A244413 -id:A244413 - cp's OEIS frontend

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

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

Offset: 1

John Tromp, Dec 11 1996

This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre, Mar 06 2003
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ...), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - Philippe Deléham, Mar 15 2004
Fixed point of the morphism 0->010, 1->2, 2->3, ..., n->(n+1), .... - Joerg Arndt, Apr 29 2014
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r = k mod 2^n and r != 1. - T. D. Noe, Jul 12 2007
The following relation holds: 2^A007814(n)*(2*A025480(n-1)+1) = A001477(n) = n. (See functions hd, tl and cons in [Paul Tarau 2009].)
a(n) is the number of 0's at the end of n when n is written in base 2.
a(n+1) is the number of 1's at the end of n when n is written in base 2. - M. F. Hasler, Aug 25 2012
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 0). That is, A003188(n) XOR A003188(n+1) == 2^A007814(n). - Russ Cox, Dec 04 2010
The sequence is squarefree (in the sense of not containing any subsequence of the form XX) [Allouche and Shallit]. Of course it contains individual terms that are squares (such as 4). - Comment expanded by N. J. A. Sloane, Jan 28 2019
a(n) is the number of zero coefficients in the n-th Stern polynomial, A125184. - T. D. Noe, Mar 01 2011
Lemma: For n < m with r = a(n) = a(m) there exists n < k < m with a(k) > r. Proof: We have n=b2^r and m=c2^r with b < c both odd; choose an even i between them; now a(i2^r) > r and n < i2^r < m. QED. Corollary: Every finite run of consecutive integers has a unique maximum 2-adic valuation. - Jason Kimberley, Sep 09 2011
a(n-2) is the 2-adic valuation of A000166(n) for n >= 2. - Joerg Arndt, Sep 06 2014
a(n) = number of 1's in the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} p_j-th prime (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24)=3; indeed, the partition having Heinz number 24 = 2*2*2*3  is [1,1,1,2]. - Emeric Deutsch, Jun 04 2015
a(n+1) is the difference between the two largest parts in the integer partition having viabin number n (0 is assumed to be a part). Example: a(20) = 2. Indeed, we have 19 = 10011_2, leading to the Ferrers board of the partition [3,1,1]. For the definition of viabin number see the comment in A290253. - Emeric Deutsch, Aug 24 2017
Apart from being squarefree, as noted above, the sequence has the property that every consecutive subsequence contains at least one number an odd number of times. - Jon Richfield, Dec 20 2018
a(n+1) is the 2-adic valuation of Sum_{e=0..n} u^e = (1 + u + u^2 + ... + u^n), for any u of the form 4k+1 (A016813). - Antti Karttunen, Aug 15 2020
{a(n)} represents the "first black hat" strategy for the game of countably infinitely many hats, with a probability of success of 1/3; cf. the Numberphile link below. - Frederic Ruget, Jun 14 2021
a(n) is the least nonnegative integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022

			2^3 divides 24, so a(24)=3.
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
  0;
  1,0;
  2,0,1,0;
  3,0,1,0,2,0,1,0;
  4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 27.
K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

