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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A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

Original entry on oeis.org

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

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Author

Keywords

Comments

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where m
Between every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive 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
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

Examples

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 2nd ed., 2001-2003; see Dim- and Dim+ on p. 98; Dividing Rulers, on pp. 436-437; The Ruler Game, pp. 469-470; Ruler Fours, Fives, ... Fifteens on p. 470.
  • L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E22.
  • A. M. Hinz, The Tower of Hanoi, in Algebras and combinatorics (Hong Kong, 1997), 277-289, Springer, Singapore, 1999.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589.
  • Andrew Schloss, "Towers of Hanoi" composition, in The Digital Domain. Elektra/Asylum Records 9 60303-2, 1983. Works by Jaffe (Finale to "Silicon Valley Breakdown"), McNabb ("Love in the Asylum"), Schloss ("Towers of Hanoi"), Mattox ("Shaman"), Rush, Moorer ("Lions are Growing") and others.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

Programs

  • Haskell
    a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
    -- Reinhard Zumkeller, Sep 27 2011
    
  • Haskell
    a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
    -- Walt Rorie-Baety, Mar 22 2013
    
  • MATLAB
    nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016
    
  • Magma
    [Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015
    
  • Maple
    A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
    # This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
    A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
    a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018
  • Mathematica
    Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
    Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
    IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
    myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
    Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)
  • PARI
    a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */
    
  • PARI
    a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */
    
  • PARI
    {a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */
    
  • PARI
    {a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */
    
  • Python
    def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017
    
  • Python
    A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020
    
  • Python
    def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022
    
  • Sage
    [valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015
    
  • Scheme
    (define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Extensions

Name edited following suggestion by David James Sycamore, Oct 05 2023

A007913 Squarefree part of n: a(n) is the smallest positive number m such that n/m is a square.

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15, 1, 17, 2, 19, 5, 21, 22, 23, 6, 1, 26, 3, 7, 29, 30, 31, 2, 33, 34, 35, 1, 37, 38, 39, 10, 41, 42, 43, 11, 5, 46, 47, 3, 1, 2, 51, 13, 53, 6, 55, 14, 57, 58, 59, 15, 61, 62, 7, 1, 65, 66, 67, 17, 69, 70, 71, 2, 73, 74, 3, 19, 77
Offset: 1

Author

R. Muller, Mar 15 1996

Keywords

Comments

Also called core(n). [Not to be confused with the squarefree kernel of n, A007947.]
Sequence read mod 4 gives A065882. - Philippe Deléham, Mar 28 2004
This is an arithmetic function and is undefined if n <= 0.
A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), lcm(A007947(b),c) = A007947(n) = "squarefree kernel" of n and bc = A019554(n) = "outer square root" of n. [Corrected by M. F. Hasler, Mar 01 2018]
If n > 1, the quantity f(n) = log(n/core(n))/log(n) satisfies 0 <= f(n) <= 1; f(n) = 0 when n is squarefree and f(n) = 1 when n is a perfect square. One can define n as being "epsilon-almost squarefree" if f(n) < epsilon. - Kurt Foster (drsardonicus(AT)earthlink.net), Jun 28 2008
a(n) is the smallest natural number m such that product of geometric mean of the divisors of n and geometric mean of the divisors of m are integers. Geometric mean of the divisors of number n is real number b(n) = Sqrt(n). a(n) = 1 for infinitely many n. a(n) = 1 for numbers from A000290: a(A000290(n)) = 1. For n = 8; b(8) = sqrt(8), a(n) = 2 because b(2) = sqrt(2); sqrt(8) * sqrt(2) = 4 (integer). - Jaroslav Krizek, Apr 26 2010
Dirichlet convolution of A010052 with the sequence of absolute values of A055615. - R. J. Mathar, Feb 11 2011
Booker, Hiary, & Keating outline a method for bounding (on the GRH) a(n) for large n using L-functions. - Charles R Greathouse IV, Feb 01 2013
According to the formula a(n) = n/A000188(n)^2, the scatterplot exhibits the straight lines y=x, y=x/4, y=x/9, ..., i.e., y=x/k^2 for all k=1,2,3,... - M. F. Hasler, May 08 2014
The Dirichlet inverse of this sequence is A008836(n) * A063659(n). - Álvar Ibeas, Mar 19 2015
a(n) = 1 if n is a square, a(n) = n if n is a product of distinct primes. - Zak Seidov, Jan 30 2016
All solutions of the Diophantine equation n*x=y^2 or, equivalently, G(n,x)=y, with G being the geometric mean, are of the form x=k^2*a(n), y=k*sqrt(n*a(n)), where k is a positive integer. - Stanislav Sykora, Feb 03 2016
If f is a multiplicative function then Sum_{d divides n} f(a(d)) is also multiplicative. For example, A010052(n) = Sum_{d divides n} mu(a(d)) and A046951(n) = Sum_{d divides n} mu(a(d)^2). - Peter Bala, Jan 24 2024

Crossrefs

See A000188, A007947, A008833, A019554, A117811 for related information, specific to n.
See A027746, A027748, A124010 for factorization data for n.
Analogous sequences: A050985, A053165, A055231.
Cf. A002734, A005117 (range of values), A059897, A069891 (partial sums), A090699, A350389.
Related to A006519 via A225546.

Programs

  • Haskell
    a007913 n = product $
                zipWith (^) (a027748_row n) (map (`mod` 2) $ a124010_row n)
    -- Reinhard Zumkeller, Jul 06 2012
    
  • Magma
    [ Squarefree(n) : n in [1..256] ]; // N. J. A. Sloane, Dec 23 2006
    
  • Maple
    A007913 := proc(n) local f,a,d; f := ifactors(n)[2] ; a := 1 ; for d in f do if type(op(2,d),'odd') then a := a*op(1,d) ; end if; end do: a; end proc: # R. J. Mathar, Mar 18 2011
    # second Maple program:
    a:= n-> mul(i[1]^irem(i[2], 2), i=ifactors(n)[2]):
    seq(a(n), n=1..100);  # Alois P. Heinz, Jul 20 2015
    seq(n / expand(numtheory:-nthpow(n, 2)), n=1..77);  # Peter Luschny, Jul 12 2022
  • Mathematica
    data = Table[Sqrt[n], {n, 1, 100}]; sp = data /. Sqrt[] -> 1; sfp = data/sp /. Sqrt[x] -> x (* Artur Jasinski, Nov 03 2008 *)
    Table[Times@@Power@@@({#[[1]],Mod[ #[[2]],2]}&/@FactorInteger[n]),{n,100}] (* Zak Seidov, Apr 08 2009 *)
    Table[{p, e} = Transpose[FactorInteger[n]]; Times @@ (p^Mod[e, 2]), {n, 100}] (* T. D. Noe, May 20 2013 *)
    Sqrt[#] /. (c_:1)*a_^(b_:0) -> (c*a^b)^2& /@ Range@100 (* Bill Gosper, Jul 18 2015 *)
  • PARI
    a(n)=core(n)
    
  • Python
    from sympy import factorint, prod
    def A007913(n):
        return prod(p for p, e in factorint(n).items() if e % 2)
    # Chai Wah Wu, Feb 03 2015
    
  • Sage
    [squarefree_part(n) for n in (1..77)] # Peter Luschny, Feb 04 2015

Formula

Multiplicative with a(p^k) = p^(k mod 2). - David W. Wilson, Aug 01 2001
a(n) modulo 2 = A035263(n); a(A036554(n)) is even; a(A003159(n)) is odd. - Philippe Deléham, Mar 28 2004
Dirichlet g.f.: zeta(2s)*zeta(s-1)/zeta(2s-2). - R. J. Mathar, Feb 11 2011
a(n) = n/( Sum_{k=1..n} floor(k^2/n)-floor((k^2 -1)/n) )^2. - Anthony Browne, Jun 06 2016
a(n) = rad(n)/a(n/rad(n)), where rad = A007947. This recurrence relation together with a(1) = 1 generate the sequence. - Velin Yanev, Sep 19 2017
From Peter Munn, Nov 18 2019: (Start)
a(k*m) = A059897(a(k), a(m)).
a(n) = n / A008833(n).
(End)
a(A225546(n)) = A225546(A006519(n)). - Peter Munn, Jan 04 2020
From Amiram Eldar, Mar 14 2021: (Start)
Theorems proven by Copil and Panaitopol (2007):
Lim sup_{n->oo} a(n+1)-a(n) = oo.
Lim inf_{n->oo} a(n+1)-a(n) = -oo.
Sum_{k=1..n} 1/a(k) ~ c*sqrt(n) + O(log(n)), where c = zeta(3/2)/zeta(3) (A090699). (End)
a(n) = A019554(n)^2/n. - Jianing Song, May 08 2022
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/30 = 0.328986... . - Amiram Eldar, Oct 25 2022
a(n) = A007947(A350389(n)). - Amiram Eldar, Jan 20 2024

Extensions

More terms from Michael Somos, Nov 24 2001
Definition reformulated by Daniel Forgues, Mar 24 2009

A002162 Decimal expansion of the natural logarithm of 2.

