cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A019565 The squarefree numbers ordered lexicographically by their prime factorization (with factors written in decreasing order). a(n) = Product_{k in I} prime(k+1), where I is the set of indices of nonzero binary digits in n = Sum_{k in I} 2^k.

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 15, 30, 7, 14, 21, 42, 35, 70, 105, 210, 11, 22, 33, 66, 55, 110, 165, 330, 77, 154, 231, 462, 385, 770, 1155, 2310, 13, 26, 39, 78, 65, 130, 195, 390, 91, 182, 273, 546, 455, 910, 1365, 2730, 143, 286, 429, 858, 715, 1430, 2145, 4290
Offset: 0

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Keywords

Comments

A permutation of the squarefree numbers A005117. The missing positive numbers are in A013929. - Alois P. Heinz, Sep 06 2014
From Antti Karttunen, Apr 18 & 19 2017: (Start)
Because a(n) toggles the parity of n there are neither fixed points nor any cycles of odd length.
Conjecture: there are no finite cycles of any length. My grounds for this conjecture: any finite cycle in this sequence, if such cycles exist at all, must have at least one member that occurs somewhere in A285319, the terms that seem already to be quite rare. Moreover, any such a number n should satisfy in addition to A019565(n) < n also that A048675^{k}(n) is squarefree, not just for k=0, 1 but for all k >= 0. As there is on average a probability of only 6/(Pi^2) = 0.6079... that any further term encountered on the trajectory of A048675 is squarefree, the total chance that all of them would be squarefree (which is required from the elements of A019565-cycles) is soon minuscule, especially as A048675 is not very tightly bounded (many trajectories seem to skyrocket, at least initially). I am also assuming that usually there is no significant correlation between the binary expansions of n and A048675(n) (apart from their least significant bits), or, for that matter, between their prime factorizations.
See also the slightly stronger conjecture in A285320, which implies that there would neither be any two-way infinite cycles.
If either of the conjectures is false (there are cycles), then certainly neither sequence A285332 nor its inverse A285331 can be a permutation of natural numbers. (End)
The conjecture made in A087207 (see also A288569) implies the two conjectures mentioned above. A further constraint for cycles is that in any A019565-trajectory which starts from a squarefree number (A005117), every other term is of the form 4k+2, while every other term is of the form 6k+3. - Antti Karttunen, Jun 18 2017
The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever x and y do not have a 1-bit in the same position, i.e., when A004198(x,y) = 0. See also A283475. - Antti Karttunen, Oct 31 2019
The above identity becomes unconditional if binary exclusive OR, A003987(.,.), is substituted for addition, and A059897(.,.), a multiplicative equivalent of A003987, is substituted for multiplication. This gives us a(A003987(x,y)) = A059897(a(x), a(y)). - Peter Munn, Nov 18 2019
Also the Heinz number of the binary indices of n, where the Heinz number of a sequence (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and a number's binary indices (A048793) are the positions of 1's in its reversed binary expansion. - Gus Wiseman, Dec 28 2022

Examples

			5 = 2^2+2^0, e_1 = 2, e_2 = 0, prime(2+1) = prime(3) = 5, prime(0+1) = prime(1) = 2, so a(5) = 5*2 = 10.
From _Philippe Deléham_, Jun 03 2015: (Start)
This sequence regarded as a triangle withs rows of lengths 1, 1, 2, 4, 8, 16, ...:
   1;
   2;
   3,  6;
   5, 10, 15, 30;
   7, 14, 21, 42, 35,  70, 105, 210;
  11, 22, 33, 66, 55, 110, 165, 330, 77, 154, 231, 462, 385, 770, 1155, 2310;
  ...
(End)
From _Peter Munn_, Jun 14 2020: (Start)
The initial terms are shown below, equated with the product of their prime factors to exhibit the lexicographic order. We start with 1, since 1 is factored as the empty product and the empty list is first in lexicographic order.
   n     a(n)
   0     1 = .
   1     2 = 2.
   2     3 = 3.
   3     6 = 3*2.
   4     5 = 5.
   5    10 = 5*2.
   6    15 = 5*3.
   7    30 = 5*3*2.
   8     7 = 7.
   9    14 = 7*2.
  10    21 = 7*3.
  11    42 = 7*3*2.
  12    35 = 7*5.
(End)
		

Crossrefs

Row 1 of A285321.
Equivalent sequences for k-th-power-free numbers: A101278 (k=3), A101942 (k=4), A101943 (k=5), A054842 (k=10).
Cf. A109162 (iterates).
Cf. also A048675 (a left inverse), A087207, A097248, A260443, A054841.
Cf. A285315 (numbers for which a(n) < n), A285316 (for which a(n) > n).
Cf. A276076, A276086 (analogous sequences for factorial and primorial bases), A334110 (terms squared).
For partial sums see A288570.
A003961, A003987, A004198, A059897, A089913, A331590, A334747 are used to express relationships between sequence terms.
Column 1 of A329332.
Even bisection (which contains the odd terms): A332382.
A160102 composed with A052330, and subsequence of the latter.
Related to A000079 via A225546, to A057335 via A122111, to A008578 via A336322.
Least prime index of a(n) is A001511.
Greatest prime index of a(n) is A029837 or A070939.
Taking prime indices gives A048793, reverse A272020, row sums A029931.
A112798 lists prime indices, length A001222, sum A056239.