T. D. Noe, Table of n, a(n) for n = 1..10000
Joerg Arndt, Subset-lex: did we miss an order?, arXiv:1405.6503 [math.CO], 2014.
K. Atanassov, On Some of Smarandache's Problems
Alain Connes, Caterina Consani, and Henri Moscovici, Zeta zeros and prolate wave operators, arXiv:2310.18423 [math.NT], 2023.
Dario T. de Castro, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.
Mathieu Guay-Paquet and Jeffrey Shallit, Avoiding Squares and Overlaps Over the Natural Numbers, (2009) Discrete Math., 309 (2009), 6245-6254.
Mathieu Guay-Paquet and Jeffrey Shallit, Avoiding Squares and Overlaps Over the Natural Numbers, arXiv:0901.1397 [math.CO], 2009.
M. Hassani, Equations and inequalities involving v_p(n!), J. Inequ. Pure Appl. Math. 6 (2005) vol. 2, #29.
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 61. Book's website
R. Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) 463-535, doi, Section 3.2.3.
Clark Kimberling, Affinely recursive sets and orderings of languages, Discrete Math., 274 (2004), 147-160.
Francis Laclé, 2-adic parity explorations of the 3n+ 1 problem, hal-03201180v2 [cs.DM], 2021.
Shuo Li, Palindromic length sequence of the ruler sequence and of the period-doubling sequence, arXiv:2007.08317 [math.CO], 2020.
Nicolas Mallet, Trial for a proof of the Syracuse conjecture, arXiv preprint arXiv:1507.05039 [math.GM], 2015.
S. Mazzanti, Plain Bases for Classes of Primitive Recursive Functions, Mathematical Logic Quarterly, 48 (2002).
Matthew Andres Moreno, Luis Zaman, and Emily Dolson, Structured Downsampling for Fast, Memory-efficient Curation of Online Data Streams, arXiv:2409.06199 [cs.DS], 2024. See p. 5.
Sascha Mücke, Coding Nuggets Faster QUBO Brute-Force Solving, TU Dortmund Univ. (Germany 2023).
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Numberphile, Hat Problems - Numberphile
Giovanni Pighizzini, Limited Automata: Properties, Complexity and Variants, International Conference on Descriptional Complexity of Formal Systems (DCFS 2019) Descriptional Complexity of Formal Systems, Lecture Notes in Computer Science (LNCS, Vol. 11612) Springer, Cham, 57-73.
Simon Plouffe, On the values of the functions ... [zeta and Gamma] ..., arXiv preprint arXiv:1310.7195 [math.NT], 2013.
A. Postnikov (MIT) and B. Sagan, What power of two divides a weighted Catalan number?, arXiv:math/0601339 [math.CO], 2006.
Lara Pudwell and Eric Rowland, Avoiding fractional powers over the natural numbers, arXiv:1510.02807 [math.CO] (2015). Also Electronic Journal of Combinatorics, Volume 25(2) (2018), #P2.27. See Section 2.
Ville Salo, Decidability and Universality of Quasiminimal Subshifts, arXiv preprint arXiv:1411.6644 [math.DS], 2014.
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2014.
F. Smarandache, Only Problems, Not Solutions!.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Paul Tarau, A Groupoid of Isomorphic Data Transformations. Calculemus 2009, 8th International Conference, MKM 2009, pp. 170-185, Springer, LNAI 5625.
P. M. B. Vitanyi, An optimal simulation of counter machines, SIAM J. Comput, 14:1(1985), 1-33.
Eric Weisstein's World of Mathematics, Binary, Binary Carry Sequence, and Double-Free Set.
Wikipedia, P-adic order.
Index entries for sequences that are fixed points of mappings
Index entries for sequences related to binary expansion of n

Cf. A011371 (partial sums), A094267 (first differences), A001511 (bisection), A346070 (mod 4).
Bisection of A050605 and |A088705|. Pairwise sums are A050603 and A136480. Difference of A285406 and A281264.
This is Guy Steele's sequence GS(1, 4) (see A135416). Cf. A053398(1,n). Column/row 1 of table A050602.
Cf. A007949 (3-adic), A235127 (4-adic), A112765 (5-adic), A122841 (6-adic), A214411 (7-adic), A244413 (8-adic), A122840 (10-adic).
Cf. A000079, A002487, A006519, A051064, A055457, A063787, A215366, A220466.
Cf. A086463 (Dgf at s=2).

a007814 n = if m == 0 then 1 + a007814 n' else 0
            where (n', m) = divMod n 2
-- Reinhard Zumkeller, Jul 05 2013, May 14 2011, Apr 08 2011

a007814 n | odd n = 0 | otherwise = 1 + a007814 (n `div` 2)
--  Walt Rorie-Baety, Mar 22 2013

