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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A230097 Indices of records in A159918.

Original entry on oeis.org

0, 1, 3, 5, 11, 21, 39, 45, 75, 155, 181, 627, 923, 1241, 2505, 3915, 5221, 6475, 11309, 15595, 19637, 31595, 44491, 69451, 113447, 185269, 244661, 357081, 453677, 908091, 980853, 2960011, 2965685, 5931189, 11862197, 20437147, 22193965, 43586515, 57804981, 157355851
Offset: 1

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Author

N. J. A. Sloane, Oct 11 2013

Keywords

Comments

The records themselves are not so interesting: 0, 1, 2, 3, 5, 6, 7, 8, 9, 10, 13, 14, 15, 16, 17, 18, 19, 20, ... (A357304).
Lindström mentions that the record value 34 in A159918 is first reached at n = 980853.

Crossrefs

Programs

  • Haskell
    a230097 n = a230097_list !! (n-1)
    a230097_list = 0 : f 0 0 where
       f i m = if v > m then i : f (i + 1) v else f (i + 1) m
               where v = a159918 i
    -- Reinhard Zumkeller, Oct 12 2013
    (Python 3.10+)
    from itertools import count, islice
    def A230097_gen(): # generator of terms
        c = -1
        for n in count(0):
            if (m := (n**2).bit_count())>c:
                yield n
                c = m
    A230097_list = list(islice(A230097_gen(),20)) # Chai Wah Wu, Oct 01 2022

Formula

Lindström shows that lim sup wt(m^2)/log_2 m = 2.

Extensions

a(19)-a(40) from Reinhard Zumkeller, Oct 12 2013

A285717 a(n) = A007814(n) + A159918(n) = A007814(n) + A000120(n^2).

Original entry on oeis.org

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

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Author

Antti Karttunen, Apr 27 2017

Keywords

Crossrefs

Programs

  • Mathematica
    Table[IntegerExponent[n, 2] + DigitCount[n^2, 2, 1], {n, 120}] (* Indranil Ghosh, Apr 27 2017 *)
  • PARI
    a(n) = valuation(n, 2) + hammingweight(n^2); \\ Indranil Ghosh, Apr 27 2017
    
  • Python
    import math
    def a(n): return int(math.log(n - (n & n - 1), 2)) + bin(n**2)[2:].count("1") # Indranil Ghosh, Apr 27 2017
  • Scheme
    (define (A285717 n) (+ (A007814 n) (A159918 n)))
    (define (A285717 n) (A007814 (* n (A201555 n))))
    

Formula

a(n) = A007814(n) + A159918(n) = A007814(n) + A000120(n^2).
a(n) = A007814(n*A201555(n)).

A000290 The squares: a(n) = n^2.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500
Offset: 0