Programs

  • Haskell
    a019565 n = product $ zipWith (^) a000040_list (a030308_row n)
    -- Reinhard Zumkeller, Apr 27 2013
    
  • Maple
    a:= proc(n) local i, m, r; m:=n; r:=1;
          for i while m>0 do if irem(m,2,'m')=1
            then r:=r*ithprime(i) fi od; r
        end:
    seq(a(n), n=0..60);  # Alois P. Heinz, Sep 06 2014
  • Mathematica
    Do[m=1;o=1;k1=k;While[ k1>0, k2=Mod[k1, 2];If[k2\[Equal]1, m=m*Prime[o]];k1=(k1-k2)/ 2;o=o+1];Print[m], {k, 0, 55}] (* Lei Zhou, Feb 15 2005 *)
    Table[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2], {n, 0, 55}]  (* Michael De Vlieger, Aug 27 2016 *)
    b[0] := {1}; b[n_] := Flatten[{ b[n - 1], b[n - 1] * Prime[n] }];
      a = b[6] (* Fred Daniel Kline, Jun 26 2017 *)
  • PARI
    a(n)=factorback(vecextract(primes(logint(n+!n,2)+1),n))  \\ M. F. Hasler, Mar 26 2011, updated Aug 22 2014, updated Mar 01 2018
    
  • Python
    from operator import mul
    from functools import reduce
    from sympy import prime
    def A019565(n):
        return reduce(mul,(prime(i+1) for i,v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1
    # Chai Wah Wu, Dec 25 2014
    
  • Scheme
    (define (A019565 n) (let loop ((n n) (i 1) (p 1)) (cond ((zero? n) p) ((odd? n) (loop (/ (- n 1) 2) (+ 1 i) (* p (A000040 i)))) (else (loop (/ n 2) (+ 1 i) p))))) ;; (Requires only the implementation of A000040 for prime numbers.) - Antti Karttunen, Apr 20 2017

Formula

G.f.: Product_{k>=0} (1 + prime(k+1)*x^2^k), where prime(k)=A000040(k). - Ralf Stephan, Jun 20 2003
a(n) = f(n, 1, 1) with f(x, y, z) = if x > 0 then f(floor(x/2), y*prime(z)^(x mod 2), z+1) else y. - Reinhard Zumkeller, Mar 13 2010
For all n >= 0: A048675(a(n)) = n; A013928(a(n)) = A064273(n). - Antti Karttunen, Jul 29 2015
a(n) = a(2^x)*a(2^y)*a(2^z)*... = prime(x+1)*prime(y+1)*prime(z+1)*..., where n = 2^x + 2^y + 2^z + ... - Benedict W. J. Irwin, Jul 24 2016
From Antti Karttunen, Apr 18 2017 and Jun 18 2017: (Start)
a(n) = A097248(A260443(n)), a(A005187(n)) = A283475(n), A108951(a(n)) = A283477(n).
A055396(a(n)) = A001511(n), a(A087207(n)) = A007947(n). (End)
a(2^n - 1) = A002110(n). - Michael De Vlieger, Jul 05 2017
a(n) = A225546(A000079(n)). - Peter Munn, Oct 31 2019
From Peter Munn, Mar 04 2022: (Start)
a(2n) = A003961(a(n)); a(2n+1) = 2*a(2n).
a(x XOR y) = A059897(a(x), a(y)) = A089913(a(x), a(y)), where XOR denotes bitwise exclusive OR (A003987).
a(n+1) = A334747(a(n)).
a(x+y) = A331590(a(x), a(y)).
a(n) = A336322(A008578(n+1)).
(End)

Extensions

Definition corrected by Klaus-R. Löffler, Aug 20 2014
New name from Peter Munn, Jun 14 2020

A029931 If 2n = Sum 2^e_i, a(n) = Sum e_i.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 11, 12, 13, 14, 14, 15, 16, 17, 15, 16, 17, 18, 18, 19, 20, 21, 7, 8, 9, 10, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16
Offset: 0

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Comments

Write n in base 2, n = sum b(i)*2^(i-1), then a(n) = sum b(i)*i. - Benoit Cloitre, Jun 09 2002
May be regarded as a triangular array read by rows, giving weighted sum of compositions in standard order. The standard order of compositions is given by A066099. - Franklin T. Adams-Watters, Nov 06 2006
Sum of all positive integer roots m_i of polynomial {m,k} - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015
Also the sum of binary indices of n, where a binary index of n (A048793) is any position of a 1 in its reversed binary expansion. For example, the binary indices of 11 are {1,2,4}, so a(11) = 7. - Gus Wiseman, May 22 2024

Examples

			14 = 8+4+2 so a(7) = 3+2+1 = 6.
Composition number 11 is 2,1,1; 1*2+2*1+3*1 = 7, so a(11) = 7.
The triangle starts:
  0
  1
  2 3
  3 4 5 6
The reversed binary expansion of 18 is (0,1,0,0,1) with 1's at positions {2,5}, so a(18) = 2 + 5 = 7. - _Gus Wiseman_, Jul 22 2019
		

Crossrefs

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A022290 (Fibonacci), A059590 (factorials), A073642, A089625 (primes), A116549, A326031.
Cf. A001793 (row sums), A011782 (row lengths), A059867, A066099, A124757.
Row sums of A048793 and A272020.
Contains exactly A000009(n) copies of n.
For length instead of sum we have A000120, complement A023416.
For minimum instead of sum we have A001511, opposite A000012.
For maximum instead of sum we have A029837 or A070939, opposite A070940.
For product instead of sum we have A096111.
The reverse version is A230877, row sums of A371572.
The reverse complement is A359359, row sums of A371571.
The complement is A359400, row sums of A368494.
Numbers k such that a(k) is prime are A372689.
A014499 lists binary indices of prime numbers.
A019565 gives Heinz number of binary indices, inverse A048675.
A372471 lists binary indices of primes, row-sums A372429.