[Valuation(n, 2): n in [1..120]]; // Bruno Berselli, Aug 05 2013

ord := proc(n) local i,j; if n=0 then return 0; fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end proc: seq(ord(n), n=1..111);
A007814 := n -> padic[ordp](n,2): seq(A007814(n), n=1..111); # Peter Luschny, Nov 26 2010

Table[IntegerExponent[n, 2], {n, 64}] (* Eric W. Weisstein *)
IntegerExponent[Range[64], 2] (* Eric W. Weisstein, Feb 01 2024 *)
p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (* Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004 *)
Nest[Join[ #, ReplacePart[ #, Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (* N. J. Gunther, May 23 2009 *)
Nest[ Flatten[# /. a_Integer -> {0, a + 1}] &, {0}, 7] (* Robert G. Wilson v, Jan 17 2011 *)

PARI
```
A007814(n)=valuation(n,2);
```

import math
def a(n): return int(math.log(n - (n & n - 1), 2)) # Indranil Ghosh, Apr 18 2017

def A007814(n): return (~n & n-1).bit_length() # Chai Wah Wu, Jul 01 2022

sapply(1:100,function(x) sum(gmp::factorize(x)==2)) # Christian N. K. Anderson, Jun 20 2013

(define (A007814 n) (let loop ((n n) (e 0)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

a(n) = A001511(n) - 1.
a(2*n) = A050603(2*n) = A001511(n).
a(n) = A091090(n-1) + A036987(n-1) - 1.
a(n) = 0 if n is odd, otherwise 1 + a(n/2). - Reinhard Zumkeller, Aug 11 2001
Sum_{k=1..n} a(k) = n - A000120(n). - Benoit Cloitre, Oct 19 2002
G.f.: A(x) = Sum_{k>=1} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Apr 10 2002
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Franklin T. Adams-Watters, Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a(k) = c(0) + c(0)*c(1) + c(0)*c(1)*c(2) + ... + c(0)*c(1)...c(n-1); a(k+1) = b(0) + b(0)*b(1) + b(0)*b(1)*b(2) + ... + b(0)*b(1)...b(n-1). - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
Sum_{k=1..n} (-1)^A000120(n-k)*a(k) = (-1)^(A000120(n)-1)*(A000120(n) - A000035(n)). - Vladimir Shevelev, Mar 17 2009
a(A001147(n) + A057077(n-1)) = a(2*n). - Vladimir Shevelev, Mar 21 2009
For n>=1, a(A004760(n+1)) = a(n). - Vladimir Shevelev, Apr 15 2009
2^(a(n)) = A006519(n). - Philippe Deléham, Apr 22 2009
a(n) = A063787(n) - A000120(n). - Gary W. Adamson, Jun 04 2009
a(C(n,k)) = A000120(k) + A000120(n-k) - A000120(n). - Vladimir Shevelev, Jul 19 2009
a(n!) = n - A000120(n). - Vladimir Shevelev, Jul 20 2009
v_{2}(n) = Sum_{r>=1} (r / 2^(r+1)) Sum_{k=0..2^(r+1)-1} e^(2(k*Pi*i(n+2^r))/(2^(r+1))). - A. Neves, Sep 28 2010, corrected Oct 04 2010
a(n) mod 2 = A096268(n-1). - Robert G. Wilson v, Jan 18 2012
a(A005408(n)) = 1; a(A016825(n)) = 3; A017113(a(n)) = 5; A051062(a(n)) = 7; a(n) = (A037227(n)-1)/2. - Reinhard Zumkeller, Jun 30 2012
a((2*n-1)*2^p) = p, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 04 2013
a(n) = A067255(n,1). - Reinhard Zumkeller, Jun 11 2013
a(n) = log_2(n - (n AND n-1)). - Gary Detlefs, Jun 13 2014
a(n) = 1 + A000120(n-1) - A000120(n), where A000120 is the Hamming weight function. - Stanislav Sykora, Jul 14 2014
A053398(n,k) = a(A003986(n-1,k-1)+1); a(n) = A053398(n,1) = A053398(n,n) = A053398(2*n-1,n) = Min_{k=1..n} A053398(n,k). - Reinhard Zumkeller, Aug 04 2014
a((2*x-1)*2^n) = a((2*y-1)*2^n) for positive n, x and y. - Juri-Stepan Gerasimov, Aug 04 2016
a(n) = A285406(n) - A281264(n). - Ralf Steiner, Apr 18 2017
a(n) = A000005(n)/(A000005(2*n) - A000005(n)) - 1. - conjectured by Velin Yanev, Jun 30 2017, proved by Nicholas Stearns, Sep 11 2017
Equivalently to above formula, a(n) = A183063(n) / A001227(n), i.e., a(n) is the number of even divisors of n divided by number of odd divisors of n. - Franklin T. Adams-Watters, Oct 31 2018
a(n)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Feb 16 2019
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Jul 11 2020
a(n) = 2*Sum_{j=1..floor(log_2(n))} frac(binomial(n, 2^j)*2^(j-1)/n). - Dario T. de Castro, Jul 08 2022
a(n) = A070939(n) - A070939(A030101(n)). - Andrew T. Porter, Dec 16 2022
a(n) = floor((gcd(n, 2^n)^(n+1) mod (2^(n+1)-1)^2)/(2^(n+1)-1)) (see Lemma 3.4 from Mazzanti's 2002 article). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 - A088705(n). - Chai Wah Wu, Sep 18 2024