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Author

Keywords

Comments

To test if a number is a square, see Cohen, p. 40. - N. J. A. Sloane, Jun 19 2011
Zero followed by partial sums of A005408 (odd numbers). - Jeremy Gardiner, Aug 13 2002
Begin with n, add the next number, subtract the previous number and so on ending with subtracting a 1: a(n) = n + (n+1) - (n-1) + (n+2) - (n-2) + (n+3) - (n-3) + ... + (2n-1) - 1 = n^2. - Amarnath Murthy, Mar 24 2004
Sum of two consecutive triangular numbers A000217. - Lekraj Beedassy, May 14 2004
Numbers with an odd number of divisors: {d(n^2) = A048691(n); for the first occurrence of 2n + 1 divisors, see A071571(n)}. - Lekraj Beedassy, Jun 30 2004
See also A000037.
First sequence ever computed by electronic computer, on EDSAC, May 06 1949 (see Renwick link). - Russ Cox, Apr 20 2006
Numbers k such that the imaginary quadratic field Q(sqrt(-k)) has four units. - Marc LeBrun, Apr 12 2006
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A006881(k)^(n-1)); a(n) = A000005(A000400(n-1)) = A000005(A011557(n-1)) = A000005(A001023(n-1)) = A000005(A001024(n-1)). - Reinhard Zumkeller, Mar 04 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
Numbers a such that a^1/2 + b^1/2 = c^1/2 and a^2 + b = c. - Cino Hilliard, Feb 07 2008 (this comment needs clarification, Joerg Arndt, Sep 12 2013)
Numbers k such that the geometric mean of the divisors of k is an integer. - Ctibor O. Zizka, Jun 26 2008
Equals row sums of triangle A143470. Example: 36 = sum of row 6 terms: (23 + 7 + 3 + 1 + 1 + 1). - Gary W. Adamson, Aug 17 2008
Equals row sums of triangles A143595 and A056944. - Gary W. Adamson, Aug 26 2008
Number of divisors of 6^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Denominators of Lyman spectrum of hydrogen atom. Numerators are A005563. A000290-A005563 = A000012. - Paul Curtz, Nov 06 2008
a(n) is the number of all partitions of the sum 2^2 + 2^2 + ... + 2^2, (n-1) times, into powers of 2. - Valentin Bakoev, Mar 03 2009
a(n) is the maximal number of squares that can be 'on' in an n X n board so that all the squares turn 'off' after applying the operation: in any 2 X 2 sub-board, a square turns from 'on' to 'off' if the other three are off. - Srikanth K S, Jun 25 2009
Zero together with the numbers k such that 2 is the number of perfect partitions of k. - Juri-Stepan Gerasimov, Sep 26 2009
Totally multiplicative sequence with a(p) = p^2 for prime p. - Jaroslav Krizek, Nov 01 2009
Satisfies A(x)/A(x^2), A(x) = A173277: (1, 4, 13, 32, 74, ...). - Gary W. Adamson, Feb 14 2010
Positive members are the integers with an odd number of odd divisors and an even number of even divisors. See also A120349, A120359, A181792, A181793, A181795. - Matthew Vandermast, Nov 14 2010
Besides the first term, this sequence is the denominator of Pi^2/6 = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + ... . - Mohammad K. Azarian, Nov 01 2011
Partial sums give A000330. - Omar E. Pol, Jan 12 2013
Drmota, Mauduit, and Rivat proved that the Thue-Morse sequence along the squares is normal; see A228039. - Jonathan Sondow, Sep 03 2013
a(n) can be decomposed into the sum of the four numbers [binomial(n, 1) + binomial(n, 2) + binomial(n-1, 1) + binomial(n-1, 2)] which form a "square" in Pascal's Triangle A007318, or the sum of the two numbers [binomial(n, 2) + binomial(n+1, 2)], or the difference of the two numbers [binomial(n+2, 3) - binomial(n, 3)]. - John Molokach, Sep 26 2013
In terms of triangular tiling, the number of equilateral triangles with side length 1 inside an equilateral triangle with side length n. - K. G. Stier, Oct 30 2013
Number of positive roots in the root systems of type B_n and C_n (when n > 1). - Tom Edgar, Nov 05 2013
Squares of squares (fourth powers) are also called biquadratic numbers: A000583. - M. F. Hasler, Dec 29 2013
For n > 0, a(n) is the largest integer k such that k^2 + n is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n is a multiple of k + n is given by k = n^(2*m). - Derek Orr, Sep 03 2014
For n > 0, a(n) is the number of compositions of n + 5 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
a(n), for n >= 3, is also the number of all connected subtrees of a cycle graph, having n vertices. - Viktar Karatchenia, Mar 02 2016
On every sequence of natural continuous numbers with an even number of elements, the summatory of the second half of the sequence minus the summatory of the first half of the sequence is always a square. Example: Sequence from 61 to 70 has an even number of elements (10). Then 61 + 62 + 63 + 64 + 65 = 315; 66 + 67 + 68 + 69 + 70 = 340; 340 - 315 = 25. (n/2)^2 for n = number of elements. - César Aguilera, Jun 20 2016
On every sequence of natural continuous numbers from n^2 to (n+1)^2, the sum of the differences of pairs of elements of the two halves in every combination possible is always (n+1)^2. - César Aguilera, Jun 24 2016
Suppose two circles with radius 1 are tangent to each other as well as to a line not passing through the point of tangency. Create a third circle tangent to both circles as well as the line. If this process is continued, a(n) for n > 0 is the reciprocals of the radii of the circles, beginning with the largest circle. - Melvin Peralta, Aug 18 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Numerators of the solution to the generalization of the Feynman triangle problem, with an offset of 2. If each vertex of a triangle is joined to the point (1/p) along the opposite side (measured say clockwise), then the area of the inner triangle formed by these lines is equal to (p - 2)^2/(p^2 - p + 1) times the area of the original triangle, p > 2. For example, when p = 3, the ratio of the areas is 1/7. The denominators of the ratio of the areas is given by A002061. [Cook & Wood, 2004] - Joe Marasco, Feb 20 2017
Equals row sums of triangle A004737, n >= 1. - Martin Michael Musatov, Nov 07 2017
Right-hand side of the binomial coefficient identity Sum_{k = 0..n} (-1)^(n+k+1)*binomial(n,k)*binomial(n + k,k)*(n - k) = n^2. - Peter Bala, Jan 12 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)-1} such that |A|=|B|=k and A+B contains {0,1,2,...,a(n)-1}} = n. - Michael Chu, Mar 09 2022
Number of 3-permutations of n elements avoiding the patterns 132, 213, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022
Number of intercalates in cyclic Latin squares of order 2n (cyclic Latin squares of odd order do not have intercalates). - Eduard I. Vatutin, Feb 15 2024
a(n) is the number of ternary strings of length n with at most one 0, exactly one 1, and no restriction on the number of 2's. For example, a(3)=9, consisting of the 6 permutations of the string 102 and the 3 permutations of the string 122. - Enrique Navarrete, Mar 12 2025

Examples

			For n = 8, a(8) = 8 * 15 - (1 + 3 + 5 + 7 + 9 + 11 + 13) - 7 = 8 * 15 - 49 - 7 = 64. - _Bruno Berselli_, May 04 2010
G.f. = x + 4*x^2 + 9*x^3 + 16*x^4 + 25*x^5 + 36*x^6 + 49*x^7 + 64*x^8 + 81*x^9 + ...
a(4) = 16. For n = 4 vertices, the cycle graph C4 is A-B-C-D-A. The subtrees are: 4 singles: A, B, C, D; 4 pairs: A-B, BC, C-D, A-D; 4 triples: A-B-C, B-C-D, C-D-A, D-A-B; 4 quads: A-B-C-D, B-C-D-A, C-D-A-B, D-A-B-C; 4 + 4 + 4 + 4 = 16. - _Viktar Karatchenia_, Mar 02 2016
		

References

  • G. L. Alexanderson et al., The William Lowell Putnam Mathematical Competition, Problems and Solutions: 1965-1984, "December 1967 Problem B4(a)", pp. 8(157) MAA Washington DC 1985.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Chapter XV, pp. 135-167.
  • R. P. Burn & A. Chetwynd, A Cascade Of Numbers, "The prison door problem" Problem 4 pp. 5-7; 79-80 Arnold London 1996.
  • H. Cohen, A Course in Computational Algebraic Number Theory, Springer, 1996, p. 40.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 31, 36, 38, 63.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), p. 6.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments, Chapter 6 pp. 71-2, W. H. Freeman NY 1988.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology and §8.6 Figurate Numbers, pp. 264, 290-291.
  • Alfred S. Posamentier, The Art of Problem Solving, Section 2.4 "The Long Cell Block" pp. 10-1; 12; 156-7 Corwin Press Thousand Oaks CA 1996.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 35, 52-53, 129-132, 244.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 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).
  • J. K. Strayer, Elementary Number Theory, Exercise Set 3.3 Problems 32, 33, p. 88, PWS Publishing Co. Boston MA 1996.
  • C. W. Trigg, Mathematical Quickies, "The Lucky Prisoners" Problem 141 pp. 40, 141, Dover NY 1985.
  • R. Vakil, A Mathematical Mosaic, "The Painted Lockers" pp. 127;134 Brendan Kelly Burlington Ontario 1996.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Crossrefs

Cf. A092205, A128200, A005408, A128201, A002522, A005563, A008865, A059100, A143051, A143470, A143595, A056944, A001157 (inverse Möbius transform), A001788 (binomial transform), A228039, A001105, A004159, A159918, A173277, A095794, A162395, A186646 (Pisano periods), A028338 (2nd diagonal).
A row or column of A132191.
This sequence is related to partitions of 2^n into powers of 2, as it is shown in A002577. So A002577 connects the squares and A000447. - Valentin Bakoev, Mar 03 2009
Boustrophedon transforms: A000697, A000745.
Cf. A342819.
Cf. A013661.