Programs

  • Haskell
    a029931 = sum . zipWith (*) [1..] . a030308_row
    -- Reinhard Zumkeller, Feb 28 2014
    
  • Maple
    HammingWeight := n -> add(i, i = convert(n, base, 2)):
    a := proc(n) option remember; `if`(n = 0, 0,
    ifelse(n::even, a(n/2) + HammingWeight(n/2), a(n-1) + 1)) end:
    seq(a(n), n = 0..78); # Peter Luschny, Oct 30 2021
  • Mathematica
    a[n_] := (b = IntegerDigits[n, 2]).Reverse @ Range[Length @ b]; Array[a,78,0] (* Jean-François Alcover, Apr 28 2011, after B. Cloitre *)
  • PARI
    for(n=0,100,l=length(binary(n)); print1(sum(i=1,l, component(binary(n),i)*(l-i+1)),","))
    
  • PARI
    a(n) = my(b=binary(n)); b*-[-#b..-1]~; \\ Ruud H.G. van Tol, Oct 17 2023
    
  • Python
    def A029931(n): return sum(i if j == '1' else 0 for i, j in enumerate(bin(n)[:1:-1],1)) # Chai Wah Wu, Dec 20 2022
    (C#)
    ulong A029931(ulong n) {
        ulong result = 0, counter = 1;
        while(n > 0) {
            if (n % 2 == 1)
              result += counter;
            counter++;
            n /= 2;
        }
        return result;
    } // Frank Hollstein, Jan 07 2023

Formula

a(n) = a(n - 2^L(n)) + L(n) + 1 [where L(n) = floor(log_2(n)) = A000523(n)] = sum of digits of A048794 [at least for n < 512]. - Henry Bottomley, Mar 09 2001
a(0) = 0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n) + 1, where e1(n) = A000120(n). a(n) = log_2(A029930(n)). - Ralf Stephan, Jun 19 2003
G.f.: (1/(1-x)) * Sum_{k>=0} (k+1)*x^2^k/(1+x^2^k). - Ralf Stephan, Jun 23 2003
a(n) = Sum_{k>=0} A030308(n,k)*A000027(k+1). - Philippe Deléham, Oct 15 2011
a(n) = sum of n-th row of the triangle in A213629. - Reinhard Zumkeller, Jun 17 2012
From Reinhard Zumkeller, Feb 28 2014: (Start)
a(A089633(n)) = n and a(m) != n for m < A089633(n).
a(n) = Sum_{k=1..A070939(n)} k*A030308(n,k-1). (End)
a(n) = A073642(n) + A000120(n). - Peter Kagey, Apr 04 2016

Extensions

More terms from Erich Friedman

A008585 a(n) = 3*n.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 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, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177
Offset: 0

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Keywords

Comments

If n != 1 and n^2+2 is prime then n is a member of this sequence. - Cino Hilliard, Mar 19 2007
Multiples of 3. Positive members of this sequence are the third transversal numbers (or 3-transversal numbers): Numbers of the 3rd column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 3rd column in the square array A057145. - Omar E. Pol, May 02 2008
Numbers n for which polynomial 27*x^6-2^n is factorizable. - Artur Jasinski, Nov 01 2008
1/7 in base-2 notation = 0.001001001... = 1/2^3 + 1/2^6 + 1/2^9 + ... - Gary W. Adamson, Jan 24 2009
A165330(a(n)) = 153 for n > 0; subsequence of A031179. - Reinhard Zumkeller, Sep 17 2009
A011655(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2009
A215879(a(n)) = 0. - Reinhard Zumkeller, Dec 28 2012
Moser conjectured, and Newman proved, that the terms of this sequence are more likely to have an even number of 1s in binary than an odd number. The excess is an undulating multiple of n^(log 3/log 4). See also Coquet, who refines this result. - Charles R Greathouse IV, Jul 17 2013
Integer areas of medial triangles of integer-sided triangles.
Also integer subset of A188158(n)/4.
A medial triangle MNO is formed by joining the midpoints of the sides of a triangle ABC. The area of a medial triangle is A/4 where A is the area of the initial triangle ABC. - Michel Lagneau, Oct 28 2013
From Derek Orr, Nov 22 2014: (Start)
Let b(0) = 0, and b(n) = the number of distinct terms in the set of pairwise sums {b(0), ... b(n-1)} + {b(0), ... b(n-1)}. Then b(n+1) = a(n), for n > 0.
Example: b(1) = the number of distinct sums of {0} + {0}. The only possible sum is {0} so b(1) = 1. b(2) = the number of distinct sums of {0,1} + {0,1}. The possible sums are {0,1,2} so b(2) = 3. b(3) = the number of distinct sums of {0,1,3} + {0,1,3}. The possible sums are {0, 1, 2, 3, 4, 6} so b(3) = 6. This continues and one can see that b(n+1) = a(n). (End)
Number of partitions of 6n into exactly 2 parts. - Colin Barker, Mar 23 2015
Partial sums are in A045943. - Guenther Schrack, May 18 2017
Number of edges in a maximal planar graph with n+2 vertices, n > 0 (see A008486 comments). - Jonathan Sondow, Mar 03 2018
Also numbers such that when the leftmost digit is moved to the unit's place the result is divisible by 3. - Stefano Spezia, Jul 08 2025

Examples

			G.f.: 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + 21*x^7 + ...
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.