Formula index adapted to the offset of A025480 by R. J. Mathar, Jul 20 2010

Edited by Ralf Stephan, Feb 08 2014

A010877 a(n) = n mod 8.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

The rightmost digit in the base-8 representation of n. Also, the equivalent value of the three rightmost digits in the base-2 representation of n. - Hieronymus Fischer, Jun 12 2007

Antti Karttunen, Table of n, a(n) for n = 0..65536
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,1).

Partial sums: A130486. Other related sequences A130481, A130482, A130483, A130484, A130485.

Cf. A000035, A010887, A010873, A130909, A168181, A244413, A253513.

Table[Mod[n,8],{n,0,120}]   (* Harvey P. Dale, Apr 21 2011 *)

vector(100,i,i)%8 \\ Charles R Greathouse IV, Jul 16 2011

def A010877(n): return n&7 # Chai Wah Wu, Jul 09 2022

Complex representation: a(n) = (1/8)*(1-r^n)*Sum_{k=1..7} k*Product_{m=1..7, m<>k} (1 - r^(n-m)) where r = exp(Pi/4*i) = (1+i)*sqrt(2)/2 and i=sqrt(-1).
Trigonometric representation: a(n) = 256*(sin(n*Pi/8))^2*Sum_{k=1..7} k*Product_{m=1..7, m<>k} (sin((n-m)*Pi/8))^2.
G.f.: g(x) = (Sum_{k=1..7}, k*x^k)/(1-x^8).
Also: g(x) = x(7x^8-8x^7+1)/((1-x^8)(1-x)^2). - Hieronymus Fischer, May 31 2007
a(n) = n mod 2 + 2*(floor(n/2) mod 4) = A000035(n) + 2*A010873(A004526(n)).
a(n) = n mod 4 + 4*(floor(n/4) mod 2) = A010873(n) + 4*A000035(A002265(n)).
a(n) = n mod 2 + 2*(floor(n/2) mod 2) + 4*(floor(n/4) mod 2) = A000035(n) + 2*A000035(A004526(n)) + 4*A000035(A002265(n)). - Hieronymus Fischer, Jun 12 2007
a(n) = (1/2)*(7 - (-1)^n - 2*(-1)^(b/4) - 4*(-1)^((b - 2 + 2*(-1)^(b/4))/8)) where b = 2n - 1 + (-1)^n. - Hieronymus Fischer, Jun 12 2007
General formula for period 2^k: a(n) = (1/2)*(2^k - 1 - Sum_{j=0..k-1} 2^j*(-1)^p(j,n)) where p(j,n) is defined recursively by p(0,n)=n, p(j,n) = (1/4)*(2*p(j-1,n) - 1 + (-1)^p(j-1,n)). - Hieronymus Fischer, Jun 14 2007
a(n) = floor(1234567/99999999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor(48913/2396745*8^(n+1)) mod 8. - Hieronymus Fischer, Jan 04 2013