Original entry on oeis.org

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

Keywords

Comments

Newton calculated the first 16 terms of this sequence.
Area bounded by y = tan x, y = cot x, y = 0. - Clark Kimberling, Jun 26 2020
Choose four values independently and uniformly at random from the unit interval [0,1]. Sort them, and label them a,b,c,d from least to greatest (so that a b^2+c^2. - Akiva Weinberger, Dec 02 2024
Define the trihyperboloid to be the intersection of the three solid hyperboloids x^2+y^2-z^2<1, x^2-y^2+z^2<1, and -x^2+y^2+z^2<1. This fits perfectly within the cube [-1,1]^3. Then this is the ratio of the volume of the trihyperboloid to its bounding cube. - Akiva Weinberger, Dec 02 2024

Examples

			0.693147180559945309417232121458176568075500134360255254120680009493393...
		

References

  • G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
  • Calvin C. Clawson, Mathematical Mysteries: The Beauty and Magic of Numbers, Springer, 2013. See p. 227.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 250.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3, 2.21, 6.2, and 7.2.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 25 and appendix A, equations 25:14:3 and A:7:3 at pages 232, 670.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.

Crossrefs

Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.

Programs

  • Mathematica
    RealDigits[N[Log[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
    RealDigits[Log[2],10,120][[1]] (* Harvey P. Dale, Jan 25 2024 *)
  • PARI
    { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ Harry J. Smith, Apr 21 2009

Formula

log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = ((Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2)+396)/576. - Alexander R. Povolotsky, Jan 14 2011
From Alexander R. Povolotsky, Dec 16 2008: (Start)
log(2) = 105*(319/44100 - Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7))).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63). (End)
log(2) = Sum_{k>=1} A191907(2,k)/k. - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03 2015: (Start)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
From Peter Bala, Oct 30 2016: (Start)
Asymptotic expansions:
for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
log(2) = Sum_{n>=1} A006519(n) / ((1 + 2^A006519(n)) * A000265(n) * (1 + A000265(n))). - Nicolas Nagel, Mar 19 2018
From Amiram Eldar, Jul 02 2020: (Start)
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Sum_{k>=1} 1/A002939(k).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
From Peter Bala, Nov 14 2020: (Start)
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - Peter Bala, Jan 14 2022
log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - Bernard Schott, Jul 11 2022
From Peter Bala, Oct 22 2023: (Start)
log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2 = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
From Peter Bala, Nov 13 2023: (Start)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
= 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
= 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
log(2) = K_{n>=0} (n^2 + [n=0])/1, where K is the Gauss notation for an infinite continued fraction. In the expanded form, log(2) = 1/(1 + 1/(1 + 4/(1 + 9/1 + 16/(1 + 25/(1 + ... (see Clawson at p. 227). - Stefano Spezia, Jul 01 2024
log(2) = lim_{n->oo} Sum_{k=1..n} 1/(n + k) = lim_{x->0} (2^x - 1)/x = lim_{x->0} (2^x - 2^(-x))/(2*x) (see Finch). - Stefano Spezia, Oct 19 2024
From Colin Linzer, Nov 08 2024: (Start)
log(2) = Integral_{t=0...oo} (1 - tanh(t)) dt.
log(2) = Integral_{t=0...1} arctanh(t) dt.
log(2) = (1/2) * Integral_{t=-1...1} |arctanh(t)| dt. (End)
log(2) = 1 + Sum_{n >= 1} (-1)^n/(n*(4*n^2 - 1)) = 1/2 + (1/2)*Sum_{n >= 1} 1/(n*(4*n^2 - 1)). - Peter Bala, Jan 07 2025
log(2) = Integral_{x=0..1} Integral_{y=0..1} 1/((1 - x*y)*(1 + x)*(1 + y)) dy dx. - Kritsada Moomuang, May 22 2025

A204892 Least k such that n divides s(k)-s(j) for some j in [1,k), where s(k)=prime(k).

Original entry on oeis.org

2, 3, 3, 4, 4, 5, 7, 5, 5, 6, 6, 7, 10, 7, 7, 8, 8, 9, 13, 9, 9, 10, 16, 10, 16, 10, 10, 11, 11, 12, 19, 12, 20, 12, 12, 13, 22, 13, 13, 14, 14, 15, 24, 15, 15, 16, 25, 16, 26, 16, 16, 17, 29, 17, 30, 17, 17, 18, 18, 19, 31, 19, 32, 19, 19, 20, 33, 20, 20, 21
Offset: 1

Author

Clark Kimberling, Jan 20 2012

Keywords

Comments

Suppose that (s(i)) is a strictly increasing sequence in the set N of positive integers. For i in N, let r(h) be the residue of s(i+h)-s(i) mod n, for h=1,2,...,n+1. There are at most n distinct residues r(h), so that there must exist numbers h and h' such that r(h)=r(h'), where 0<=h
Corollary: for each n, there are infinitely many pairs (j,k) such that n divides s(k)-s(j), and this result holds if s is assumed unbounded, rather than strictly increasing.
Guide to related sequences:
...
s(n)=prime(n), primes
... k(n), j(n): A204892, A204893
... s(k(n)),s(j(n)): A204894, A204895
... s(k(n))-s(j(n)): A204896, A204897
s(n)=prime(n+1), odd primes
... k(n), j(n): A204900, A204901
... s(k(n)),s(j(n)): A204902, A204903
... s(k(n))-s(j(n)): A109043(?), A000034(?)
s(n)=prime(n+2), primes >=5
... k(n), j(n): A204908, A204909
... s(k(n)),s(j(n)): A204910, A204911
... s(k(n))-s(j(n)): A109043(?), A000034(?)
s(n)=prime(n)*prime(n+1) product of consecutive primes
... k(n), j(n): A205146, A205147
... s(k(n)),s(j(n)): A205148, A205149
... s(k(n))-s(j(n)): A205150, A205151
s(n)=(prime(n+1)+prime(n+2))/2: averages of odd primes
... k(n), j(n): A205153, A205154
... s(k(n)),s(j(n)): A205372, A205373
... s(k(n))-s(j(n)): A205374, A205375
s(n)=2^(n-1), powers of 2
... k(n), j(n): A204979, A001511(?)
... s(k(n)),s(j(n)): A204981, A006519(?)
... s(k(n))-s(j(n)): A204983(?), A204984
s(n)=2^n, powers of 2
... k(n), j(n): A204987, A204988
... s(k(n)),s(j(n)): A204989, A140670(?)
... s(k(n))-s(j(n)): A204991, A204992
s(n)=C(n+1,2), triangular numbers
... k(n), j(n): A205002, A205003
... s(k(n)),s(j(n)): A205004, A205005
... s(k(n))-s(j(n)): A205006, A205007
s(n)=n^2, squares
... k(n), j(n): A204905, A204995
... s(k(n)),s(j(n)): A204996, A204997
... s(k(n))-s(j(n)): A204998, A204999
s(n)=(2n-1)^2, odd squares
... k(n), j(n): A205378, A205379
... s(k(n)),s(j(n)): A205380, A205381
... s(k(n))-s(j(n)): A205382, A205383
s(n)=n(3n-1), pentagonal numbers
... k(n), j(n): A205138, A205139
... s(k(n)),s(j(n)): A205140, A205141
... s(k(n))-s(j(n)): A205142, A205143
s(n)=n(2n-1), hexagonal numbers
... k(n), j(n): A205130, A205131
... s(k(n)),s(j(n)): A205132, A205133
... s(k(n))-s(j(n)): A205134, A205135
s(n)=C(2n-2,n-1), central binomial coefficients
... k(n), j(n): A205010, A205011
... s(k(n)),s(j(n)): A205012, A205013
... s(k(n))-s(j(n)): A205014, A205015
s(n)=(1/2)C(2n,n), (1/2)*(central binomial coefficients)
... k(n), j(n): A205386, A205387
... s(k(n)),s(j(n)): A205388, A205389
... s(k(n))-s(j(n)): A205390, A205391
s(n)=n(n+1), oblong numbers
... k(n), j(n): A205018, A205028
... s(k(n)),s(j(n)): A205029, A205030
... s(k(n))-s(j(n)): A205031, A205032
s(n)=n!, factorials
... k(n), j(n): A204932, A204933
... s(k(n)),s(j(n)): A204934, A204935
... s(k(n))-s(j(n)): A204936, A204937
s(n)=n!!, double factorials
... k(n), j(n): A204982, A205100
... s(k(n)),s(j(n)): A205101, A205102
... s(k(n))-s(j(n)): A205103, A205104
s(n)=3^n-2^n
... k(n), j(n): A205000, A205107
... s(k(n)),s(j(n)): A205108, A205109
... s(k(n))-s(j(n)): A205110, A205111
s(n)=Fibonacci(n+1)
... k(n), j(n): A204924, A204925
... s(k(n)),s(j(n)): A204926, A204927
... s(k(n))-s(j(n)): A204928, A204929
s(n)=Fibonacci(2n-1)
... k(n), j(n): A205442, A205443
... s(k(n)),s(j(n)): A205444, A205445
... s(k(n))-s(j(n)): A205446, A205447
s(n)=Fibonacci(2n)
... k(n), j(n): A205450, A205451
... s(k(n)),s(j(n)): A205452, A205453
... s(k(n))-s(j(n)): A205454, A205455
s(n)=Lucas(n)
... k(n), j(n): A205114, A205115
... s(k(n)),s(j(n)): A205116, A205117
... s(k(n))-s(j(n)): A205118, A205119
s(n)=n*(2^(n-1))
... k(n), j(n): A205122, A205123
... s(k(n)),s(j(n)): A205124, A205125
... s(k(n))-s(j(n)): A205126, A205127
s(n)=ceiling[n^2/2]
... k(n), j(n): A205394, A205395
... s(k(n)),s(j(n)): A205396, A205397
... s(k(n))-s(j(n)): A205398, A205399
s(n)=floor[(n+1)^2/2]
... k(n), j(n): A205402, A205403
... s(k(n)),s(j(n)): A205404, A205405
... s(k(n))-s(j(n)): A205406, A205407