Crossrefs

Row / column 3 of A004247 and of A325820.
Cf. A016957, A057145, A139600, A139606, A001651 (complement), A032031 (partial products), A190944 (binary), A061819 (base 4).

Programs

Formula

G.f.: 3*x/(1-x)^2. - R. J. Mathar, Oct 23 2008
a(n) = A008486(n), n > 0. - R. J. Mathar, Oct 28 2008
G.f.: A(x) - 1, where A(x) is the g.f. of A008486. - Gennady Eremin, Feb 20 2021
a(n) = Sum_{k=0..inf} A030308(n,k)*A007283(k). - Philippe Deléham, Oct 17 2011
E.g.f.: 3*x*exp(x). - Ilya Gutkovskiy, May 18 2016
From Guenther Schrack, May 18 2017: (Start)
a(3*k) = a(a(k)) = A008591(n).
a(3*k+1) = a(a(k) + 1) = a(A016777(n)) = A017197(n).
a(3*k+2) = a(a(k) + 2) = a(A016789(n)) = A017233(n). (End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A005811 Number of runs in binary expansion of n (n>0); number of 1's in Gray code for n.

Original entry on oeis.org

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

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Comments

Starting with a(1) = 0 mirror all initial 2^k segments and increase by one.
a(n) gives the net rotation (measured in right angles) after taking n steps along a dragon curve. - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
This sequence generates A082410: (0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) and A014577; identical to the latter except starting 1, 1, 0, ...; by writing a "1" if a(n+1) > a(n); if not, write "0". E.g., A014577(2) = 0, since a(3) < a(2), or 1 < 2. - Gary W. Adamson, Sep 20 2003
Starting with 1 = partial sums of A034947: (1, 1, -1, 1, 1, -1, -1, 1, 1, 1, ...). - Gary W. Adamson, Jul 23 2008
The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.
Can be used as a binomial transform operator: Let a(n) = the n-th term in any S(n); then extract 2^k strings, adding the terms. This results in the binomial transform of S(n). Say S(n) = 1, 3, 5, ...; then we obtain the strings: (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), ...; = the binomial transform of (1, 3, 5, ...) = (1, 4, 12, 32, 80, ...). Example: the 8-bit string has a sum of 32 with a distribution of (1, 3, 3, 1) or one 1, three 3's, three 5's, and one 7; as expected. - Gary W. Adamson, Jun 21 2012
Considers all positive odd numbers as nodes of a graph. Two nodes are connected if and only if the sum of the two corresponding odd numbers is a power of 2. Then a(n) is the distance between 2n + 1 and 1. - Jianing Song, Apr 20 2019

Examples

			Considered as a triangle with 2^k terms per row, the first few rows are:
  1
  2, 1
  2, 3, 2, 1
  2, 3, 4, 3, 2, 3, 2, 1
  ...
The n-th row becomes right half of next row; left half is mirrored terms of n-th row increased by one. - _Gary W. Adamson_, Jun 20 2012
		

References

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

Crossrefs

Cf. A037834 (-1), A088748 (+1), A246960 (mod 4), A034947 (first differences), A000975 (indices of record highs), A173318 (partial sums).
Partial sums of A112347. Recursion depth of A035327.

Programs

  • Haskell
    import Data.List (group)
    a005811 0 = 0
    a005811 n = length $ group $ a030308_row n
    a005811_list = 0 : f [1] where
       f (x:xs) = x : f (xs ++ [x + x `mod` 2, x + 1 - x `mod` 2])
    -- Reinhard Zumkeller, Feb 16 2013, Mar 07 2011
    
  • Maple
    A005811 := proc(n)
        local i, b, ans;
        if n = 0 then
            return 0 ;
        end if;
        ans := 1;
        b := convert(n, base, 2);
        for i from nops(b)-1 to 1 by -1 do
            if b[ i+1 ]<>b[ i ] then
                ans := ans+1
            fi
        od;
        return ans ;
    end proc:
    seq(A005811(i), i=1..50) ;
    # second Maple program:
    a:= n-> add(i, i=Bits[Split](Bits[Xor](n, iquo(n, 2)))):
    seq(a(n), n=0..100);  # Alois P. Heinz, Feb 01 2023
  • Mathematica
    Table[ Length[ Length/@Split[ IntegerDigits[ n, 2 ] ] ], {n, 1, 255} ]
    a[n_] := DigitCount[BitXor[n, Floor[n/2]], 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 11 2024 *)
  • PARI
    a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2))
    
  • PARI
    a(n)=if(n<1,0,a(n\2)+(a(n\2)+n)%2) \\ Benoit Cloitre, Jan 20 2014
    
  • PARI
    a(n) = hammingweight(bitxor(n, n>>1));  \\ Gheorghe Coserea, Sep 03 2015
    