Formula section re-edited for better readability by Hieronymus Fischer

A235127 Greatest k such that 4^k divides n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1

Offset: 1

Tom Edgar, Jan 03 2014

			Since 4^2 divides 32 and 4^3 does not, we have a(32) = 2. Likewise, since no positive power of 4 divides 9, a(9) = 0.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A004526, A007814, A007949, A122841, A115362, A234957, A244413.

IntegerExponent[Range@ 105, 4] (* Michael De Vlieger, Nov 18 2017 *)

A235127(n) = valuation(n,4); \\ Antti Karttunen, Nov 18 2017

def A235127(n): return (~n&n-1).bit_length()>>1 # Chai Wah Wu, Jul 08 2022

n=100 #change n for more terms
[valuation(i,4) for i in [1..n]]

a(n) = valuation(n,4).
G.f.: Sum_{k>=1} x^(4^k)/(1 - x^(4^k)). - Ilya Gutkovskiy, Jan 28 2017
a(n) = A004526(A007814(n)). - Antti Karttunen, Nov 18 2017
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/3. - Amiram Eldar, Jan 17 2022

More terms from Antti Karttunen, Nov 18 2017

A168181 Characteristic function of numbers that are not multiples of 8.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0

Offset: 0

Reinhard Zumkeller, Nov 30 2009

Multiplicative with a(p^e) = (if p=2 then A019590(e) else 1), p prime and e>0.

Period 8 Repeat: [0, 1, 1, 1, 1, 1, 1, 1]. - Wesley Ivan Hurt, Jun 21 2014

			G.f. = x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^9 + x^10 + x^11 + ...

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for characteristic functions
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,1).

Cf. A168185, A145568, A168184, A168182, A109720, A097325, A011558, A166486, A011655, A000035, A010877, A244413, A253513.

[Sign(n mod 8) : n in [0..100]]; // Wesley Ivan Hurt, Jun 21 2014

with(numtheory); A168181:=n->signum(n mod 8); seq(A168181(n), n=0..100); # Wesley Ivan Hurt, Jun 21 2014

Table[Sign[Mod[n,8]], {n, 0, 100}] (* Wesley Ivan Hurt, Jun 21 2014 *)

a(n)=n%8 > 0 \\ Felix Fröhlich, Aug 11 2014

def A168181(n): return int(bool(n&7)) # Chai Wah Wu, Jul 09 2022

a(n+8) = a(n);
a(n) = A000007(A010877(n));
a(A047592(n)) = 1; a(A008590(n)) = 0;
A033440(n) = Sum_{k=0..n} a(k)*(n-k).
Dirichlet g.f. (1-1/8^s)*zeta(s). - R. J. Mathar, Feb 19 2011
For the general case: the characteristic function of numbers that are not multiples of m is a(n) = floor((n-1)/m) - floor(n/m) + 1, m,n > 0. - Boris Putievskiy, May 08 2013
a(n) = sign(n mod 8). - Wesley Ivan Hurt, Jun 21 2014
a(n) = sign( 1 - floor(cos(Pi*n/4)) ). - Wesley Ivan Hurt, Jun 21 2014
Euler transform of length 8 sequence [ 1, 0, 0, 0, 0, 0, -1, 1]. - Michael Somos, Jun 24 2014
Moebius transform is length 8 sequence [ 1, 0, 0, 0, 0, 0, 0, -1]. - Michael Somos, Jun 24 2014
G.f.: x * (1 - x^7) / ((1 - x) * (1 - x^8)). - Michael Somos, Jun 24 2014
a(n) = 1-A253513(n). - Antti Karttunen, Oct 08 2017

