo. (End)

An incorrect g.f. was deleted by N. J. A. Sloane, Sep 13 2009

Examples added by Hieronymus Fischer, Jun 06 2012

A054861 Greatest k such that 3^k divides n!.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 2, 2, 4, 4, 4, 5, 5, 5, 6, 6, 6, 8, 8, 8, 9, 9, 9, 10, 10, 10, 13, 13, 13, 14, 14, 14, 15, 15, 15, 17, 17, 17, 18, 18, 18, 19, 19, 19, 21, 21, 21, 22, 22, 22, 23, 23, 23, 26, 26, 26, 27, 27, 27, 28, 28, 28, 30, 30, 30, 31, 31, 31, 32, 32, 32, 34, 34, 34, 35, 35

Offset: 0

Henry Bottomley, May 22 2000

Also the number of trailing zeros in the base-3 representation of n!. - Hieronymus Fischer, Jun 18 2007
Also the highest power of 6 dividing n!. - Hieronymus Fischer, Aug 14 2007
A column of A090622. - Alois P. Heinz, Oct 05 2012
The 'missing' values are listed in A096346. - Stanislav Sykora, Jul 16 2014

			a(100) = 48.
a(10^3) = 498.
a(10^4) = 4996.
a(10^5) = 49995.
a(10^6) = 499993.
a(10^7) = 4999994.
a(10^8) = 49999990.
a(10^9) = 499999993.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
S-C Liu and J. C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, Eq. (5).
A. M. Oller-Marcen and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8

Cf. A011371 (for analog involving powers of 2). See also A027868.
Cf. A004128 (for a(3n)).
Cf. A000142, A007949, A053735, A054895, A054899, A067080, A096396, A098844, A132027.

[Valuation(Factorial(n), 3): n in [0..80]]; // Bruno Berselli, Aug 05 2013

A054861 := proc(n)
    (n - convert(convert(n, base, 3), `+`))/2 ;
end proc:
seq(A054861(n),n=0..1000) ; # Robert Israel, Jul 17 2014

(Plus@@Floor[#/3^Range[Length[IntegerDigits[#,3]]-1]]&)/@Range[0,100] (* Peter J. C. Moses, Apr 07 2012 *)
FoldList[Plus, 0, IntegerExponent[Range[100], 3]] (* T. D. Noe, Apr 10 2012 *)
Table[IntegerExponent[n!,3],{n,0,80}] (* Harvey P. Dale, Feb 05 2015 *)

a(n)=my(s);while(n\=3,s+=n);s \\ Charles R Greathouse IV, Jul 25 2011

a(n)=(n - vecsum(digits(n,3)))\2; \\ Gheorghe Coserea, Jan 01 2018

def A054861(n):
    A004128 = lambda n: A004128(n//3) + n if n > 0 else 0
    return A004128(n//3)
[A054861(i) for i in (0..76)]  # Peter Luschny, Nov 16 2012

a(n) = floor(n/3) + floor(n/9) + floor(n/27) + floor(n/81) + ... .
a(n) = (n - A053735(n))/2.
a(n+1) = Sum_{k=1..n} A007949(k). - Benoit Cloitre, Mar 24 2002
From Hieronymus Fischer, Jun 18, Jun 25 and Aug 14 2007: (Start)
G.f.: (1/(1-x))*Sum_{k>0} x^(3^k)/(1-x^(3^k)).
a(n) = Sum_{k=3..n} Sum_{j>=3, j|k} (floor(log_3(j)) - floor(log_3(j-1))).
G.f.: L[b(k)](x)/(1-x), where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k) = 1, if k>1 is a power of 3, otherwise b(k)=0.
G.f.: (1/(1-x))*Sum_{k>0} c(k)*x^k, where c(k) = Sum_{j>1, j|k} (floor(log_3(j)) - floor(log_3(j-1))).
Recurrence:
a(n) = floor(n/3) + a(floor(n/3));
a(3*n) = n + a(n);
a(n*3^m) = n*(3^m-1)/2 + a(n).
a(k*3^m) = k*(3^m-1)/2, for 0 <= k < 3, m >= 0.
Asymptotic behavior:
a(n) = n/2 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/2; equality holds for powers of 3.
a(n) >= (n-2)/2 - floor(log_3(n)); equality holds for n = 3^m - 1, m > 0.
lim inf (n/2 - a(n)) = 1/2 for n->oo.
lim sup (n/2 - log_3(n) - a(n)) = 0 for n->oo.
lim sup (a(n+1) - a(n) - log_3(n)) = 0 for n->oo. (End)
a(n) = A007949(n!). - R. J. Mathar, Sep 03 2016
From R. J. Mathar, Jul 08 2021: (Start)
a(n) = A122841(n!).
Partial sums of A007949. (End)
a(n) = A007949(A000142(n)). - David A. Corneth, Nov 02 2023

Examples added by Hieronymus Fischer, Jun 06 2012

New name by David A. Corneth, Nov 02 2023

A098844 a(1)=1, a(n) = n*a(floor(n/2)).