  • Python
    def a(n): return bin(n^(n>>1))[2:].count("1") # Indranil Ghosh, Apr 29 2017

Formula

a(2^k + i) = a(2^k - i + 1) + 1 for k >= 0 and 0 < i <= 2^k. - Reinhard Zumkeller, Aug 14 2001
a(2n+1) = 2a(n) - a(2n) + 1, a(4n) = a(2n), a(4n+2) = 1 + a(2n+1).
a(j+1) = a(j) + (-1)^A014707(j). - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^2^k/(1+x^2^(k+1)). - Ralf Stephan, May 02 2003
Delete the 0, make subsets of 2^n terms; and reverse the terms in each subset to generate A088696. - Gary W. Adamson, Oct 19 2003
a(0) = 0, a(2n) = a(n) + [n odd], a(2n+1) = a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(n) = Sum_{k=1..n} (-1)^((k/2^A007814(k)-1)/2) = Sum_{k=1..n} (-1)^A025480(k-1). - Ralf Stephan, Oct 29 2003
a(n) = A069010(n) + A033264(n). - Ralf Stephan, Oct 29 2003
a(0) = 0 then a(n) = a(floor(n/2)) + (a(floor(n/2)) + n) mod 2. - Benoit Cloitre, Jan 20 2014
a(n) = A037834(n) + 1.
a(n) = A000120(A003188(n)). - Amiram Eldar, Jul 11 2024

Extensions

Additional description from Wouter Meeussen

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.

Original entry on oeis.org

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

Views

Author

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

Keywords

Comments

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

Examples

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

References

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

Crossrefs

Cf. A001511 (first differences), A122247 (partial sums), A055938 (complement).

Programs

  • Haskell
    a005187 n = a005187_list !! n
    a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
    -- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011
    
  • Magma
    [n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019
    
  • Maple
    A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
    seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014
  • Mathematica
    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 *)
  • PARI
    {a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */
    
  • PARI
    a(n)=my(s=n);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Apr 07 2012
    
  • PARI
    a(n)=2*n-hammingweight(n) \\ Charles R Greathouse IV, Jan 07 2013
    
  • Python
    def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021
  • Sage
    @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
    

Formula

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

A005836 Numbers whose base-3 representation contains no 2.

Original entry on oeis.org

0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121, 243, 244, 246, 247, 252, 253, 255, 256, 270, 271, 273, 274, 279, 280, 282, 283, 324, 325, 327, 328, 333, 334, 336, 337, 351, 352
Offset: 1

Views

Author

Keywords

Comments

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.
This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001
In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010
Complement of A074940. - Reinhard Zumkeller, Mar 23 2003
Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003
Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003
A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.
Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006
Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009
Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011
Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011
It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011
Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015
Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019

Examples

			12 is a term because 12 = 110_3.
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
   0
   1
   3,  4
   9, 10, 12, 13
  27, 28, 30, 31, 36, 37, 39, 40
  81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121
... - _Philippe Deléham_, Jun 06 2015
		

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E10, pp. 317-323.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A039966 (characteristic function).
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Row 3 of array A104257.
Summary of increasing sequences avoiding arithmetic progressions of specified lengths (the second of each pair is obtained by adding 1 to the first):
3-term AP: A005836 (>=0), A003278 (>0);
4-term AP: A005839 (>=0), A005837 (>0);
5-term AP: A020654 (>=0), A020655 (>0);
6-term AP: A020656 (>=0), A005838 (>0);
7-term AP: A020657 (>=0), A020658 (>0);
8-term AP: A020659 (>=0), A020660 (>0);
9-term AP: A020661 (>=0), A020662 (>0);
10-term AP: A020663 (>=0), A020664 (>0).
See also A000452.

Programs

  • Haskell
    a005836 n = a005836_list !! (n-1)
    a005836_list = filter ((== 1) . a039966) [0..]
    -- Reinhard Zumkeller, Jun 09 2012, Sep 29 2011
    
  • Julia
    function a(n)
        m, r, b = n, 0, 1
        while m > 0
            m, q = divrem(m, 2)
            r += b * q
            b *= 3
        end
    r end; [a(n) for n in 0:57] |> println # Peter Luschny, Jan 03 2021
  • Maple
    t := (j, n) -> add(binomial(n,k)^j, k=0..n):
    for i from 1 to 400 do
        if(t(4,i) mod 3 <>0) then print(i) fi
    od; # Gary Detlefs, Nov 28 2011
    # alternative Maple program:
    a:= proc(n) option remember: local k, m:
    if n=1 then 0 elif n=2 then 1 elif n>2 then k:=floor(log[2](n-1)): m:=n-2^k: procname(m)+3^k: fi: end proc:
    seq(a(n), n=1.. 20); # Paul Weisenhorn, Mar 22 2020
    # third Maple program:
    a:= n-> `if`(n=1, 0, irem(n-1, 2, 'q')+3*a(q+1)):
    seq(a(n), n=1..100);  # Alois P. Heinz, Jan 26 2022
  • Mathematica
    Table[FromDigits[IntegerDigits[k, 2], 3], {k, 60}]
    Select[Range[0, 400], DigitCount[#, 3, 2] == 0 &] (* Harvey P. Dale, Jan 04 2012 *)
    Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)
    FromDigits[#,3]&/@Tuples[{0,1},7] (* Harvey P. Dale, May 10 2019 *)
  • PARI
    A=vector(100);for(n=2,#A,A[n]=if(n%2,3*A[n\2+1],A[n-1]+1));A \\ Charles R Greathouse IV, Jul 24 2012
    