Programs

Formula

G.f.: x*(1 + x) / (1 - x)^3.
E.g.f.: exp(x)*(x + x^2).
Dirichlet g.f.: zeta(s-2).
a(n) = a(-n).
Multiplicative with a(p^e) = p^(2*e). - David W. Wilson, Aug 01 2001
Sum of all matrix elements M(i, j) = 2*i/(i+j) (i, j = 1..n). a(n) = Sum_{i = 1..n} Sum_{j = 1..n} 2*i/(i + j). - Alexander Adamchuk, Oct 24 2004
a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 2. - Miklos Kristof, Mar 09 2005
From Pierre CAMI, Oct 22 2006: (Start)
a(n) is the sum of the odd numbers from 1 to 2*n - 1.
a(0) = 0, a(1) = 1, then a(n) = a(n-1) + 2*n - 1. (End)
For n > 0: a(n) = A130064(n)*A130065(n). - Reinhard Zumkeller, May 05 2007
a(n) = Sum_{k = 1..n} A002024(n, k). - Reinhard Zumkeller, Jun 24 2007
Left edge of the triangle in A132111: a(n) = A132111(n, 0). - Reinhard Zumkeller, Aug 10 2007
Binomial transform of [1, 3, 2, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = binomial(n+1, 2) + binomial(n, 2).
This sequence could be derived from the following general formula (cf. A001286, A000330): n*(n+1)*...*(n+k)*(n + (n+1) + ... + (n+k))/((k+2)!*(k+1)/2) at k = 0. Indeed, using the formula for the sum of the arithmetic progression (n + (n+1) + ... + (n+k)) = (2*n + k)*(k + 1)/2 the general formula could be rewritten as: n*(n+1)*...*(n+k)*(2*n+k)/(k+2)! so for k = 0 above general formula degenerates to n*(2*n + 0)/(0 + 2) = n^2. - Alexander R. Povolotsky, May 18 2008
From a(4) recurrence formula a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) and a(1) = 1, a(2) = 4, a(3) = 9. - Artur Jasinski, Oct 21 2008
The recurrence a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) is satisfied by all k-gonal sequences from a(3), with a(0) = 0, a(1) = 1, a(2) = k. - Jaume Oliver Lafont, Nov 18 2008
a(n) = floor(n*(n+1)*(Sum_{i = 1..n} 1/(n*(n+1)))). - Ctibor O. Zizka, Mar 07 2009
Product_{i >= 2} 1 - 2/a(i) = -sin(A063448)/A063448. - R. J. Mathar, Mar 12 2009
a(n) = A002378(n-1) + n. - Jaroslav Krizek, Jun 14 2009
a(n) = n*A005408(n-1) - (Sum_{i = 1..n-2} A005408(i)) - (n-1) = n*A005408(n-1) - a(n-1) - (n-1). - Bruno Berselli, May 04 2010
a(n) == 1 (mod n+1). - Bruno Berselli, Jun 03 2010
a(n) = a(n-1) + a(n-2) - a(n-3) + 4, n > 2. - Gary Detlefs, Sep 07 2010
a(n+1) = Integral_{x >= 0} exp(-x)/( (Pn(x)*exp(-x)*Ei(x) - Qn(x))^2 +(Pi*exp(-x)*Pn(x))^2 ), with Pn the Laguerre polynomial of order n and Qn the secondary Laguerre polynomial defined by Qn(x) = Integral_{t >= 0} (Pn(x) - Pn(t))*exp(-t)/(x-t). - Groux Roland, Dec 08 2010
Euler transform of length-2 sequence [4, -1]. - Michael Somos, Feb 12 2011
A162395(n) = -(-1)^n * a(n). - Michael Somos, Mar 19 2011
a(n) = A004201(A000217(n)); A007606(a(n)) = A000384(n); A007607(a(n)) = A001105(n). - Reinhard Zumkeller, Feb 12 2011
Sum_{n >= 1} 1/a(n)^k = (2*Pi)^k*B_k/(2*k!) = zeta(2*k) with Bernoulli numbers B_k = -1, 1/6, 1/30, 1/42, ... for k >= 0. See A019673, A195055/10 etc. [Jolley eq 319].
Sum_{n>=1} (-1)^(n+1)/a(n)^k = 2^(k-1)*Pi^k*(1-1/2^(k-1))*B_k/k! [Jolley eq 320] with B_k as above.
A007968(a(n)) = 0. - Reinhard Zumkeller, Jun 18 2011
A071974(a(n)) = n; A071975(a(n)) = 1. - Reinhard Zumkeller, Jul 10 2011
a(n) = A199332(2*n - 1, n). - Reinhard Zumkeller, Nov 23 2011
For n >= 1, a(n) = Sum_{d|n} phi(d)*psi(d), where phi is A000010 and psi is A001615. - Enrique Pérez Herrero, Feb 29 2012
a(n) = A000217(n^2) - A000217(n^2 - 1), for n > 0. - Ivan N. Ianakiev, May 30 2012
a(n) = (A000217(n) + A000326(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = A162610(n, n) = A209297(n, n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(A000217(n)) = Sum_{i = 1..n} Sum_{j = 1..n} i*j, for n > 0. - Ivan N. Ianakiev, Apr 20 2013
a(n) = A133280(A000217(n)). - Ivan N. Ianakiev, Aug 13 2013
a(2*a(n)+2*n+1) = a(2*a(n)+2*n) + a(2*n+1). - Vladimir Shevelev, Jan 24 2014
a(n+1) = Sum_{t1+2*t2+...+n*tn = n} (-1)^(n+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*4^(t1)*7^(t2)*8^(t3+...+tn). - Mircea Merca, Feb 27 2014
a(n) = floor(1/(1-cos(1/n)))/2 = floor(1/(1-n*sin(1/n)))/6, n > 0. - Clark Kimberling, Oct 08 2014
a(n) = ceiling(Sum_{k >= 1} log(k)/k^(1+1/n)) = -Zeta'[1+1/n]. Thus any exponent greater than 1 applied to k yields convergence. The fractional portion declines from A073002 = 0.93754... at n = 1 and converges slowly to 0.9271841545163232... for large n. - Richard R. Forberg, Dec 24 2014
a(n) = Sum_{j = 1..n} Sum_{i = 1..n} ceiling((i + j - n + 1)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = Product_{j = 1..n-1} 2 - 2*cos(2*j*Pi/n). - Michel Marcus, Jul 24 2015
From Ilya Gutkovskiy, Jun 21 2016: (Start)
Product_{n >= 1} (1 + 1/a(n)) = sinh(Pi)/Pi = A156648.
Sum_{n >= 0} 1/a(n!) = BesselI(0, 2) = A070910. (End)
a(n) = A028338(n, n-1), n >= 1 (second diagonal). - Wolfdieter Lang, Jul 21 2017
For n >= 1, a(n) = Sum_{d|n} sigma_2(d)*mu(n/d) = Sum_{d|n} A001157(d)*A008683(n/d). - Ridouane Oudra, Apr 15 2021
a(n) = Sum_{i = 1..2*n-1} ceiling(n - i/2). - Stefano Spezia, Apr 16 2021
From Richard L. Ollerton, May 09 2021: (Start) For n >= 1,
a(n) = Sum_{k=1..n} psi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} psi(gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)). (End)
a(n) = (A005449(n) + A000326(n))/3. - Klaus Purath, May 13 2021
Let T(n) = A000217(n), then a(T(n)) + a(T(n+1)) = T(a(n+1)). - Charlie Marion, Jun 27 2022
a(n) = Sum_{k=1..n} sigma_1(k) + Sum_{i=1..n} (n mod i). - Vadim Kataev, Dec 07 2022
a(n^2) + a(n^2+1) + ... + a(n^2+n) + 4*A000537(n) = a(n^2+n+1) + ... + a(n^2+2n). In general, if P(k,n) = the n-th k-gonal number, then P(2k,n^2) + P(2k,n^2+1) + ... + P(2k,n^2+n) + 4*(k-1)*A000537(n) = P(2k,n^2+n+1) + ... + P(2k,n^2+2n). - Charlie Marion, Apr 26 2024
Sum_{n>=1} 1/a(n) = A013661. - Alois P. Heinz, Oct 19 2024
a(n) = 1 + 3^3*((n-1)/(n+1))^2 + 5^3*((n-1)*(n-2)/((n+1)*(n+2)))^2 + 7^3*((n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)))^2 + ... for n >= 1. - Peter Bala, Dec 09 2024