Examples

			Let s(k)=prime(k).  As in A204890, the ordering of differences s(k)-s(j), follows from the arrangement shown here:
k...........1..2..3..4..5...6...7...8...9
s(k)........2..3..5..7..11..13..17..19..23
...
s(k)-s(1)......1..3..5..9..11..15..17..21..27
s(k)-s(2).........2..4..8..10..14..16..20..26
s(k)-s(3)............2..6..8...12..14..18..24
s(k)-s(4)...............4..6...10..12..16..22
...
least (k,j) such that 1 divides s(k)-s(j) for some j is (2,1), so a(1)=2.
least (k,j) such that 2 divides s(k)-s(j): (3,2), so a(2)=3.
least (k,j) such that 3 divides s(k)-s(j): (3,1), so a(3)=3.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := s[n] = Prime[n]; z1 = 400; z2 = 50;
    Table[s[n], {n, 1, 30}]          (* A000040 *)
    u[m_] := u[m] = Flatten[Table[s[k] - s[j],
       {k, 2, z1}, {j, 1, k - 1}]][[m]]
    Table[u[m], {m, 1, z1}]          (* A204890 *)
    v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]
    w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]
    d[n_] := d[n] = First[Delete[w[n],
       Position[w[n], 0]]]
    Table[d[n], {n, 1, z2}]          (* A204891 *)
    k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]
    m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]
    j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2
    Table[k[n], {n, 1, z2}]          (* A204892 *)
    Table[j[n], {n, 1, z2}]          (* A204893 *)
    Table[s[k[n]], {n, 1, z2}]       (* A204894 *)
    Table[s[j[n]], {n, 1, z2}]       (* A204895 *)
    Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204896 *)
    Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204897 *)
    (* Program 2: generates A204892 and A204893 rapidly *)
    s = Array[Prime[#] &, 120];
    lk = Table[NestWhile[# + 1 &, 1, Min[Table[Mod[s[[#]] - s[[j]], z], {j, 1, # - 1}]] =!= 0 &], {z, 1, Length[s]}]
    Table[NestWhile[# + 1 &, 1, Mod[s[[lk[[j]]]] - s[[#]], j] =!= 0 &], {j, 1, Length[lk]}]
    (* Peter J. C. Moses, Jan 27 2012 *)
  • PARI
    a(n)=forprime(p=n+2,,forstep(k=p%n,p-1,n,if(isprime(k), return(primepi(p))))) \\ Charles R Greathouse IV, Mar 20 2013

A001316 Gould's sequence: a(n) = Sum_{k=0..n} (binomial(n,k) mod 2); number of odd entries in row n of Pascal's triangle (A007318); a(n) = 2^A000120(n).

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32, 32, 64, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32
Offset: 0

Keywords

Comments

Also called Dress's sequence.
This sequence might be better called Glaisher's sequence, since James Glaisher showed that odd binomial coefficients are counted by 2^A000120(n) in 1899. - Eric Rowland, Mar 17 2017 [However, the name "Gould's sequence" is deeply entrenched in the literature. - N. J. A. Sloane, Mar 17 2017] [Named after the American mathematician Henry Wadsworth Gould (b. 1928). - Amiram Eldar, Jun 19 2021]
All terms are powers of 2. The first occurrence of 2^k is at n = 2^k - 1; e.g., the first occurrence of 16 is at n = 15. - Robert G. Wilson v, Dec 06 2000
a(n) is the highest power of 2 dividing binomial(2n,n) = A000984(n). - Benoit Cloitre, Jan 23 2002
Also number of 1's in n-th row of triangle in A070886. - Hans Havermann, May 26 2002. Equivalently, number of live cells in generation n of a one-dimensional cellular automaton, Rule 90, starting with a single live cell. - Ben Branman, Feb 28 2009. Ditto for Rule 18. - N. J. A. Sloane, Aug 09 2014. This is also the odd-rule cellular automaton defined by OddRule 003 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015
Also number of numbers k, 0<=k<=n, such that (k OR n) = n (bitwise logical OR): a(n) = #{k : T(n,k)=n, 0<=k<=n}, where T is defined as in A080098. - Reinhard Zumkeller, Jan 28 2003
To construct the sequence, start with 1 and use the rule: If k >= 0 and a(0),a(1),...,a(2^k-1) are the first 2^k terms, then the next 2^k terms are 2*a(0),2*a(1),...,2*a(2^k-1). - Benoit Cloitre, Jan 30 2003
Also, numerator((2^k)/k!). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
The odd entries in Pascal's triangle form the Sierpiński Gasket (a fractal). - Amarnath Murthy, Nov 20 2004
Row sums of Sierpiński's Gasket A047999. - Johannes W. Meijer, Jun 05 2011
Fixed point of the morphism "1" -> "1,2", "2" -> "2,4", "4" -> "4,8", ..., "2^k" -> "2^k,2^(k+1)", ... starting with a(0) = 1; 1 -> 12 -> 1224 -> = 12242448 -> 122424482448488(16) -> ... . - Philippe Deléham, Jun 18 2005
a(n) = number of 1's of stage n of the one-dimensional cellular automaton with Rule 90. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 01 2006
a(33)..a(63) = A117973(1)..A117973(31). - Stephen Crowley, Mar 21 2007
Or the number of solutions of the equation: A000120(x) + A000120(n-x) = A000120(n). - Vladimir Shevelev, Jul 19 2009
For positive n, a(n) equals the denominator of the permanent of the n X n matrix consisting entirely of (1/2)'s. - John M. Campbell, May 26 2011
Companions to A001316 are A048896, A105321, A117973, A151930 and A191488. They all have the same structure. We observe that for all these sequences a((2*n+1)*2^p-1) = C(p)*A001316(n), p >= 0. If C(p) = 2^p then a(n) = A001316(n), if C(p) = 1 then a(n) = A048896(n), if C(p) = 2^p+2 then a(n) = A105321(n+1), if C(p) = 2^(p+1) then a(n) = A117973(n), if C(p) = 2^p-2 then a(n) = (-1)*A151930(n) and if C(p) = 2^(p+1)+2 then a(n) = A191488(n). Furthermore for all a(2^p - 1) = C(p). - Johannes W. Meijer, Jun 05 2011
a(n) = number of zeros in n-th row of A219463 = number of ones in n-th row of A047999. - Reinhard Zumkeller, Nov 30 2012
This is the Run Length Transform of S(n) = {1,2,4,8,16,...} (cf. A000079). The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product). - N. J. A. Sloane, Sep 05 2014
A105321(n+1) = a(n+1) + a(n). - Reinhard Zumkeller, Nov 14 2014
a(n) = A261363(n,n) = number of distinct terms in row n of A261363 = number of odd terms in row n+1 of A261363. - Reinhard Zumkeller, Aug 16 2015
From Gary W. Adamson, Aug 26 2016: (Start)
A production matrix for the sequence is lim_{k->infinity} M^k, the left-shifted vector of M:
1, 0, 0, 0, 0, ...
2, 0, 0, 0, 0, ...
0, 1, 0, 0, 0, ...
0, 2, 0, 0, 0, ...
0, 0, 1, 0, 0, ...
0, 0, 2, 0, 0, ...
0, 0, 0, 1, 0, ...
...
The result is equivalent to the g.f. of Apr 06 2003: Product_{k>=0} (1 + 2*z^(2^k)). (End)
Number of binary palindromes of length n for which the first floor(n/2) symbols are themselves a palindrome (Ji and Wilf 2008). - Jeffrey Shallit, Jun 15 2017