A249344 A(n,k) = exponent of the largest power of n-th prime which divides k, square array read by antidiagonals.

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Antti Karttunen, Oct 28 2014

Square array A(n,k), where n = row, k = column, read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ... (transpose of array A060175).
A(n,k) is the (p_n)-adic valuation of k, where p_n is the n-th prime, A000040(n).
Each row is effectively a ruler function, s, with s(1) = 0. - Peter Munn, Apr 30 2022

			The top-left corner of the array:
  0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...
  0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, ...
  0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, ...
  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, ...
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, ...
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, ...
  ...
A(1,8) = 3, because 2^3 is the largest power of 2 (= p_1 = A000040(1)) that divides 8.
a(2,9) = 2, because 3^2 is the largest power of 3 (= p_2) that divides 9.
a(3,15) = 1, because 5^1 is the largest power of 5 (= p_3) that divides 15.

Transpose: A060175.
Row 1: A007814.
Row 2: A007949.
Row 3: A112765.
Row 4: A214411.
Cf. also A000040, A002260, A004736, A085604, A090622, A115627, A249421.
Completely additive sequences where more than one prime is mapped to 1, all other primes to 0: A065339, A083025, A087436, A169611.
Ruler functions, s, with s(1) = 0 that are not rows here: A122840, A122841, A235127, A244413.

A[n_, k_] := IntegerExponent[k, Prime[n]]; Table[A[k, n - k + 1], {n, 1, 15}, {k, 1, n}] // Flatten (* Amiram Eldar, Oct 01 2023 *)

a(n, k) = valuation(k, prime(n)); \\ Michel Marcus, Jun 24 2017

from sympy import prime
def a(n, k):
    p=prime(n)
    i=z=0
    while p**i<=k:
        if k%(p**i)==0: z=i
        i+=1
    return z
for n in range(1, 10): print([a(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Jun 24 2017

(define (A249344 n) (A249344bi (A002260 n) (A004736 n)))
(define (A249344bi row col) (let ((p (A000040 row))) (let loop ((n col) (i 0)) (cond ((not (zero? (modulo n p))) i) (else (loop (/ n p) (+ i 1)))))))

Row n, as a sequence, is completely additive with A(n, prime(n)) = 1, A(n, prime(m)) = 0 for m <> n. - Peter Munn, Apr 30 2022

Sum_{k=1..m} A(n,k) ~ (1/(prime(n)-1)) * m. - Amiram Eldar, Oct 01 2023

A090617 Exponent of highest power of 8 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 13, 13, 13, 13, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 18, 18, 18, 18, 19, 19, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24

Offset: 0

Henry Bottomley, Dec 06 2003

nonn

			a(8)=2 since 8! = 40320 = 8^2 * 630.

Cf. A011371, A054861, A090616, A027868, A054896.