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010

A000120 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n).

Original entry on oeis.org

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

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Author

Keywords

Comments

The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021]
a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002
To construct the sequence, start with 0 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 a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003
An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003
Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006
a(n) is the number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006
a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006
The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006
Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009
For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009
Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009
a(k*m) <= a(k) * a(m). - Robert Israel, Sep 03 2023
The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017
Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012
From Daniel Forgues, Mar 13 2015: (Start)
Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015
The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015
a(n) is also the largest part of 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 20 2017
a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020

Examples

			Using the formula a(n) = a(floor(n / floor_pow4(n))) + a(n mod floor_pow4(n)):
  a(4) = a(1) + a(0) = 1,
  a(8) = a(2) + a(0) = 1,
  a(13) = a(3) + a(1) = 2 + 1 = 3,
  a(23) = a(1) + a(7) = 1 + a(1) + a(3) = 1 + 1 + 2 = 4.
_Gary W. Adamson_ points out (Jun 03 2009) that this can be written as a triangle:
  0,
  1,
  1,2,
  1,2,2,3,
  1,2,2,3,2,3,3,4,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
  1,2,2,3,2,3,...
where the rows converge to A063787.
From _Joerg Arndt_, Nov 09 2012: (Start)
Connection to the compositions of n as lists of parts (see comment):
[ #]:   a(n)  composition
[ 0]:   [0]   1 1 1 1 1
[ 1]:   [1]   1 1 1 2
[ 2]:   [1]   1 1 2 1
[ 3]:   [2]   1 1 3
[ 4]:   [1]   1 2 1 1
[ 5]:   [2]   1 2 2
[ 6]:   [2]   1 3 1
[ 7]:   [3]   1 4
[ 8]:   [1]   2 1 1 1
[ 9]:   [2]   2 1 2
[10]:   [2]   2 2 1
[11]:   [3]   2 3
[12]:   [2]   3 1 1
[13]:   [3]   3 2
[14]:   [3]   4 1
[15]:   [4]   5
(End)
		

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 119.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - N. J. A. Sloane, Aug 03 2012
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws. W.H. Freeman, 1991, p. 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).

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000788, A000069, A001969, A023416, A059015, A007088.
Partial sums see A000788. For run lengths see A131534. See also A001792, A010062.
Number of 0's in n: A023416 and A080791.
a(n) = n - A011371(n).
Sum of digits of n written in bases 2-16: this sequence, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
This is Guy Steele's sequence GS(3, 4) (see A135416).
Cf. A230952 (boustrophedon transform).
Cf. A070939 (length of binary representation of n).