Examples

			Has a natural structure as a triangle:
  1,
  2,
  2,4,
  2,4,4,8,
  2,4,4,8,4,8,8,16,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32,64,
  ...
The rows converge to A117973.
From _Omar E. Pol_, Jun 07 2009: (Start)
Also, triangle begins:
   1;
   2,2;
   4,2,4,4;
   8,2,4,4,8,4,8,8;
  16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16;
  32,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32;
  64,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,...
(End)
G.f. = 1 + 2*x + 2*x^2 + 4*x^3 + 2*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 2*x^8 + ... - _Michael Somos_, Aug 26 2015
		

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, p. 75ff.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 145-151.
  • James W. L. Glaisher, On the residue of a binomial-theorem coefficient with respect to a prime modulus, Quarterly Journal of Pure and Applied Mathematics, Vol. 30 (1899), pp. 150-156.
  • H. W. Gould, Exponential Binomial Coefficient Series. Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sep 1961.
  • Olivier Martin, Andrew M. Odlyzko, and Stephen Wolfram, Algebraic properties of cellular automata, Comm. Math. Physics, Vol. 93 (1984), pp. 219-258. Reprinted in Theory and Applications of Cellular Automata, S Wolfram, Ed., World Scientific, 1986, pp. 51-90 and in Cellular Automata and Complexity: Collected Papers of Stephen Wolfram, Addison-Wesley, 1994, pp. 71-113
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991, page 383.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Andrew Wuensche, Exploring Discrete Dynamics, Luniver Press, 2011. See Fig. 2.3.

Crossrefs

Equals left border of triangle A166548. - Gary W. Adamson, Oct 16 2009
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.
For partial sums see A006046. For first differences see A151930.
This is the numerator of 2^n/n!, while A049606 gives the denominator.
If we subtract 1 from the terms we get a pair of essentially identical sequences, A038573 and A159913.
A163000 and A163577 count binomial coefficients with 2-adic valuation 1 and 2. A275012 gives a measure of complexity of these sequences. - Eric Rowland, Mar 15 2017
Cf. A286575 (run-length transform), A368655 (binomial transform), also A037445.

Programs

  • Haskell
    import Data.List (transpose)
    a001316 = sum . a047999_row  -- Reinhard Zumkeller, Nov 24 2012
    a001316_list = 1 : zs where
       zs = 2 : (concat $ transpose [zs, map (* 2) zs])
    -- Reinhard Zumkeller, Aug 27 2014, Sep 16 2011
    (Sage, Python)
    from functools import cache
    @cache
    def A001316(n):
        if n <= 1: return n+1
        return A001316(n//2) << n%2
    print([A001316(n) for n in range(88)])  # Peter Luschny, Nov 19 2012
    
  • Maple
    A001316 := proc(n) local k; add(binomial(n,k) mod 2, k=0..n); end;
    S:=[1]; S:=[op(S),op(2*s)]; # repeat ad infinitum!
    a := n -> 2^add(i,i=convert(n,base,2)); # Peter Luschny, Mar 11 2009
  • Mathematica
    Table[ Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ], {n, 0, 100} ]
    Nest[ Join[#, 2#] &, {1}, 7] (* Robert G. Wilson v, Jan 24 2006 and modified Jul 27 2014 *)
    Map[Function[Apply[Plus,Flatten[ #1]]], CellularAutomaton[90,{{1},0},100]] (* Produces counts of ON cells. N. J. A. Sloane, Aug 10 2009 *)
    ArrayPlot[CellularAutomaton[90, {{1}, 0}, 20]] (* Illustration of first 20 generations. - N. J. A. Sloane, Aug 14 2014 *)
    Table[2^(RealDigits[n - 1, 2][[1]] // Total), {n, 1, 100}] (* Gabriel C. Benamy, Dec 08 2009 *)
    CoefficientList[Series[Exp[2*x], {x, 0, 100}], x] // Numerator (* Jean-François Alcover, Oct 25 2013 *)
    Count[#,?OddQ]&/@Table[Binomial[n,k],{n,0,90},{k,0,n}] (* _Harvey P. Dale, Sep 22 2015 *)
    2^DigitSum[Range[0, 100], 2] (* Paolo Xausa, Jul 31 2025 *)
  • PARI
    {a(n) = if( n<0, 0, numerator(2^n / n!))};
    
  • PARI
    A001316(n)=1<M. F. Hasler, May 03 2009
    
  • PARI
    a(n)=2^hammingweight(n) \\ Charles R Greathouse IV, Jan 04 2013
    
  • Python
    def A001316(n):
        return 2**bin(n)[2:].count("1") # Indranil Ghosh, Feb 06 2017
    
  • Python
    def A001316(n): return 1<Karl-Heinz Hofmann, Aug 01 2025
    
  • Python
    import numpy # (version >= 2.0.0)
    n_up_to = 2**22
    A000079 = 1 << numpy.arange(n_up_to.bit_length())
    A001316 = A000079[numpy.bitwise_count(numpy.arange(n_up_to))]
    print(A001316[0:100]) # Karl-Heinz Hofmann, Aug 01 2025
    
  • Scheme
    (define (A001316 n) (let loop ((n n) (z 1)) (cond ((zero? n) z) ((even? n) (loop (/ n 2) z)) (else (loop (/ (- n 1) 2) (* z 2)))))) ;; Antti Karttunen, May 29 2017

Formula

a(n) = 2^A000120(n).
a(0) = 1; for n > 0, write n = 2^i + j where 0 <= j < 2^i; then a(n) = 2*a(j).
a(n) = 2*a(n-1)/A006519(n) = A000079(n)*A049606(n)/A000142(n).
a(n) = A038573(n) + 1.
G.f.: Product_{k>=0} (1+2*z^(2^k)). - Ralf Stephan, Apr 06 2003
a(n) = Sum_{i=0..2*n} (binomial(2*n, i) mod 2)*(-1)^i. - Benoit Cloitre, Nov 16 2003
a(n) mod 3 = A001285(n). - Benoit Cloitre, May 09 2004
a(n) = 2^n - 2*Sum_{k=0..n} floor(binomial(n, k)/2). - Paul Barry, Dec 24 2004
a(n) = Product_{k=0..log_2(n)} 2^b(n, k), b(n, k) = coefficient of 2^k in binary expansion of n. - Paul D. Hanna
Sum_{k=0..n-1} a(k) = A006046(n).
a(n) = n/2 + 1/2 + (1/2)*Sum_{k=0..n} (-(-1)^binomial(n,k)). - Stephen Crowley, Mar 21 2007
G.f. for a(n)/A156769(n): (1/2)*z^(1/2)*sinh(2*z^(1/2)). - Johannes W. Meijer, Feb 20 2009
Equals infinite convolution product of [1,2,0,0,0,0,0,0,0] aerated (A000079 - 1) times, i.e., [1,2,0,0,0,0,0,0,0] * [1,0,2,0,0,0,0,0,0] * [1,0,0,0,2,0,0,0,0]. - Mats Granvik, Gary W. Adamson, Oct 02 2009
a(n) = f(n, 1) with f(x, y) = if x = 0 then y otherwise f(floor(x/2), y*(1 + x mod 2)). - Reinhard Zumkeller, Nov 21 2009
a(n) = 2^(number of 1's in binary form of (n-1)). - Gabriel C. Benamy, Dec 08 2009
a((2*n+1)*2^p-1) = (2^p)*a(n), p >= 0. - Johannes W. Meijer, Jun 05 2011
a(n) = A000120(A001317(n)). - Reinhard Zumkeller, Nov 24 2012
a(n) = A226078(n,1). - Reinhard Zumkeller, May 25 2013
a(n) = lcm(n!, 2^n) / n!. - Daniel Suteu, Apr 28 2017
a(n) = A061142(A005940(1+n)). - Antti Karttunen, May 29 2017
a(0) = 1, a(2*n) = a(n), a(2*n+1) = 2*a(n). - Daniele Parisse, Feb 15 2024
a(n*m) <= a(n)^A000120(m). - Joe Amos, Mar 27 2025

Extensions

Additional comments from Henry Bottomley, Mar 12 2001
Further comments from N. J. A. Sloane, May 30 2009

A003188 Decimal equivalent of Gray code for n.