Table[IntegerExponent[n!,8],{n,0,80}] (* Harvey P. Dale, Mar 21 2013 *)

a(n) = valuation(n!, 8); \\ Michel Marcus, Jul 10 2022

a(n) = A090622(n, 8) = floor(A011371(n)/3) = floor((floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ...)/3).

a(n) = A244413(n!) . - R. J. Mathar, Jul 08 2021

A309891 a(n) is the total number of trailing zeros in the representations of n over all bases b >= 2.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 1, 5, 3, 3, 1, 6, 1, 3, 3, 8, 1, 6, 1, 6, 3, 3, 1, 9, 3, 3, 5, 6, 1, 7, 1, 10, 3, 3, 3, 11, 1, 3, 3, 9, 1, 7, 1, 6, 6, 3, 1, 13, 3, 6, 3, 6, 1, 9, 3, 9, 3, 3, 1, 12, 1, 3, 6, 14, 3, 7, 1, 6, 3, 7, 1, 15, 1, 3, 6, 6, 3, 7, 1, 13, 8, 3, 1, 12

Offset: 1

Rémy Sigrist, Aug 21 2019

a(n) depends only on the prime signature of n.

a(n) is the sum of the k-adic valuations of n for k >= 2. - Friedjof Tellkamp, Jan 25 2025

			For n = 12: 12 has 2 trailing zeros in base 2 (1100), 1 trailing zero in bases 3, 4, 6 and 12 (110, 30, 20, 10) and no trailing zero in other bases, hence a(12) = 1*2 + 4*1 = 6.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000
Index entries for sequences computed from exponents in factorization of n

Cf. A001221, A006218, A007814, A007949, A033093, A056239, A112765, A122841, A169594, A214411, A235127, A244413, A286561, A293514, A369180.
Cf. A111003.
First differences of A078632.

Table[DivisorSum[n, IntegerExponent[n, #] &, # > 1 &], {n, 84}] (* Jon Maiga, Aug 25 2019 *)

a(n) = sumdiv(n, d, if (d>1, valuation(n,d), 0))

a(n) = {if(n == 1, return(0)); my(f = factor(n)[, 2], res = 0, t = 2, of = f, nf = f >> 1, nd(v) = prod(i = 1, #v, v[i] + 1)); while(Set(of) != [0], res += (nd(of) - nd(nf)) * (t-1); of = nf; t++; nf = f \ t); res} \\ David A. Corneth, Aug 22 2019

a(n) = Sum_{d|n, d>1} A286561(n,d), where A286561 gives the d-valuation of n.
a(p) = 1 for any prime number p.
a(p^k) = A006218(k) for any k >= 0 and any prime number p.
a(n) = 2^A001221(n) - 1 for any squarefree number n.
a(n) = 3 for any semiprime number n.
a(m*n) >= a(m) + a(n).
a(n) >= A007814(n) + A007949(n) + A235127(n) + A112765(n) + A122841(n) + A214411(n) + A244413(n).
a(n) = A056239(A293514(n)). - Antti Karttunen, Aug 22 2019
a(n) <= A033093(n). - Michel Marcus, Aug 22 2019
a(n) = A169594(n) - 1. - Jon Maiga, Aug 25 2019
From Friedjof Tellkamp, Feb 27 2024: (Start)
G.f.: Sum_{k>=2, j>=1} x^(k^j)/(1-x^(k^j)).
Dirichlet g.f.: zeta(s) * Sum_{k>=1} (zeta(k*s) - 1).
Sum_{n>=1} a(n)/n^2 = Pi^2/8 (A111003). (End)

A025673 Exponent of 8 (value of j) in n-th number of form 5^i*8^j.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 4, 1, 2, 3, 0, 4, 1, 5, 2, 3, 0, 4, 1, 5, 2, 6, 3, 0, 4, 1, 5, 2, 6, 3, 0, 7, 4, 1, 5, 2, 6, 3, 0, 7, 4, 1, 8, 5, 2, 6, 3, 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 10, 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 10, 0, 7, 4, 11, 1, 8, 5, 2, 9, 6, 3, 10, 0, 7, 4, 11, 1

Offset: 1

David W. Wilson

nonn

Cf. A025623, A025653, A244413.

Differs from A025658 at a(2805).