Programs

  • Fortran
    c See link in A139351
    
  • Haskell
    import Data.Bits (Bits, popCount)
    a000120 :: (Integral t, Bits t) => t -> Int
    a000120 = popCount
    a000120_list = 0 : c [1] where c (x:xs) = x : c (xs ++ [x,x+1])
    -- Reinhard Zumkeller, Aug 26 2013, Feb 19 2012, Jun 16 2011, Mar 07 2011
    
  • Haskell
    a000120 = concat r
        where r = [0] : (map.map) (+1) (scanl1 (++) r)
    -- Luke Palmer, Feb 16 2014
    
  • Magma
    [Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020
    
  • Magma
    [&+Intseq(n, 2):n in [0..104]]; // Marius A. Burtea, Jan 22 2020
  • Maple
    A000120 := proc(n) local w,m,i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
    A000120 := n -> add(i, i=convert(n,base,2)): # Peter Luschny, Feb 03 2011
    with(Bits): p:=n->ilog2(n-And(n,n-1)): seq(p(binomial(2*n,n)),n=0..200) # Gary Detlefs, Jan 27 2019
  • Mathematica
    Table[DigitCount[n, 2, 1], {n, 0, 105}]
    Nest[Flatten[# /. # -> {#, # + 1}] &, {0}, 7] (* Robert G. Wilson v, Sep 27 2011 *)
    Table[Plus @@ IntegerDigits[n, 2], {n, 0, 104}]
    Nest[Join[#, # + 1] &, {0}, 7] (* IWABUCHI Yu(u)ki, Jul 19 2012 *)
    Log[2, Nest[Join[#, 2#] &, {1}, 14]] (* gives 2^14 term, Carlos Alves, Mar 30 2014 *)
  • PARI
    {a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};
    
  • PARI
    {a(n) = if( n<0, 0, subst(Pol(binary(n)), x ,1))};
    
  • PARI
    {a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    a(n)=my(v=binary(n));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Jun 24 2011
    
  • PARI
    a(n)=norml2(binary(n)) \\ better use {A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020
    
  • PARI
    a(n)=hammingweight(n) \\ Michel Marcus, Oct 19 2013
    (Common Lisp) (defun floor-to-power (n pow) (declare (fixnum pow)) (expt pow (floor (log n pow)))) (defun enabled-bits (n) (if (< n 4) (n-th n (list 0 1 1 2)) (+ (enabled-bits (floor (/ n (floor-to-power n 4)))) (enabled-bits (mod n (floor-to-power n 4)))))) ; Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
    
  • Python
    def A000120(n): return bin(n).count('1') # Chai Wah Wu, Sep 03 2014
    
  • Python
    import numpy as np
    A000120 = np.array([0], dtype="uint8")
    for bitrange in range(25): A000120 = np.append(A000120, np.add(A000120, 1))
    print([A000120[n] for n in range(0, 105)]) # Karl-Heinz Hofmann, Nov 07 2022
    
  • Python
    def A000120(n): return n.bit_count() # Requires Python 3.10 or higher. - Pontus von Brömssen, Nov 08 2022
    
  • Python
    # Also see links.
    
  • SageMath
    def A000120(n):
        if n <= 1: return Integer(n)
        return A000120(n//2) + n%2
    [A000120(n) for n in range(105)]  # Peter Luschny, Nov 19 2012
    
  • SageMath
    def A000120(n) : return sum(n.digits(2)) # Eric M. Schmidt, Apr 26 2013
    
  • Scala
    (0 to 127).map(Integer.bitCount()) // _Alonso del Arte, Mar 05 2019
    

Formula

a(0) = 0, a(2*n) = a(n), a(2*n+1) = a(n) + 1.
a(0) = 0, a(2^i) = 1; otherwise if n = 2^i + j with 0 < j < 2^i, a(n) = a(j) + 1.
G.f.: Product_{k >= 0} (1 + y*x^(2^k)) = Sum_{n >= 0} y^a(n)*x^n. - N. J. A. Sloane, Jun 04 2009
a(n) = a(n-1) + 1 - A007814(n) = log_2(A001316(n)) = 2n - A005187(n) = A070939(n) - A023416(n). - Henry Bottomley, Apr 04 2001; corrected by Ralf Stephan, Apr 15 2002
a(n) = log_2(A000984(n)/A001790(n)). - Benoit Cloitre, Oct 02 2002
For n > 0, a(n) = n - Sum_{k=1..n} A007814(k). - Benoit Cloitre, Oct 19 2002
a(n) = n - Sum_{k>=1} floor(n/2^k) = n - A011371(n). - Benoit Cloitre, Dec 19 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^(2^k)/(1+x^(2^k)). - Ralf Stephan, Apr 19 2003
a(0) = 0, a(n) = a(n - 2^floor(log_2(n))) + 1. Examples: a(6) = a(6 - 2^2) + 1 = a(2) + 1 = a(2 - 2^1) + 1 + 1 = a(0) + 2 = 2; a(101) = a(101 - 2^6) + 1 = a(37) + 1 = a(37 - 2^5) + 2 = a(5 - 2^2) + 3 = a(1 - 2^0) + 4 = a(0) + 4 = 4; a(6275) = a(6275 - 2^12) + 1 = a(2179 - 2^11) + 2 = a(131 - 2^7) + 3 = a(3 - 2^1) + 4 = a(1 - 2^0) + 5 = 5; a(4129) = a(4129 - 2^12) + 1 = a(33 - 2^5) + 2 = a(1 - 2^0) + 3 = 3. - Hieronymus Fischer, Jan 22 2006
A fixed point of the mapping 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, ... With f(i) = floor(n/2^i), a(n) is the number of odd numbers in the sequence f(0), f(1), f(2), f(3), f(4), f(5), ... - Philippe Deléham, Jan 04 2004
When read mod 2 gives the Morse-Thue sequence A010060.
Let floor_pow4(n) denote n rounded down to the next power of four, floor_pow4(n) = 4 ^ floor(log4 n). Then a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(n) = a(floor(n / floor_pow4(n))) + a(n % floor_pow4(n)). - Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
a(n) = n - Sum_{k=2..n} Sum_{j|n, j >= 2} (floor(log_2(j)) - floor(log_2(j-1))). - Hieronymus Fischer, Jun 18 2007
a(n) = A138530(n, 2) for n > 1. - Reinhard Zumkeller, Mar 26 2008
a(A077436(n)) = A159918(A077436(n)); a(A000290(n)) = A159918(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = A063787(n) - A007814(n). - Gary W. Adamson, Jun 04 2009
a(n) = A007814(C(2n, n)) = 1 + A007814(C(2n-1, n)). - Vladimir Shevelev, Jul 20 2009
For odd m >= 1, a((4^m-1)/3) = a((2^m+1)/3) + (m-1)/2 (mod 2). - Vladimir Shevelev, Sep 03 2010
a(n) - a(n-1) = { 1 - a(n-1) if and only if A007814(n) = a(n-1), 1 if and only if A007814(n) = 0, -1 for all other A007814(n) }. - Brent Spillner (spillner(AT)acm.org), Sep 01 2010
a(A001317(n)) = 2^a(n). - Vladimir Shevelev, Oct 25 2010
a(n) = A139351(n) + A139352(n) = Sum_k {A030308(n, k)}. - Philippe Deléham, Oct 14 2011
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = Sum_{j = 1..m+1} (floor(n/2^j + 1/2) - floor(n/2^j)), where m = floor(log_2(n)).
General formulas for the number of digits >= d in the base p representation of n, where 1 <= d < p: a(n) = Sum_{j = 1..m+1} (floor(n/p^j + (p-d)/p) - floor(n/p^j)), where m=floor(log_p(n)); g.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = A213629(n, 1) for n > 0. - Reinhard Zumkeller, Jul 04 2012
a(n) = A240857(n,n). - Reinhard Zumkeller, Apr 14 2014
a(n) = log_2(C(2*n,n) - (C(2*n,n) AND C(2*n,n)-1)). - Gary Detlefs, Jul 10 2014
Sum_{n >= 1} a(n)/2n(2n+1) = (gamma + log(4/Pi))/2 = A344716, where gamma is Euler's constant A001620; see Sondow 2005, 2010 and Allouche, Shallit, Sondow 2007. - Jonathan Sondow, Mar 21 2015
For any integer base b >= 2, the sum of digits s_b(n) of expansion base b of n is the solution of this recurrence relation: s_b(n) = 0 if n = 0 and s_b(n) = s_b(floor(n/b)) + (n mod b). Thus, a(n) satisfies: a(n) = 0 if n = 0 and a(n) = a(floor(n/2)) + (n mod 2). This easily yields a(n) = Sum_{i = 0..floor(log_2(n))} (floor(n/2^i) mod 2). From that one can compute a(n) = n - Sum_{i = 1..floor(log_2(n))} floor(n/2^i). - Marek A. Suchenek, Mar 31 2016
Sum_{k>=1} a(k)/2^k = 2 * Sum_{k >= 0} 1/(2^(2^k)+1) = 2 * A051158. - Amiram Eldar, May 15 2020
Sum_{k>=1} a(k)/(k*(k+1)) = A016627 = log(4). - Bernard Schott, Sep 16 2020
a(m*(2^n-1)) >= n. Equality holds when 2^n-1 >= A000265(m), but also in some other cases, e.g., a(11*(2^2-1)) = 2 and a(19*(2^3-1)) = 3. - Pontus von Brömssen, Dec 13 2020
G.f.: A(x) satisfies A(x) = (1+x)*A(x^2) + x/(1-x^2). - Akshat Kumar, Nov 04 2023