Original entry on oeis.org

0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 25, 27, 26, 30, 31, 29, 28, 20, 21, 23, 22, 18, 19, 17, 16, 48, 49, 51, 50, 54, 55, 53, 52, 60, 61, 63, 62, 58, 59, 57, 56, 40, 41, 43, 42, 46, 47, 45, 44, 36, 37, 39, 38, 34, 35, 33, 32, 96, 97, 99, 98, 102, 103, 101
Offset: 0

Keywords

Comments

Inverse of sequence A006068 considered as a permutation of the nonnegative integers, i.e., A006068(A003188(n)) = n = A003188(A006068(n)). - Howard A. Landman, Sep 25 2001
Restricts to a permutation of each {2^(i - 1) .. 2^i - 1}. - Jason Kimberley, Apr 02 2012
a(n) mod 2 = floor(((n + 1) mod 4) / 2), see also A021913. - Reinhard Zumkeller, Apr 28 2012
Invented by Emile Baudot (1845-1903), originally called a "cyclic-permuted" code. Gray codes are named after Frank Gray, who patented their use for shaft encoders in 1953. [F. Gray, "Pulse Code Communication", U.S. Patent 2,632,058, March 17, 1953.] - Robert G. Wilson v, Jun 22 2014
For n >= 2, let G_n be the graph whose vertices are labeled as 0,1,...,2^n-1, and two vertices are adjacent if and only if their binary expansions differ in exactly one bit, then a(0),a(1),...,a(2^n-1),a(0) is a Hamilton cycle in G_n. - Jianing Song, Jun 01 2022

Examples

			For n = 13, the binary reflected Gray code representation of n is '1011' and 1011_2 = 11_10. So, a(13) = 11. - _Indranil Ghosh_, Jan 23 2017
		

References

  • M. Gardner, Mathematical Games, Sci. Amer. Vol. 227 (No. 2, Feb. 1972), p. 107.
  • M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(2*A003714(n)) = 3*A003714(n) for all n. - Antti Karttunen, Apr 26 1999
Cf. A014550 (in binary), A055975 (first differences), A048724 (even bisection), A065621 (odd bisection).

Programs

  • C
    int a(int n) { return n ^ (n>>1); }
    
  • Haskell
    import Data.Bits (xor, shiftR)
    a003188 n = n `xor` (shiftR n 1) :: Integer
    -- Reinhard Zumkeller, May 26 2013, Apr 28 2012
    
  • Magma
    // A recursive algorithm
    N := 10; s := [[]];
    for n in [1..N] do
    for j in [#s..1 by -1] do
       Append(~s,Append(s[j],1));
       Append(~s[j],0);
    end for;
    end for;
    [SequenceToInteger(b,2):b in s]; // Jason Kimberley, Apr 02 2012
    
  • Magma
    // A direct algorithm
    I2B := func< i | [b eq 1: b in IntegerToSequence(i,2)]>;
    B2I := func< s | SequenceToInteger([b select 1 else 0:b in s],2)>;
    [B2I(Xor(I2B(i),I2B(i div 2)cat[false])):i in [1..127]]; //Jason Kimberley, Apr 02 2012
    
  • Maple
    with(combinat); graycode(6); # to produce first 64 terms
    printf(cat(` %.6d`$64), op(map(convert, graycode(6), binary))); lprint(); # to produce binary strings
    # alternative:
    read("transforms"):
    A003188 := proc(n)
        XORnos(n,floor(n/2)) ;
    end proc: # R. J. Mathar, Mar 09 2015
    # another Maple program:
    a:= n-> Bits[Xor](n, iquo(n, 2)):
    seq(a(n), n=0..70);  # Alois P. Heinz, Aug 16 2020
  • Mathematica
    f[n_] := BitXor[n, Floor[n/2]]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 09 2010 *)
  • PARI
    a(n)=bitxor(n,n>>1);
    
  • PARI
    a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2)*2^valuation(k,2))
    
  • Python
    def A003188(n):
        return int(bin(n^(n//2))[2:],2) # Indranil Ghosh, Jan 23 2017
    
  • Python
    def A003188(n): return n^ n>>1 # Chai Wah Wu, Jun 29 2022
    
  • R
    maxn <- 63 # by choice
    a <- 1
    for(n in 1:maxn){ a[2*n  ] <- 2*a[n] + (n%%2 != 0)
                      a[2*n+1] <- 2*a[n] + (n%%2 == 0)}
    (a <- c(0,a))
    # Yosu Yurramendi, Apr 10 2020
    (C#)
    static uint a(this uint n) => (n >> 1) ^ n; // Frank Hollstein, Mar 12 2021

Formula

a(n) = 2*a(floor(n/2)) + A021913(n - 1). - Henry Bottomley, Apr 05 2001
a(n) = n XOR floor(n/2), where XOR is the binary exclusive OR operator. - Paul D. Hanna, Jun 04 2002
G.f.: (1/(1-x)) * Sum_{k>=0} 2^k*x^2^k/(1 + x^2^(k+1)). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n) = 2a(n) + [n odd], a(2n + 1) = 2a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(0) = 0, a(n) = 2 a(floor(n/2)) + mod(floor((n + 1)/2), 2).
a(n) = Sum_{k=1..n} 2^A007814(k) * (-1)^((k/2^A007814(k) - 1)/2). - Ralf Stephan, Oct 29 2003
a(0) = 0, a(n + 1) = a(n) XOR 2^A007814(n) - Jaume Simon Gispert (jaume(AT)nuem.com), Sep 11 2004
Inverse of sequence A006068. - Philippe Deléham, Apr 29 2005
a(n) = a(n-1) XOR A006519(n). - Franklin T. Adams-Watters, Jul 18 2011
From Mikhail Kurkov, Sep 27 2023: (Start)
a(2^m + k) = a(2^m - k - 1) + 2^m for 0 <= k < 2^m, m >= 0.
a(n) = a(A053645(A054429(n))) + A053644(n) for n > 0.
a(n) = A063946(a(A053645(n)) + A053644(n)) for n > 0. (End)

A000118 Number of ways of writing n as a sum of 4 squares; also theta series of four-dimensional cubic lattice Z^4.

Original entry on oeis.org

1, 8, 24, 32, 24, 48, 96, 64, 24, 104, 144, 96, 96, 112, 192, 192, 24, 144, 312, 160, 144, 256, 288, 192, 96, 248, 336, 320, 192, 240, 576, 256, 24, 384, 432, 384, 312, 304, 480, 448, 144, 336, 768, 352, 288, 624, 576, 384, 96, 456, 744, 576, 336, 432, 960, 576, 192
Offset: 0

Keywords

Comments

a^2 + b^2 + c^2 + d^2 is one of Ramanujan's 54 universal quaternary quadratic forms. - Michael Somos, Apr 01 2008
a(n) is also the number of quaternions q = a + bi + cj + dk, where a, b, c, d are integers, such that a^2 + b^2 + c^2 + d^2 = n (i.e., so that n is the norm of q). These are Lipschitz integer quaternions. - Rick L. Shepherd, Mar 27 2009
Number 5 and 35 of the 126 eta-quotients listed in Table 1 of Williams 2012. - Michael Somos, Nov 10 2018
This is the convolution square of A004018. - Pierre Abbat, May 15 2023

Examples

			G.f. = 1 + 8*q + 24*q^2 + 32*q^3 + 24*q^4 + 48*q^5 + 96*q^6 + 64*q^7 + 24*q^8 + ...
a(1)=8 counts 1 = 1^2 + 0^2 + 0^2 + 0^2 = 0^2 + 1^2 + 0^2 + 0^2 = 0^2 + 0^2 + 1^2 + 0^2 = 0^2 + 0^2 + 0^2 + 1^2 and 4 more sums where 1^2 is replaced by (-1)^2. - _R. J. Mathar_, May 16 2023
		

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996, ch. 8, pp. 231-2.
  • J. H. Conway and N. J. A. Sloane, Sphere Packing, Lattices and Groups, Springer-Verlag, p. 108, Eq. (49).
  • N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 78, Eq. (32.28). See also top of p. 94.
  • E. Freitag and R. Busam, Funktionentheorie 1, 4. Auflage, Springer, 2006, p. 392.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 314, Theorem 386.
  • Carlos J. Moreno and Samuel S. Wagstaff, Jr., Sums of Squares of integers, Chapman & Hall/CRC, 2006, p. 29.
  • S. Ramanujan, Collected Papers, Chap. 20, Cambridge Univ. Press 1927 (Proceedings of the Camb. Phil. Soc., 19 (1917) 11-21).