Take[Sort[Flatten[Table[{5^i 8^j,j},{i,0,20},{j,0,20}],1]][[;;,2]],120] (* Harvey P. Dale, Nov 10 2024 *)
With[{max = 10^11}, IntegerExponent[Sort[Flatten[Table[5^i*8^j, {i, 0, Log[5, max]}, {j, 0, Log[8, max/5^i]}]]], 8]] (* Amiram Eldar, Jul 09 2025 *)

a(n) = A244413(A025623(n)). - Amiram Eldar, Jul 09 2025

A025674 Exponent of 8 (value of j) in n-th number of form 6^i*8^j.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 7, 1, 2, 3, 4, 5, 6, 0, 7, 1, 8, 2, 3, 4, 5, 6, 0, 7, 1, 8, 2, 9, 3, 4, 5, 6, 0, 7, 1, 8, 2, 9, 3, 10, 4, 5, 6, 0, 7, 1, 8, 2, 9, 3, 10, 4, 11, 5, 6, 0, 7, 1, 8, 2, 9, 3, 10, 4, 11, 5, 12, 6, 0, 7, 1, 8, 2, 9

Offset: 1

David W. Wilson

nonn

Cf. A025627, A025661, A244413.

Differs from A025650 at a(1881).

With[{max = 10^12}, SortBy[Flatten[Table[{i, j}, {i, 0, Log[6, max]}, {j, 0, Log[8, max/6^i]}], 1], 6^#[[1]] * 8^#[[2]] &][[;; , 2]]] (* Amiram Eldar, Jul 09 2025 *)

a(n) = A244413(A025627(n)/6^A025661(n)). - Amiram Eldar, Jul 09 2025

A054897 a(n) = Sum_{k>0} floor(n/8^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12

Offset: 0

Henry Bottomley, May 23 2000

nonn

Different from the highest power of 8 dividing n!, A090617.

			a(100) = 13.
a(10^3) = 141.
a(10^4) = 1427.
a(10^5) = 14284.
a(10^6) = 142855.
a(10^7) = 1428569.
a(10^8) = 14285710.
a(10^9) = 142857138.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

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

Cf. A054899, A067080, A098844, A132032.

m:=8;
function a(n) // a = A054897
  if n eq 0 then return n;
  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=8; While[s=Floor[n/p]; t=t+s; s>0, p *= 8]; t, {n,0,100}]

def A054897(n): return (n-sum(int(d) for d in oct(n)[2:]))//7 # Chai Wah Wu, Jul 09 2022

m=8 # a = A054897
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/8) + floor(n/64) + floor(n/512) + floor(n/4096) + ....
a(n) = (n - A053829(n))/7.
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence:
a(n) = floor(n/8) + a(floor(n/8));
a(8*n) = n + a(n);
a(n*8^m) = n*(8^m-1)/7 + a(n).
a(k*8^m) = k*(8^m-1)/7, for 0 <= k < 8, m >= 0.
Asymptotic behavior:
a(n) = n/7 + O(log(n)),
a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= (n-1)/7; equality holds for powers of 8.
a(n) >= (n-7)/7 - floor(log_8(n)); equality holds for n=8^m-1, m>0.
lim inf (n/7 - a(n)) = 1/7, for n -> oo.
lim sup (n/7 - log_8(n) - a(n)) = 0, for n -> oo.
lim sup (a(n+1) - a(n) - log_8(n)) = 0, for n -> oo.
G.f.: g(x) = ( Sum_{k>0} x^(8^k)/(1-x^(8^k)) )/(1-x). (End)
Partial sums of A244413. - R. J. Mathar, Jul 08 2021

Examples added by Hieronymus Fischer, Jun 06 2012

cp's OEIS Frontend

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

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A010877 a(n) = n mod 8.

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A235127 Greatest k such that 4^k divides n.

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A168181 Characteristic function of numbers that are not multiples of 8.

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A249344 A(n,k) = exponent of the largest power of n-th prime which divides k, square array read by antidiagonals.

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A090617 Exponent of highest power of 8 dividing n!.

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