  • PARI
    is(n)=while(n,if(n%3>1,return(0));n\=3);1 \\ Charles R Greathouse IV, Mar 07 2013
    
  • PARI
    a(n) = fromdigits(binary(n-1),3);  \\ Gheorghe Coserea, Jun 15 2018
    
  • Python
    def A005836(n):
        return int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015
    

Formula

a(n) = A005823(n)/2 = A003278(n)-1 = A033159(n)-2 = A033162(n)-3.
Numbers n such that the coefficient of x^n is > 0 in prod (k >= 0, 1 + x^(3^k)). - Benoit Cloitre, Jul 29 2003
a(n+1) = Sum_{k=0..m} b(k)* 3^k and n = Sum( b(k)* 2^k ).
a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0.
a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2). - Ralf Stephan, Apr 27 2003
G.f.: (x/(1-x)) * Sum_{k>=0} 3^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003
a(n) = Sum_{k = 1..n-1} (1 + 3^A007814(k)) / 2. - Philippe Deléham, Jul 09 2005
From Reinhard Zumkeller, Mar 02 2008: (Start)
A081603(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 2 then y else if x mod 3 = 2 then f(y+1, y+1) else f(floor(x/3), y). (End)
With offset a(0) = 0: a(n) = Sum_{k>=0} A030308(n,k)*3^k. - Philippe Deléham, Oct 15 2011
a(2^n) = A003462(n). - Philippe Deléham, Jun 06 2015
We have liminf_{n->infinity} a(n)/n^(log(3)/log(2)) = 1/2 and limsup_{n->infinity} a(n)/n^(log(3)/log(2)) = 1. - Gheorghe Coserea, Sep 13 2015
a(2^k+m) = a(m) + 3^k with 1 <= m <= 2^k and 1 <= k, a(1)=0, a(2)=1. - Paul Weisenhorn, Mar 22 2020
Sum_{n>=2} 1/a(n) = 2.682853110966175430853916904584699374821677091415714815171756609672281184705... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022
A065361(a(n)) = n-1. - Rémy Sigrist, Feb 06 2023
a(n) ≍ n^k, where k = log 3/log 2 = 1.5849625007. (I believe the constant varies from 1/2 to 1.) - Charles R Greathouse IV, Mar 29 2024

Extensions

Offset corrected by N. J. A. Sloane, Mar 02 2008
Edited by the Associate Editors of the OEIS, Apr 07 2009

A008586 Multiples of 4.

Original entry on oeis.org

0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228
Offset: 0

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Author

Keywords

Comments

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 14 ).
A000466(n), a(n) and A053755(n) are Pythagorean triples. - Zak Seidov, Jan 16 2007
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-3) is equal to the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Number of n-permutations (n>=1) of 5 objects u, v, z, x, y with repetition allowed, containing n-1 u's. Example: if n=1 then n-1 = zero (0) u, a(1)=4 because we have v, z, x, y. If n=2 then n-1 = one (1) u, a(2)=8 because we have vu, zu, xu, yu, uv, uz, ux, uy. A038231 formatted as a triangular array: diagonal: 4, 8, 12, 16, 20, 24, 28, 32, ... - Zerinvary Lajos, Aug 06 2008
For n > 0: numbers having more even than odd divisors: A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012
A214546(a(n)) < 0 for n > 0. - Reinhard Zumkeller, Jul 20 2012
A090418(a(n)) = 0 for n > 0. - Reinhard Zumkeller, Aug 06 2012
Terms are the differences of consecutive centered square numbers (A001844). - Mihir Mathur, Apr 02 2013
a(n)*Pi = nonnegative zeros of the cycloid generated by a circle of radius 2 rolling along the positive x-axis from zero. - Wesley Ivan Hurt, Jul 01 2013
Apart from the initial term, number of vertices of minimal path on an n-dimensional cubic lattice (n>1) of side length 2, until a self-avoiding walk gets stuck. A004767 + 1. - Matthew Lehman, Dec 23 2013
The number of orbits of Aut(Z^7) as function of the infinity norm n of the representative lattice point of the orbit, when the cardinality of the orbit is equal to 2688. - Philippe A.J.G. Chevalier, Dec 29 2015
First differences of A001844. - Robert Price, May 13 2016
Numbers k such that Fibonacci(k) is a multiple of 3 (A033888). - Bruno Berselli, Oct 17 2017

Crossrefs

Number of orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008585, A005843, A001477, A000217.

Programs

Formula

a(n) = A008574(n), n>0. - R. J. Mathar, Oct 28 2008
a(n) = Sum_{k>=0} A030308(n,k)*2^(k+2). - Philippe Deléham, Oct 17 2011
a(n+1) = A000290(n+2) - A000290(n). - Philippe Deléham, Mar 31 2013
G.f.: 4*x/(1-x)^2. - David Wilding, Jun 21 2014
E.g.f.: 4*x*exp(x). - Stefano Spezia, May 18 2021

A030101 a(n) is the number produced when n is converted to binary digits, the binary digits are reversed and then converted back into a decimal number.