Crossrefs

Row d=4 of A122141 and of A319574, 4th column of A286815.
For number of solutions to a^2+b^2+c^2+k*d^2=n for k=1, 2, 3, 4, 5, 6, 7, 8, 12, see A000118, A236928, A236926, A236923, A236930, A236931, A236932, A236927, A236933.

Programs

  • Haskell
    a000118 0 = 1
    a000118 n = 8 * a046897 n  -- Reinhard Zumkeller, Aug 12 2015
    
  • Julia
    # JacobiTheta3 is defined in A000122.
    A000118List(len) = JacobiTheta3(len, 4)
    A000118List(57) |> println # Peter Luschny, Mar 12 2018
    
  • MATLAB
    a(n) = 8 * sum(find(mod(n,1:n)==0 & mod(1:n,4))) + (n==0) % David Mellinger, Aug 04 2025
  • Magma
    A := Basis( ModularForms( Gamma0(4), 2), 57); A[1] + 8*A[2]; /* Michael Somos, Aug 21 2014 */
    
  • Maple
    (add(q^(m^2),m=-10..10))^4; seq(coeff(%,q,n), n=0..50);
    # Alternative:
    A000118list := proc(len) series(JacobiTheta3(0, x)^4, x, len+1);
    seq(coeff(%, x, j), j=0..len-1) end: A000118list(57); # Peter Luschny, Oct 02 2018
  • Mathematica
    Table[SquaresR[4, n], {n, 0, 46}]
    a[ n_] :=  SeriesCoefficient[ EllipticTheta[ 3, 0, q]^4, {q, 0, n}]; (* Michael Somos, Jun 12 2014 *)
    a[ n_] := If[ n < 1, Boole[ n == 0], 8 Sum[ If[ Mod[ d, 4] > 0, d, 0], {d, Divisors @ n }]]; (* Michael Somos, Feb 20 2015 *)
    QP = QPochhammer; CoefficientList[QP[-q]^8/QP[q^2]^4 + O[q]^60, q] (* Jean-François Alcover, Nov 24 2015 *)
  • PARI
    {a(n) = if( n<1, n==0, 8 * sumdiv( n, d, if( d%4, d)))}; /* Michael Somos, Apr 01 2003 */
    
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x^2 + A)^5 / (eta(x + A)^2 * eta(x^4 + A)^2))^4, n))}; /* Michael Somos, Apr 01 2008 */
    
  • PARI
    q='q+O('q^66); Vec((eta(q^2)^5/(eta(q)^2*eta(q^4)^2))^4) /* Joerg Arndt, Apr 08 2013 */
    
  • PARI
    a(n) = 8*sigma(n) - if (n % 4, 0, 32*sigma(n/4)); \\ Michel Marcus, Jul 13 2016
    
  • Python
    from sympy import divisors
    def a(n): return 1 if n==0 else 8*sum(d for d in divisors(n) if d%4 != 0)
    print([a(n) for n in range(57)]) # Michael S. Branicky, Jan 08 2021
    
  • Python
    from sympy import divisor_sigma
    def A000118(n): return 1 if n == 0 else 8*divisor_sigma(n) if n % 2 else 24*divisor_sigma(int(bin(n)[2:].rstrip('0'),2)) # Chai Wah Wu, Jun 27 2022
    
  • Sage
    A = ModularForms( Gamma0(4), 2, prec=57) . basis(); A[0] + 8*A[1]; # Michael Somos, Jun 12 2014
    
  • Sage
    Q = DiagonalQuadraticForm(ZZ, [1]*4)
    Q.representation_number_list(60) # Peter Luschny, Jun 20 2014
    

Formula

G.f.: theta_3(q)^4 = (Product_{n>=1} (1-q^(2n))*(1+q^(2n-1))^2)^4 = eta(-q)^8/eta(q^2)^4; eta = Dedekind's function.
a(n) = 8*sigma(n) - 32*sigma(n/4) for n > 0, where the latter term is 0 if n is not a multiple of 4.
Euler transform of period 4 sequence [8, -12, 8, -4, ...]. - Michael Somos, Dec 16 2002
G.f. A(x) satisfies 0 = f(A(x), A(x^3), A(x^9)) where f(u, v, w) = v^4 - 30*u*v^2*w + 12*u*v*w*(u + 9*w) - u*w*(u^2 + 9*w*u + 81*w^2). - Michael Somos, Nov 02 2006
G.f. is a period 1 Fourier series which satisfies f(-1/(4*t)) = 4*(t/i)^2*f(t) where q = exp(2*Pi*i*t). - Michael Somos, Jan 25 2008
For n > 0, a(n)/8 is multiplicative and a(p^n)/8 = 1 + p + p^2 + ... + p^n for p an odd prime, a(2^n)/8 = 1 + 2 for n > 0.
a(n) = 8*A000203(n/A006519(n))*(2 + (-1)^n). - Benoit Cloitre, May 16 2002
G.f.: 1 + 8*Sum_{k>0} x^k / (1 + (-x)^k)^2 = 1 + 8*Sum_{k>0} k * x^k / (1 + (-x)^k).
G.f. = s(2)^20/(s(1)*s(4))^8, where s(k) := subs(q=q^k, eta(q)), where eta(q) is Dedekind's function, cf. A010815. [Fine]
Fine gives another explicit formula for a(n) in terms of the divisors of n.
a(n) = 8*A046897(n), n > 0. - Ralf Stephan, Apr 02 2003
A096727(n) = (-1)^n * a(n). a(2*n) = A004011(n). a(2*n + 1) = A005879(n).
Dirichlet g.f.: Sum_{n>=1} a(n)/n^s = 8*(1-4^(1-s))*zeta(s)*zeta(s-1). [Ramanu. J. 7 (2003) 95-127, eq (3.2)]. - R. J. Mathar, Jul 02 2012
Average value is (Pi^2/2)*n + O(sqrt(n)). - Charles R Greathouse IV, Feb 17 2015
From Wolfdieter Lang, Jan 14 2016: (Start)
For n >= 1: a(n) = 8*Sum_{d | n} b(d)*d, with b(d) = 1 if d/4 is not an integer else 0. See, e.g., the Freitag-Busam reference, p. 392.
For n >= 1: a(n) = 8*sigma(n) if n is odd else 24*sigma(m(n)), where m(n) is the largest odd divisor of n (see A000265), and sigma is given in A000203. See the Moreno-Wagstaff reference, Theorem 2. 6 (Jacobi), p. 29. (End)
a(n) = (8/n)*Sum_{k=1..n} A186690(k)*a(n-k), a(0) = 1. - Seiichi Manyama, May 27 2017

A002326 Multiplicative order of 2 mod 2n+1.

Original entry on oeis.org

1, 2, 4, 3, 6, 10, 12, 4, 8, 18, 6, 11, 20, 18, 28, 5, 10, 12, 36, 12, 20, 14, 12, 23, 21, 8, 52, 20, 18, 58, 60, 6, 12, 66, 22, 35, 9, 20, 30, 39, 54, 82, 8, 28, 11, 12, 10, 36, 48, 30, 100, 51, 12, 106, 36, 36, 28, 44, 12, 24, 110, 20, 100, 7, 14, 130, 18, 36, 68, 138, 46, 60, 28
Offset: 0