A007088 The binary numbers (or binary words, or binary vectors, or binary expansion of n): numbers written in base 2.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111
Offset: 0

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Keywords

Comments

List of binary numbers. (This comment is to assist people searching for that particular phrase. - N. J. A. Sloane, Apr 08 2016)
Or, numbers that are sums of distinct powers of 10.
Or, numbers having only digits 0 and 1 in their decimal representation.
Complement of A136399; A064770(a(n)) = a(n). - Reinhard Zumkeller, Dec 30 2007
From Rick L. Shepherd, Jun 25 2009: (Start)
Nonnegative integers with no decimal digit > 1.
Thus nonnegative integers n in base 10 such that kn can be calculated by normal addition (i.e., n + n + ... + n, with k n's (but not necessarily k + k + ... + k, with n k's)) or multiplication without requiring any carry operations for 0 <= k <= 9. (End)
For n > 1: A257773(a(n)) = 10, numbers that are Belgian-k for k=0..9. - Reinhard Zumkeller, May 08 2015
For any integer n>=0, find the binary representation and then interpret as decimal representation giving a(n). - Michael Somos, Nov 15 2015
N is in this sequence iff A007953(N) = A101337(N). A028897 is a left inverse. - M. F. Hasler, Nov 18 2019
For n > 0, numbers whose largest decimal digit is 1. - Stefano Spezia, Nov 15 2023

Examples

			a(6)=110 because (1/2)*((1-(-1)^6)*10^0 + (1-(-1)^3)*10^1 + (1-(-1)^1)*10^2) = 10 + 100.
G.f. = x + 10*x^2 + 11*x^3 + 100*x^4 + 101*x^5 + 110*x^6 + 111*x^7 + 1000*x^8 + ...
.
  000    The numbers < 2^n can be regarded as vectors with
  001    a fixed length n if padded with zeros on the left
  010    side. This represents the n-fold Cartesian product
  011    over the set {0, 1}. In the example on the left,
  100    n = 3. (See also the second Python program.)
  101    Binary vectors in this format can also be seen as a
  110    representation of the subsets of a set with n elements.
  111    - _Peter Luschny_, Jan 22 2024
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 21.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §2.8 Binary, Octal, Hexadecimal, p. 64.
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991, p. 383.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000120 (Hammingweight: sum of bits), A000788 (partial sums of A000120), A000069 (A000120 is odd), A001969 (A000120 is even), A023416 (number of bits 0), A059015 (partial sums). Bisections A099820 and A099821.
Cf. A028897 (convert binary to decimal).