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 13, 3, 11, 7, 15, 1, 17, 9, 25, 5, 21, 13, 29, 3, 19, 11, 27, 7, 23, 15, 31, 1, 33, 17, 49, 9, 41, 25, 57, 5, 37, 21, 53, 13, 45, 29, 61, 3, 35, 19, 51, 11, 43, 27, 59, 7, 39, 23, 55, 15, 47, 31, 63, 1, 65, 33, 97, 17, 81, 49, 113, 9, 73, 41, 105, 25, 89, 57
Offset: 0

Views

Author

Keywords

Comments

As with decimal reversal, initial zeros are ignored; otherwise, the reverse of 1 would be 1000000... ad infinitum.
Numerators of the binary van der Corput sequence. - Eric Rowland, Feb 12 2008
It seems that in most cases A030101(x) = A000265(x) and that if A030101(x) <> A000265(x), the next time A030101(y) = A000265(x), A030101(x) = A000265(y). Also, it seems that if a pair of values exist at one index, they will exist at any index where one of them exist. It also seems like the greater of the pair always shows up on A000265 first. - Dylan Hamilton, Aug 04 2010
The number of occasions A030101(n) = A000265(n) before n = 2^k is A053599(k) + 1. For n = 0..2^19, the sequences match less than 1% of the time. - Andrew Woods, May 19 2012
For n > 0: a(a(n)) = n if and only if n is odd; a(A006995(n)) = A006995(n). - Juli Mallett, Nov 11 2010, corrected: Reinhard Zumkeller, Oct 21 2011
n is binary palindromic if and only if a(n) = n. - Reinhard Zumkeller, corrected: Jan 17 2012, thanks to Hieronymus Fischer, who pointed this out; Oct 21 2011
Given any n > 1, the set of numbers A030109(i) = (A030101(i) - 1)/2 for indexes i ranging from 2^n to 2^(n + 1) - 1 is a permutation of the set of consecutive integers {0, 1, 2, ..., 2^n - 1}. This is important in the standard FFT algorithms (starting or ending bit-reversal permutation). - Stanislav Sykora, Mar 15 2012
Row n of A030308 gives the binary digits of a(n), prepended with zero at even positions. - Reinhard Zumkeller, Jun 17 2012
The binary van der Corput sequence is the infinite sequence of fractions { A030101(n)/A062383(n), n = 0, 1, 2, 3, ... }, and begins 0, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16, 5/16, 13/16, 3/16, 11/16, 7/16, 15/16, 1/32, 17/32, 9/32, 25/32, 5/32, 21/32, 13/32, 29/32, 3/32, 19/32, 11/32, 27/32, 7/32, 23/32, 15/32, 31/32, 1/64, 33/64, 17/64, 49/64, ... - N. J. A. Sloane, Dec 01 2019
Record highs occur at n = A209492(m) (for n>=1) with values a(n) = A224195(m) (for n>=3). - Bill McEachen, Aug 02 2023

Examples

			a(100) = 19 because 100 (base 10) = 1100100 (base 2) and R(1100100 (base 2)) = 10011 (base 2) = 19 (base 10).
		

References

  • Hlawka E. The theory of uniform distribution. Academic Publishers, Berkhamsted, 1984. See pp. 93, 94 for the van der Corput sequence. - N. J. A. Sloane, Dec 01 2019

Crossrefs

Cf. A055944 (reverse and add), A178225, A273258.
Cf. A056539, A057889 (bijective variants), A224195, A209492.

Programs

  • Haskell
    a030101 = f 0 where
       f y 0 = y
       f y x = f (2 * y + b) x'  where (x', b) = divMod x 2
    -- Reinhard Zumkeller, Mar 18 2014, Oct 21 2011
    
  • J
    ([: #. [: |. #:)"0 NB. Stephen Makdisi, May 07 2018
    
  • Magma
    A030101:=func; // Jason Kimberley, Sep 19 2011
    
  • Maple
    A030101 := proc(n)
        convert(n,base,2) ;
        ListTools[Reverse](%) ;
        add(op(i,%)*2^(i-1),i=1..nops(%)) ;
    end proc: # R. J. Mathar, Mar 10 2015
    # second Maple program:
    a:= proc(n) local m, r; m:=n; r:=0;
          while m>0 do r:=r*2+irem(m, 2, 'm') od; r
        end:
    seq(a(n), n=0..80);  # Alois P. Heinz, Nov 17 2015
  • Mathematica
    Table[FromDigits[Reverse[IntegerDigits[i, 2]], 2], {i, 0, 80}]
    bitRev[n_] := Switch[Mod[n, 4], 0, bitRev[n/2], 1, 2 bitRev[(n + 1)/2] - bitRev[(n - 1)/4], 2, bitRev[n/2], 3, 3 bitRev[(n - 1)/2] - 2 bitRev[(n - 3)/4]]; bitRev[0] = 0; bitRev[1] = 1; bitRev[3] = 3; Array[bitRev, 80, 0] (* Robert G. Wilson v, Mar 18 2014 *)
  • PARI
    a(n)=if(n<1,0,subst(Polrev(binary(n)),x,2))
    
  • PARI
    a(n) = fromdigits(Vecrev(binary(n)), 2); \\ Michel Marcus, Nov 10 2017
    
  • Python
    def a(n): return int(bin(n)[2:][::-1], 2) # Indranil Ghosh, Apr 24 2017
    