Keywords

Comments

In other words, least m > 0 such that 2n+1 divides 2^m-1.
Number of riffle shuffles of 2n+2 cards required to return a deck to initial state. A riffle shuffle replaces a list s(1), s(2), ..., s(m) with s(1), s((i/2)+1), s(2), s((i/2)+2), ... a(1) = 2 because a riffle shuffle of [1, 2, 3, 4] requires 2 iterations [1, 2, 3, 4] -> [1, 3, 2, 4] -> [1, 2, 3, 4] to restore the original order.
Concerning the complexity of computing this sequence, see for example Bach and Shallit, p. 115, exercise 8.
It is not difficult to prove that if 2n+1 is a prime then 2n is a multiple of a(n). But the converse is not true. Indeed, one can prove that a(2^(2t-1))=4t. Thus if n=2^(2t-1), where, for any m > 0, t=2^(m-1) then 2n is a multiple of a(n) while 2n+1 is a Fermat number which, as is well known, is not always a prime. It is an interesting problem to describe all composite numbers for which 2n is divisible by a(n). - Vladimir Shevelev, May 09 2008
For an algorithm of calculation of a(n) see author's comment in A179680. - Vladimir Shevelev, Jul 21 2010
From V. Raman, Sep 18 2012, Dec 10 2012: (Start)
If 2n+1 is prime, then the polynomial (x^(2n+1)+1)/(x+1) factors into 2n/a(n) polynomials of the same degree a(n) over GF(2).
If (x^(2n+1)+1)/(x+1) is irreducible over GF(2), then 2n+1 is prime, and 2 is a primitive root (mod 2n+1) (cf. A001122).
For all n > 0, a(n) is the degree of the largest irreducible polynomial factor for the polynomial (x^(2n+1)+1)/(x+1) over GF(2). (End)
a(n) is a factor of phi(2n+1) (A000010(2n+1)). - Douglas Boffey, Oct 21 2013
Conjecture: if p is an odd prime then a((p^3-1)/2) = p * a((p^2-1)/2). Because otherwise a((p^3-1)/2) < p * a((p^2-1)/2) iff a((p^3-1)/2) = a((p-1)/2) for a prime p. Equivalently p^3 divides 2^(p-1)-1, but no such prime p is known. - Thomas Ordowski, Feb 10 2014
A generalization of the previous conjecture: For each k>=2, if p is an odd prime then a(((p^(k+1))-1)/2) = p * a((p^k-1)/2). Computer testing of this generalized conjecture shows that there is no counterexample for k and p both up to 1000. - Ahmad J. Masad, Oct 17 2020

Examples

			From _Vladimir Shevelev_, Oct 03 2017: (Start)
Our algorithm for the calculation of a(n) in the author's comment in A179680 (see also the Sage program below) could be represented in the form of a "finite continued fraction". For example let n = 8, 2*n+1 = 17. We have
    1 + 17
    ------- + 17
       2
    ------------- + 17
           2
    ------------------- + 17
              2
    -------------------------- = 1
                 32
Here the denominators are the A006519 of the numerators: A006519(1+17) = 2, A006519(9+17) = 2, A006519(13+17) = 2, A006519(15+17) = 32. Summing the exponents of these powers of 2, we obtain the required result: a(8) = 1 + 1 + 1 + 5 = 8. Indeed, we have (((1*32 - 17)*2 - 17)*2 - 17)*2 - 17 = 1. So 32*2*2*2 - 1 == 0 (mod 17), 2^8 - 1 == 0 (mod 17). In the general case, note that all "partial fractions" (which indeed are integers) are odd residues modulo 2*n+1 in the interval [1, 2*n-1]. It is easy to prove that the first 1 appears not later than in the n-th step. (End)
		

References

  • E. Bach and Jeffrey Shallit, Algorithmic Number Theory, I.
  • T. Folger, "Shuffling Into Hyperspace," Discover, 1991 (vol 12, no 1), pages 66-67.
  • M. Gardner, "Card Shuffles," Mathematical Carnival chapter 10, pages 123-138. New York: Vintage Books, 1977.
  • L. Lunelli and M. Lunelli, Tavola di congruenza a^n == 1 mod K per a=2,5,10, Atti Sem. Mat. Fis. Univ. Modena 10 (1960/61), 219-236 (1961).
  • J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, p. 146, Exer. 21.3
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A024222, A006694 (number of cyclotomic cosets).
Cf. A014664 (order of 2 mod n-th prime).
Cf. A001122 (primes for which 2 is a primitive root).
Cf. A216838 (primes for which 2 is not a primitive root).
Bisections give A274298, A274299.
Partial sums: A359147.

Programs

  • GAP
    List([0..100],n->OrderMod(2,2*n+1)); # Muniru A Asiru, Feb 01 2019
    
  • Haskell
    import Data.List (findIndex)
    import Data.Maybe (fromJust)
    a002326 n = (+ 1) $ fromJust $
                findIndex ((== 0) . (`mod` (2 * n + 1))) $ tail a000225_list
    -- Reinhard Zumkeller, Apr 22 2013
    
  • Magma
    [ 1 ] cat [ Modorder(2, 2*n+1): n in [1..72] ]; // Klaus Brockhaus, Dec 03 2008
    
  • Maple
    a := n -> `if`(n=0, 1, numtheory:-order(2, 2*n+1)):
    seq(a(n), n=0..72);
  • Mathematica
    Table[MultiplicativeOrder[2, 2*n + 1], {n, 0, 100}] (* Robert G. Wilson v, Apr 05 2011 *)
  • PARI
    a(n)=if(n<0,0,znorder(Mod(2,2*n+1))) /* Michael Somos, Mar 31 2005 */
    
  • Python
    from sympy import n_order
    [n_order(2, 2*n+1) for n in range(73)] # Hermann Stamm-Wilbrandt, Jul 27 2021
  • Sage
    # From Peter Luschny, Oct 06 2017: (Start)
    [Mod(2,n).multiplicative_order() for n in (0..145) if gcd(n,2) == 1]
    # Algorithm from Vladimir Shevelev as described in A179680 and presented in Example.
    def A002326VS(n):
        s, m, N = 0, 1, 2*n + 1
        while True:
            k = N + m
            v = valuation(k, 2)
            s += v
            m = k >> v
            if m == 1: break
        return s
    [A002326VS(n) for n in (0..72)] # (End)
    

Formula

a((3^n-1)/2) = A025192(n). - Vladimir Shevelev, May 09 2008
Bisection of A007733: a(n) = A007733(2*n+1). - Max Alekseyev, Jun 11 2009
a((b(n)-1)/2) = n for odd n and even n such that b(n/2) != b(n), where b(n) = A005420(n). - Thomas Ordowski, Jan 11 2014
Note that a(2^n-1) = n+1 and a(2^n) = 2*(n+1). - Thomas Ordowski, Jan 16 2014
a(n) = A056239(A292239(n)) = A048675(A292265(n)). - Antti Karttunen, Oct 04 2017

Extensions

More terms from David W. Wilson, Jan 13 2000
More terms from Benoit Cloitre, Apr 11 2003

A228351 Triangle read by rows in which row n lists the compositions (ordered partitions) of n (see Comments lines for definition).

Original entry on oeis.org

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

Author

Omar E. Pol, Aug 30 2013

Keywords

Comments

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is (list-)reversed co-lexicographic. - Joerg Arndt, Sep 02 2013
Dropping the "(list-)reversed" in the comment above gives A228525.
The equivalent sequence for partitions is A026792.
This sequence lists (without repetitions) all finite compositions, in such a way that, if [P_1, ..., P_r] denotes the composition occupying the n-th position in the list, then (((2*n/2^(P_1)-1)/2^(P_2)-1)/...)/2^(P_r)-1 = 0. - Lorenzo Sauras Altuzarra, Jan 22 2020
The k-th composition in the list is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, and taking first differences. Reversing again gives A066099, which is described as the standard ordering. Both sequences define a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, Apr 01 2020
It follows from the previous comment that A000120(k) is the length of the k-th composition that is listed by this sequence (recall that A000120(k) is the number of 1's in the binary expansion of k). - Lorenzo Sauras Altuzarra, Sep 29 2020

Examples

			Illustration of initial terms:
-----------------------------------
n  j     Diagram     Composition j
-----------------------------------
.         _
1  1     |_|         1;
.         _ _
2  1     |_  |       2,
2  2     |_|_|       1, 1;
.         _ _ _
3  1     |_    |     3,
3  2     |_|_  |     1, 2,
3  3     |_  | |     2, 1,
3  4     |_|_|_|     1, 1, 1;
.         _ _ _ _
4  1     |_      |   4,
4  2     |_|_    |   1, 3,
4  3     |_  |   |   2, 2,
4  4     |_|_|_  |   1, 1, 2,
4  5     |_    | |   3, 1,
4  6     |_|_  | |   1, 2, 1,
4  7     |_  | | |   2, 1, 1,
4  8     |_|_|_|_|   1, 1, 1, 1;
.
Triangle begins:
[1];
[2],[1,1];
[3],[1,2],[2,1],[1,1,1];
[4],[1,3],[2,2],[1,1,2],[3,1],[1,2,1],[2,1,1],[1,1,1,1];
[5],[1,4],[2,3],[1,1,3],[3,2],[1,2,2],[2,1,2],[1,1,1,2],[4,1],[1,3,1],[2,2,1],[1,1,2,1],[3,1,1],[1,2,1,1],[2,1,1,1],[1,1,1,1,1];
...
For example [1,2] occupies the 5th position in the corresponding list of compositions and indeed (2*5/2^1-1)/2^2-1 = 0. - _Lorenzo Sauras Altuzarra_, Jan 22 2020
12 --binary expansion--> [1,1,0,0] --reverse--> [0,0,1,1] --positions of 1's--> [3,4] --prepend 0--> [0,3,4] --first differences--> [3,1]. - _Lorenzo Sauras Altuzarra_, Sep 29 2020
		