Programs

  • Haskell
    a007088 0 = 0
    a007088 n = 10 * a007088 n' + m where (n',m) = divMod n 2
    -- Reinhard Zumkeller, Jan 10 2012
    
  • Maple
    A007088 := n-> convert(n, binary): seq(A007088(n), n=0..50); # R. J. Mathar, Aug 11 2009
  • Mathematica
    Table[ FromDigits[ IntegerDigits[n, 2]], {n, 0, 39}]
    Table[Sum[ (Floor[( Mod[f/2 ^n, 2])])*(10^n) , {n, 0, Floor[Log[2, f]]}], {f, 1, 100}] (* José de Jesús Camacho Medina, Jul 24 2014 *)
    FromDigits/@Tuples[{1,0},6]//Sort (* Harvey P. Dale, Aug 10 2017 *)
  • PARI
    {a(n) = subst( Pol( binary(n)), x, 10)}; /* Michael Somos, Jun 07 2002 */
    
  • PARI
    {a(n) = if( n<=0, 0, n%2 + 10*a(n\2))}; /* Michael Somos, Jun 07 2002 */
    
  • PARI
    a(n)=fromdigits(binary(n),10) \\ Charles R Greathouse IV, Apr 08 2015
    
  • Python
    def a(n): return int(bin(n)[2:])
    print([a(n) for n in range(40)]) # Michael S. Branicky, Jan 10 2021
    
  • Python
    from itertools import product
    n = 4
    for p in product([0, 1], repeat=n): print(''.join(str(x) for x in p))
    # Peter Luschny, Jan 22 2024

Formula

a(n) = Sum_{i=0..m} d(i)*10^i, where Sum_{i=0..m} d(i)*2^i is the base 2 representation of n.
a(n) = (1/2)*Sum_{i>=0} (1-(-1)^floor(n/2^i))*10^i. - Benoit Cloitre, Nov 20 2001
a(n) = A097256(n)/9.
a(2n) = 10*a(n), a(2n+1) = a(2n)+1.
G.f.: 1/(1-x) * Sum_{k>=0} 10^k * x^(2^k)/(1+x^(2^k)) - for sequence as decimal integers. - Franklin T. Adams-Watters, Jun 16 2006
a(A000290(n)) = A001737(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = Sum_{k>=0} A030308(n,k)*10^k. - Philippe Deléham, Oct 19 2011
For n > 0: A054055(a(n)) = 1. - Reinhard Zumkeller, Apr 25 2012
a(n) = Sum_{k=0..floor(log_2(n))} floor((Mod(n/2^k, 2)))*(10^k). - José de Jesús Camacho Medina, Jul 24 2014

A004159 Sum of digits of n^2.

Original entry on oeis.org

0, 1, 4, 9, 7, 7, 9, 13, 10, 9, 1, 4, 9, 16, 16, 9, 13, 19, 9, 10, 4, 9, 16, 16, 18, 13, 19, 18, 19, 13, 9, 16, 7, 18, 13, 10, 18, 19, 13, 9, 7, 16, 18, 22, 19, 9, 10, 13, 9, 7, 7, 9, 13, 19, 18, 10, 13, 18, 16, 16, 9, 13, 19, 27, 19, 13, 18, 25, 16, 18, 13, 10, 18, 19, 22, 18, 25, 25, 18, 13
Offset: 0

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Comments

If 3|n then 9|a(n); otherwise, a(n) == 1 (mod 3). - Jon E. Schoenfield, Jun 30 2018

Examples

			Trajectories under the map x -> a(x):
1 ->  1 ->  1 ->  1 ->  1 ->  1 ->  1 ->  1 ->  1 -> ...
2 ->  4 ->  7 -> 13 -> 16 -> 13 -> 16 -> 13 -> 16 -> ...
3 ->  9 ->  9 ->  9 ->  9 ->  9 ->  9 ->  9 ->  9 -> ...
4 ->  7 -> 13 -> 16 -> 13 -> 16 -> 13 -> 16 -> 13 -> ...
5 ->  7 -> 13 -> 16 -> 13 -> 16 -> 13 -> 16 -> 13 -> ...
6 ->  9 ->  9 ->  9 ->  9 ->  9 ->  9 ->  9 ->  9 -> ...
7 -> 13 -> 16 -> 13 -> 16 -> 13 -> 16 -> 13 -> 16 -> ...
- _R. J. Mathar_, Jul 08 2012
		

Crossrefs

Cf. A240752 (first differences), A071317 (partial sums).
Cf. A062685 (smallest square with digit sum n, or 0 if no such square exists).

Programs

  • Haskell
    a004159 = a007953 . a000290  -- Reinhard Zumkeller, Apr 12 2014
    
  • Maple
    read("transforms"):
    A004159 := proc(n)
            digsum(n^2) ;
    end proc: # R. J. Mathar, Jul 08 2012
  • Mathematica
    a004159[n_Integer] := Apply[Plus, IntegerDigits[n^2]]; Table[
    a004159[n], {n, 0, 100}] (* Michael De Vlieger, Jul 21 2014 *)
    Total[IntegerDigits[#]]&/@(Range[0,100]^2) (* Harvey P. Dale, Feb 03 2019 *)
  • PARI
    A004159(n)=sumdigits(n^2) \\ M. F. Hasler, Sep 23 2014
  • Python
    def A004159(n):
        return sum(int(d) for d in str(n*n)) # Chai Wah Wu, Sep 03 2014
    

Formula

a(n) = A007953(A000290(n)); a(A058369(n)) = A007953(A058369(n)). - Reinhard Zumkeller, Apr 25 2009
a(10n) = a(n). If n > 1 is not a multiple of 10, then a(n)=4 iff n = 10^k+1 = A062397(k), a(n)=7 iff n is in A215614={4, 5, 32, 49, 149, 1049}, and else a(n) >= 9. - M. F. Hasler, Sep 23 2014

A077436 Let B(n) be the sum of binary digits of n. This sequence contains n such that B(n) = B(n^2).

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 79, 91, 96, 112, 120, 124, 126, 127, 128, 157, 158, 159, 182, 183, 187, 192, 224, 240, 248, 252, 254, 255, 256, 279, 287, 314, 316, 317, 318, 319, 351, 364, 365, 366, 374, 375, 379, 384
Offset: 1

Views

Author

Giuseppe Melfi, Nov 30 2002

Keywords

Comments

Superset of A023758.
Hare, Laishram, & Stoll show that this sequence contains infinitely many odd numbers. In particular for each k in {12, 13, 16, 17, 18, 19, 20, ...} there are infinitely many terms in this sequence with binary digit sum k. - Charles R Greathouse IV, Aug 25 2015

Examples

			The element 79 belongs to the sequence because 79=(1001111) and 79^2=(1100001100001), so B(79)=B(79^2)
		

Crossrefs

Cf. A211676 (number of n-bit numbers in this sequence).
A261586 is a subsequence. Subsequence of A352084.