  • Sage
    def A030101(n): return Integer(bin(n).lstrip("0b")[::-1],2) if n!=0 else 0
    [A030101(n) for n in (0..78)]  # Peter Luschny, Aug 09 2012
    
  • Scala
    (0 to 127).map(n => Integer.parseInt(Integer.toString(n, 2).reverse, 2)) // Alonso del Arte, Feb 11 2020

Formula

a(n) = 0, a(2n) = a(n), a(2n+1) = a(n) + 2^(floor(log_2(n)) + 1). For n > 0, a(n) = 2*A030109(n) - 1. - Ralf Stephan, Sep 15 2003
a(n) = b(n, 0) with b(n, r) = r if n = 0, otherwise b(floor(n/2), 2*r + n mod 2). - Reinhard Zumkeller, Mar 03 2010
a(1) = 1, a(3) = 3, a(2n) = a(n), a(4n+1) = 2a(2n+1) - a(n), a(4n+3) = 3a(2n+1) - 2a(n) (as in the Project Euler problem). To prove this, expand the recurrence into binary strings and reversals. - David Applegate, Mar 16 2014, following a posting to the Sequence Fans Mailing List by Martin Møller Skarbiniks Pedersen.
Conjecture: a(n) = 2*w(n) - 2*w(A053645(n)) - 1 for n > 0, where w = A264596. - Velin Yanev, Sep 12 2017

Extensions

Edits (including correction of an erroneous date pointed out by J. M. Bergot) by Jon E. Schoenfield, Mar 16 2014
Name clarified by Antti Karttunen, Nov 09 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

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Author

Keywords

Comments

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

Examples

			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 + ...
		

References

  • 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.

Crossrefs

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

Programs

  • Haskell
    a011371 n = n - a000120 n  -- Reinhard Zumkeller, Jan 24 2014
    
  • Magma
    [Valuation(Factorial(n), 2): n in [0..80]]; // Bruno Berselli, Aug 05 2013
    
  • Maple
    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
  • Mathematica
    -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} ]
  • PARI
    {a(n) = if( n<0, 0, valuation(n!, 2))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, sum(k=1, n, n\2^k))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, n - subst( Pol( binary( n ) ), x, 1))}; /* Michael Somos, Aug 28 2007 */
    
  • PARI
    a(n)=sum(k=1,log(n+1)\log(2),n>>k) \\ Charles R Greathouse IV, Oct 03 2012
    
  • PARI
    a(n)=my(s);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Aug 09 2013
    
  • PARI
    a(n) = n - hammingweight(n); \\ Michel Marcus, Jun 05 2014
    
  • Python
    [n - bin(n)[2:].count("1") for n in range(101)] # Indranil Ghosh, Apr 09 2017
    
  • Python
    # 3.10+
    def A011371(n): return n-n.bit_count() # Chai Wah Wu, Jul 09 2022

Formula

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

Extensions

Examples added by Hieronymus Fischer, Jun 06 2012

A008588 Nonnegative multiples of 6.

Original entry on oeis.org

0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222, 228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 294, 300, 306, 312, 318, 324, 330, 336, 342, 348
Offset: 0

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Author

Keywords

Comments

For n > 3, the number of squares on the infinite 3-column half-strip chessboard at <= n knight moves from any fixed point on the short edge.
Second differences of A000578. - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
A008615(a(n)) = n. - Reinhard Zumkeller, Feb 27 2008
A157176(a(n)) = A001018(n). - Reinhard Zumkeller, Feb 24 2009
These numbers can be written as the sum of four cubes (i.e., 6*n = (n+1)^3 + (n-1)^3 + (-n)^3 + (-n)^3). - Arkadiusz Wesolowski, Aug 09 2013
A122841(a(n)) > 0 for n > 0. - Reinhard Zumkeller, Nov 10 2013
Surface area of a cube with side sqrt(n). - Wesley Ivan Hurt, Aug 24 2014
a(n) is representable as a sum of three but not two consecutive nonnegative integers, e.g., 6 = 1 + 2 + 3, 12 = 3 + 4 + 5, 18 = 5 + 6 + 7, etc. (see A138591). - Martin Renner, Mar 14 2016 (Corrected by David A. Corneth, Aug 12 2016)
Numbers with three consecutive divisors: for some k, each of k, k+1, and k+2 divide n. - Charles R Greathouse IV, May 16 2016
Numbers k for which {phi(k),phi(2k),phi(3k)} is an arithmetic progression. - Ivan Neretin, Aug 12 2016

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 81.

Crossrefs

Essentially the same as A008458.
Cf. A044102 (subsequence).

Programs

Formula

From Vincenzo Librandi, Dec 24 2010: (Start)
a(n) = 6*n = 2*a(n-1) - a(n-2).
G.f.: 6*x/(1-x)^2. (End)
a(n) = Sum_{k>=0} A030308(n,k)*6*2^k. - Philippe Deléham, Oct 24 2011
a(n) = Sum_{k=2n-1..2n+1} k. - Wesley Ivan Hurt, Nov 22 2015
From Ilya Gutkovskiy, Aug 12 2016: (Start)
E.g.f.: 6*x*exp(x).
Convolution of A010722 and A057427.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/6 = A002162*A020793. (End)
a(n) = 6 * A001477(n). - David A. Corneth, Aug 12 2016
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