Original entry on oeis.org

1, 2, 3, 8, 10, 18, 21, 64, 72, 100, 110, 216, 234, 294, 315, 1024, 1088, 1296, 1368, 2000, 2100, 2420, 2530, 5184, 5400, 6084, 6318, 8232, 8526, 9450, 9765, 32768, 33792, 36992, 38080, 46656, 47952, 51984, 53352, 80000, 82000, 88200, 90300

Offset: 1

Benoit Cloitre, Nov 03 2004

nonn

			a(10) = floor(10/2^0)*floor(10/2^1)*floor(10/2^2)*floor(10/2^3) = 10*5*2*1 = 100;
a(17) = 1088 since 17 = 10001(base 2) and so a(17) = 10001*1000*100*10*1(base 2) = 17*8*4*2*1 = 1088.

Hieronymus Fischer, Table of n, a(n) for n = 1..1000

Cf. A048651, A067080, A132027, A132028, A132029, A132030, A132019, A132026, A132038.
For formulas regarding a general parameter p (i.e., terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=3 to p=12 see A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

lst={};Do[p=n;s=1;While[p>1,p=IntegerPart[p/2];s*=p;];AppendTo[lst,s],{n,1,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, Jul 28 2009 *)

PARI
```
a(n)=if(n<2,1,n*a(floor(n/2)))
```

from math import prod
def A098844(n): return n*prod(n//2**k for k in range(1,n.bit_length()-1)) # Chai Wah Wu, Jun 07 2022

a(2^n) = 2^(n*(n+1)/2) = A006125(n+1).
From Hieronymus Fischer, Aug 13 2007: (Start)
a(n) = Product_{k=0..floor(log_2(n))} floor(n/2^k), n>=1.
Recurrence:
a(n*2^m) = n^m*2^(m(m+1)/2)*a(n).
a(n) <= n^((1+log_2(n))/2) = 2^A000217(log_2(n)); equality iff n is a power of 2.
a(n) >= c(n)*(n+1)^((1 + log_2(n+1))/2) for n != 2,
where c(n) = Product_{k=1..floor(log_2(n))} (1 - 1/2^k); equality holds iff n+1 is a power of 2.
a(n) > c*(n+1)^((1 + log_2(n+1))/2)
where c = 0.288788095086602421... (see constant A048651).
lim inf a(n)/n^((1+log_2(n))/2)=0.288788095086602421... for n-->oo.
lim sup a(n)/n^((1+log_2(n))/2) = 1 for n-->oo.
lim inf a(n)/a(n+1) = 0.288788095086602421... for n-->oo (see constant A048651).
a(n) = O(n^((1+log_2(n))/2)). (End)

Formula section edited by Hieronymus Fischer, Jun 13 2012

A132038 Decimal expansion of Product_{k>0} (1-1/10^k).

Original entry on oeis.org

8, 9, 0, 0, 1, 0, 0, 9, 9, 9, 9, 8, 9, 9, 9, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 9, 9, 9, 9, 9, 9, 9, 9, 8, 9, 9, 9, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8, 9, 9, 9, 9, 9, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 9, 9, 9

Offset: 0

Hieronymus Fischer, Aug 14 2007

			0.8900100999989990000001000...

G. C. Greubel, Table of n, a(n) for n = 0..1500
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge University Press, 2009, page 49.
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.

Cf. A000203, A027878, A048651, A067080, A098844, A100220, A132019, A132026.

digits = 105; Clear[p]; p[n_] := p[n] = RealDigits[Product[1-1/10^k , {k, 1, n}], 10, digits] // First; p[10]; p[n=20]; While[p[n] != p[n/2], n = 2*n]; p[n] (* Jean-François Alcover, Feb 17 2014 *)
RealDigits[QPochhammer[1/10], 10, 105][[1]] (* Jean-François Alcover, Nov 18 2015 *)
N[QPochhammer[1/10,1/10]] (* G. C. Greubel, Nov 30 2015 *)

prodinf(x=1,-.1^x,1) \\ Charles R Greathouse IV, Nov 16 2013

Equals exp( -Sum_{n>0} sigma_1(n)/(n*10^n) ).
Equals (1/10; 1/10){infinity}, where (a; q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 30 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(10)) * exp(log(10)/24 - Pi^2/(6*log(10))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(10))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027878(n). (End)

cp's OEIS Frontend

A324322 Numbers k such that Ld(k) == k (mod Rd(k)), where Ld(k) = A067080 and Rd(k) = A067079.

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A324321 Numbers k such that Rd(k) == k (mod Ld(k)), where Rd(k) = A067079 and Ld(k) = A067080.

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A005187 a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.

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A011371 a(n) = n minus (number of 1's in binary expansion of n). Also highest power of 2 dividing n!.

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A048651 Decimal expansion of Product_{k >= 1} (1 - 1/2^k).

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