Programs

  • Haskell
    import Data.List (elemIndices)
    import Data.Function (on)
    a077436 n = a077436_list !! (n-1)
    a077436_list = elemIndices 0
       $ zipWith ((-) `on` a000120) [0..] a000290_list
    -- Reinhard Zumkeller, Apr 12 2011
    
  • Magma
    [n: n in [0..400] | &+Intseq(n, 2) eq &+Intseq(n^2, 2)]; // Vincenzo Librandi, Aug 30 2015
    
  • Maple
    select(t -> convert(convert(t,base,2),`+`) = convert(convert(t^2,base,2),`+`), [$0..1000]); # Robert Israel, Aug 27 2015
  • Mathematica
    t={}; Do[If[DigitCount[n, 2, 1] == DigitCount[n^2, 2, 1], AppendTo[t, n]], {n, 0, 364}]; t
    f[n_] := Total@ IntegerDigits[n, 2]; Select[Range[0, 384], f@ # == f[#^2] &] (* Michael De Vlieger, Aug 27 2015 *)
  • PARI
    is(n)=hammingweight(n)==hammingweight(n^2) \\ Charles R Greathouse IV, Aug 25 2015
    
  • Python
    def ok(n): return bin(n).count('1') == bin(n**2).count('1')
    print([m for m in range(400) if ok(m)]) # Michael S. Branicky, Mar 11 2022

Formula

A159918(a(n)) = A000120(a(n)). - Reinhard Zumkeller, Apr 25 2009

Extensions

Initial 0 added by Reinhard Zumkeller, Apr 28 2012, Apr 12 2011

A001737 Squares written in base 2.

Original entry on oeis.org

0, 1, 100, 1001, 10000, 11001, 100100, 110001, 1000000, 1010001, 1100100, 1111001, 10010000, 10101001, 11000100, 11100001, 100000000, 100100001, 101000100, 101101001, 110010000, 110111001, 111100100, 1000010001, 1001000000, 1001110001, 1010100100, 1011011001
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[IntegerDigits[n^2, 2]], {n, 0, 50}] (* T. D. Noe, Aug 10 2012 *)
  • Python
    print([int(bin(k**2)[2:]) for k in range(0, 28)]) # Karl-Heinz Hofmann, Feb 01 2022

Formula

a(n) = A007088(A000290(n)). - Reinhard Zumkeller, Apr 25 2009

A201555 a(n) = C(2*n^2,n^2) = A000984(n^2), where A000984 is the central binomial coefficients.

Original entry on oeis.org

1, 2, 70, 48620, 601080390, 126410606437752, 442512540276836779204, 25477612258980856902730428600, 23951146041928082866135587776380551750, 365907784099042279561985786395502921046971688680, 90548514656103281165404177077484163874504589675413336841320
Offset: 0

Views

Author

Paul D. Hanna, Dec 02 2011

Keywords

Comments

Central coefficients of triangle A228832.

Examples

			L.g.f.: L(x) = 2*x + 70*x^2/2 + 48620*x^3/3 + 601080390*x^4/4 + ...
where exponentiation equals the g.f. of A201556:
exp(L(x)) = 1 + 2*x + 37*x^2 + 16278*x^3 + 150303194*x^4 + ... + A201556(n)*x^n + ...
		

Crossrefs

Programs

  • Mathematica
    Table[Binomial[2n^2,n^2],{n,0,10}] (* Harvey P. Dale, Dec 10 2011 *)
  • PARI
    a(n) = binomial(2*n^2,n^2)
    
  • Python
    from math import comb
    def A201555(n): return comb((m:=n**2)<<1,m) # Chai Wah Wu, Jul 08 2022

Formula

L.g.f.: ignoring initial term, equals the logarithmic derivative of A201556.
a(n) = (2*n^2)! / (n^2)!^2.
a(n) = Sum_{k=0..n^2} binomial(n^2,k)^2.
For primes p >= 5: a(p) == 2 (mod p^3), Oblath, Corollary II; a(p) == binomial(2*p,p) (mod p^6) - see Mestrovic, Section 5, equation 31. - Peter Bala, Dec 28 2014
A007814(a(n)) = A159918(n). - Antti Karttunen, Apr 27 2017, based on Vladimir Shevelev's Jul 20 2009 formula in A000984.

A231897 a(n) = smallest m such that wt(m^2) = n (where wt(i) = A000120(i)), or -1 if no such m exists.

Original entry on oeis.org

0, 1, 3, 5, 13, 11, 21, 39, 45, 75, 155, 217, 331, 181, 627, 923, 1241, 2505, 3915, 5221, 6475, 11309, 15595, 19637, 31595, 44491, 69451, 113447, 185269, 244661, 357081, 453677, 1015143, 908091, 980853, 2960011, 4568757, 2965685, 5931189, 11862197, 20437147
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2013

Keywords

Comments

Conjecture: a(n) is never -1. (It seems likely that the arguments of Lindström (1997) could be modified to establish this conjecture.)
a(n) is the smallest m such that A159918(m) = n (or -1 if ...).

Crossrefs

A089998 are the corresponding squares.

Programs

  • Haskell
    a231897 n = head [x | x <- [1..], a159918 x == n]
    -- Reinhard Zumkeller, Nov 20 2013
    
  • PARI
    a(n)=if(n,my(k); while(hammingweight(k++^2)!=n,); k, 0) \\ Charles R Greathouse IV, Aug 06 2015
    
  • Python
    def wt(n): return bin(n).count('1')
    def a(n):
        m = 2**(n//2) - 1
        while wt(m**2) != n: m += 1
        return m
    print([a(n) for n in range(32)]) # Michael S. Branicky, Feb 06 2022

Formula

a(n) = 2*A211201(n-1) + 1 for n >= 1. - Hugo Pfoertner, Feb 06 2022

Extensions

a(26)-a(40) from Reinhard Zumkeller, Nov 20 2013
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