Crossrefs

Row n has length A001792(n-1). Row sums give A001787, n >= 1.
Cf. A000120 (binary weight), A001511, A006519, A011782, A026792, A065120.
A related ranking of finite sets is A048793/A272020.
All of the following consider the k-th row to be the k-th composition, ignoring the coarser grouping by sum.
- Indices of weakly increasing rows are A114994.
- Indices of weakly decreasing rows are A225620.
- Indices of strictly decreasing rows are A333255.
- Indices of strictly increasing rows are A333256.
- Indices of reversed interval rows A164894.
- Indices of interval rows are A246534.
- Indices of strict rows are A233564.
- Indices of constant rows are A272919.
- Indices of anti-run rows are A333489.
- Row k has A124767(k) runs and A333381(k) anti-runs.
- Row k has GCD A326674(k) and LCM A333226(k).
- Row k has Heinz number A333219(k).
Equals A163510+1, termwise.
Cf. A124734 (increasing length, then lexicographic).
Cf. A296774 (increasing length, then reverse lexicographic).
Cf. A337243 (increasing length, then colexicographic).
Cf. A337259 (increasing length, then reverse colexicographic).
Cf. A296773 (decreasing length, then lexicographic).
Cf. A296772 (decreasing length, then reverse lexicographic).
Cf. A337260 (decreasing length, then colexicographic).
Cf. A108244 (decreasing length, then reverse colexicographic).
Cf. A228369 (lexicographic).
Cf. A066099 (reverse lexicographic).
Cf. A228525 (colexicographic).

Programs

  • Haskell
    a228351 n = a228351_list !! (n - 1)
    a228351_list = concatMap a228351_row [1..]
    a228351_row 0 = []
    a228351_row n = a001511 n : a228351_row (n `div` 2^(a001511 n))
    -- Peter Kagey, Jun 27 2016
    
  • Maple
    # Program computing the sequence:
    A228351 := proc(n) local c, k, L, N: L, N := [], [seq(2*r, r = 1 .. n)]: for k in N do c := 0: while k != 0 do if gcd(k, 2) = 2 then k := k/2: c := c+1: else L := [op(L), op(c)]: k := k-1: c := 0: fi: od: od: L[n]: end: # Lorenzo Sauras Altuzarra, Jan 22 2020
    # Program computing the list of compositions:
    List := proc(n) local c, k, L, M, N: L, M, N := [], [], [seq(2*r, r = 1 .. 2^n-1)]: for k in N do c := 0: while k != 0 do if gcd(k, 2) = 2 then k := k/2: c := c+1: else L := [op(L), c]: k := k-1: c := 0: fi: od: M := [op(M), L]: L := []: od: M: end: # Lorenzo Sauras Altuzarra, Jan 22 2020
  • Mathematica
    bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    Table[Differences[Prepend[bpe[n],0]],{n,0,30}] (* Gus Wiseman, Apr 01 2020 *)
  • Python
    from itertools import count, islice
    def A228351_gen(): # generator of terms
        for n in count(1):
            k = n
            while k:
                yield (s:=(~k&k-1).bit_length()+1)
                k >>= s
    A228351_list = list(islice(A228351_gen(),30)) # Chai Wah Wu, Jul 17 2023

A031443 Digitally balanced numbers: positive numbers that in base 2 have the same number of 0's as 1's.

Original entry on oeis.org

2, 9, 10, 12, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 135, 139, 141, 142, 147, 149, 150, 153, 154, 156, 163, 165, 166, 169, 170, 172, 177, 178, 180, 184, 195, 197, 198, 201, 202, 204, 209, 210, 212, 216, 225, 226, 228, 232, 240, 527, 535, 539, 541, 542, 551
Offset: 1

Keywords

Comments

Also numbers k such that the binary digital mean dm(2, k) = (Sum_{i=1..d} 2*d_i - 1) / (2*d) = 0, where d is the number of digits in the binary representation of k and d_i the individual digits. - Reikku Kulon, Sep 21 2008
From Reikku Kulon, Sep 29 2008: (Start)
Each run of values begins with 2^(2k + 1) + 2^(k + 1) - 2^k - 1. The initial values increase according to the sequence {2^(k - 1), 2^(k - 2), 2^(k - 3), ..., 2^(k - k)}.
After this, the values follow a periodic sequence of increases by successive powers of two with single odd values interspersed.
Each run ends with an odd increase followed by increases of {2^(k - k), ..., 2^(k - 2), 2^(k - 1), 2^k}, finally reaching 2^(2k + 2) - 2^(k + 1).
Similar behavior occurs in other bases. (End)
Numbers k such that A000120(k)/A070939(k) = 1/2. - Ctibor O. Zizka, Oct 15 2008
Subsequence of A053754; A179888 is a subsequence. - Reinhard Zumkeller, Jul 31 2010
A000120(a(n)) = A023416(a(n)); A037861(a(n)) = 0.
A001700 gives number of terms having length 2*n in binary representation: A001700(n-1) = #{m: A070939(a(m))=2*n}. - Reinhard Zumkeller, Jun 08 2011
The number of terms below 2^k is A079309(floor(k/2)) for k > 1. - Amiram Eldar, Nov 21 2020

Examples

			9 is a term because '1001' contains 2 '0's and 2 '1's.
		

Crossrefs

Subsequence of A053754.
Row n = 2 of A378000.
Terms of binary width n are enumerated by A001700.

Programs

  • Haskell
    -- See link, showing that Ulrich Schimkes formula provides a very efficient algorithm. Reinhard Zumkeller, Jun 15 2011
    
  • Magma
    [ n: n in [2..250] | Multiplicity({* z: z in Intseq(n,2) *}, 0) eq &+Intseq(n,2) ];  // Bruno Berselli, Jun 07 2011
    
  • Maple
    a:=proc(n) local nn, n1, n0: nn:=convert(n,base,2): n1:=add(nn[i],i=1..nops(nn)): n0:=nops(nn)-n1: if n0=n1 then n else end if end proc: seq(a(n), n = 1..240); # Emeric Deutsch, Jul 31 2008
  • Mathematica
    Select[Range[250],DigitCount[#,2,1]==DigitCount[#,2,0]&] (* Harvey P. Dale, Jul 22 2013 *)
    FromDigits[#,2]&/@DeleteCases[Flatten[Permutations/@Table[PadRight[{},2n,{1,0}],{n,5}],1],?(#[[1]]==0&)]//Sort (* _Harvey P. Dale, May 30 2016 *)
  • PARI
    for(n=1,100,b=binary(n); l=length(b); if(sum(i=1,l, component(b,i))==l/2,print1(n,",")))
    
  • PARI
    is(n)=hammingweight(n)==hammingweight(bitneg(n,#binary(n))) \\ Charles R Greathouse IV, Mar 29 2013
    
  • PARI
    is(n)=2*hammingweight(n)==exponent(n)+1 \\ Charles R Greathouse IV, Apr 18 2020
    
  • Perl
    for my $half ( 1 .. 4 ) {
      my $N = 2 * $half;  # only even widths apply
      my $vector = (1 << ($N-1)) | ((1 << ($N/2-1)) - 1);  # first key
      my $n = 1; $n *= $_ for 2 .. $N;    # N!
      my $d = 1; $d *= $_ for 2 .. $N/2;  # (N/2)!
      for (1 .. $n/($d*$d*2)) {
        print "$vector, ";
        my ($v, $d) = ($vector, 0);
        until ($v & 1 or !$v) { $d = ($d << 1)|1; $v >>= 1 }
        $vector += $d + 1 + (($v ^ ($v + 1)) >> 2);  # next key
      }
    } # Ruud H.G. van Tol, Mar 30 2014
    
  • Python
    from sympy.utilities.iterables import multiset_permutations
    A031443_list = [int('1'+''.join(p),2) for n in range(1,10) for p in multiset_permutations('0'*n+'1'*(n-1))] # Chai Wah Wu, Nov 15 2019

Formula

a(n+1) = a(n) + 2^k + 2^(m-1) - 1 + floor((2^(k+m) - 2^k)/a(n))*(2^(2*m) + 2^(m-1)) where k is the largest integer such that 2^k divides a(n) and m is the largest integer such that 2^m divides a(n)/2^k+1. - Ulrich Schimke (UlrSchimke(AT)aol.com)
A145037(a(n)) = 0. - Reikku Kulon, Oct 02 2008
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