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.

Showing 1-10 of 95 results. Next

A331759 a(n) = A115004(2n+1).

Original entry on oeis.org

1, 31, 179, 585, 1463, 3065, 5729, 9797, 15737, 24087, 35315, 50073, 69025, 92871, 122475, 158681, 202529, 254597, 315957, 387977, 471589, 568227, 678971, 805241, 948515, 1109675, 1290839, 1493127, 1717571, 1966997, 2242925, 2547277, 2881033, 3246087, 3645459
Offset: 0

Views

Author

N. J. A. Sloane, Feb 04 2020

Keywords

Crossrefs

Programs

  • Python
    from sympy import totient
    def A331759(n): return (2*n+1)**2 + sum(totient(i)*(2*n+2-i)*(4*n+4-i) for i in range(2,2*n+2)) # Chai Wah Wu, Aug 17 2021

Formula

a(n) = (2*n+1)^2 + Sum_{i=2..2n+1} (2*n+2-i)*(4*n+4-i)*phi(i). - Chai Wah Wu, Aug 17 2021

A331760 a(n) = A115004(2n)/4.

Original entry on oeis.org

2, 20, 83, 237, 534, 1054, 1892, 3114, 4880, 7327, 10530, 14730, 20066, 26689, 34882, 44902, 56850, 70946, 87644, 106986, 129470, 155457, 184920, 218596, 256675, 299292, 347261, 400590, 459615, 525217, 597887, 677507, 764657, 860340, 964291, 1077693, 1201321
Offset: 1

Views

Author

N. J. A. Sloane, Feb 04 2020

Keywords

Crossrefs

Programs

  • Python
    from sympy import totient
    def A331760(n): return n**2 + sum(totient(i)*(2*n+1-i)*(4*n+2-i) for i in range(2,2*n+1))//4 # Chai Wah Wu, Aug 17 2021

Formula

a(n) = n^2 + (Sum_{i=2..2n} (2*n+1-i)*(4*n+2-i)*phi(i))/4. - Chai Wah Wu, Aug 17 2021

A335179 a(n) = A115004(n) - A334701(n).

Original entry on oeis.org

0, 2, 7, 26, 55, 118, 203, 350, 513, 802, 1153, 1594, 2105, 2878, 3701, 4770, 5973, 7510, 9221, 11282, 13411, 16202, 19255, 22674, 26371, 31018, 35865, 41422, 47377, 54114, 61297, 69470, 77803, 87758, 98311, 109766, 121851, 136006, 150443, 166278, 182903, 201110, 220179, 241030, 262447, 286886, 312411, 339422, 367551
Offset: 1

Views

Author

N. J. A. Sloane, Jun 22 2020

Keywords

Comments

It would be nice to have a formula or recurrence.

Crossrefs

A000010 Euler totient function phi(n): count numbers <= n and prime to n.

Original entry on oeis.org

1, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 10, 22, 8, 20, 12, 18, 12, 28, 8, 30, 16, 20, 16, 24, 12, 36, 18, 24, 16, 40, 12, 42, 20, 24, 22, 46, 16, 42, 20, 32, 24, 52, 18, 40, 24, 36, 28, 58, 16, 60, 30, 36, 32, 48, 20, 66, 32, 44
Offset: 1

Views

Author

Keywords

Comments

Number of elements in a reduced residue system modulo n.
Degree of the n-th cyclotomic polynomial (cf. A013595). - Benoit Cloitre, Oct 12 2002
Number of distinct generators of a cyclic group of order n. Number of primitive n-th roots of unity. (A primitive n-th root x is such that x^k is not equal to 1 for k = 1, 2, ..., n - 1, but x^n = 1.) - Lekraj Beedassy, Mar 31 2005
Also number of complex Dirichlet characters modulo n; Sum_{k=1..n} a(k) is asymptotic to (3/Pi^2)*n^2. - Steven Finch, Feb 16 2006
a(n) is the highest degree of irreducible polynomial dividing 1 + x + x^2 + ... + x^(n-1) = (x^n - 1)/(x - 1). - Alexander Adamchuk, Sep 02 2006, corrected Sep 27 2006
a(p) = p - 1 for prime p. a(n) is even for n > 2. For n > 2, a(n)/2 = A023022(n) = number of partitions of n into 2 ordered relatively prime parts. - Alexander Adamchuk, Jan 25 2007
Number of automorphisms of the cyclic group of order n. - Benoit Jubin, Aug 09 2008
a(n+2) equals the number of palindromic Sturmian words of length n which are "bispecial", prefix or suffix of two Sturmian words of length n + 1. - Fred Lunnon, Sep 05 2010
Suppose that a and n are coprime positive integers, then by Euler's totient theorem, any factor of n divides a^phi(n) - 1. - Lei Zhou, Feb 28 2012
If m has k prime factors, (p_1, p_2, ..., p_k), then phi(m*n) = (Product_{i=1..k} phi (p_i*n))/phi(n)^(k-1). For example, phi(42*n) = phi(2*n)*phi(3*n)*phi(7*n)/phi(n)^2. - Gary Detlefs, Apr 21 2012
Sum_{n>=1} a(n)/n! = 1.954085357876006213144... This sum is referenced in Plouffe's inverter. - Alexander R. Povolotsky, Feb 02 2013 (see A336334. - Hugo Pfoertner, Jul 22 2020)
The order of the multiplicative group of units modulo n. - Michael Somos, Aug 27 2013
A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Dec 30 2016
From Eric Desbiaux, Jan 01 2017: (Start)
a(n) equals the Ramanujan sum c_n(n) (last term on n-th row of triangle A054533).
a(n) equals the Jordan function J_1(n) (cf. A007434, A059376, A059377, which are the Jordan functions J_2, J_3, J_4, respectively). (End)
For n > 1, a(n) appears to be equal to the number of semi-meander solutions for n with top arches containing exactly 2 mountain ranges and exactly 2 arches of length 1. - Roger Ford, Oct 11 2017
a(n) is the minimum dimension of a lattice able to generate, via cut-and-project, the quasilattice whose diffraction pattern features n-fold rotational symmetry. The case n=15 is the first n > 1 in which the following simpler definition fails: "a(n) is the minimum dimension of a lattice with n-fold rotational symmetry". - Felix Flicker, Nov 08 2017
Number of cyclic Latin squares of order n with the first row in ascending order. - Eduard I. Vatutin, Nov 01 2020
a(n) is the number of rational numbers p/q >= 0 (in lowest terms) such that p + q = n. - Rémy Sigrist, Jan 17 2021
From Richard L. Ollerton, May 08 2021: (Start)
Formulas for the numerous OEIS entries involving Dirichlet convolution of a(n) and some sequence h(n) can be derived using the following (n >= 1):
Sum_{d|n} phi(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k)) [see P. H. van der Kamp link] = Sum_{d|n} h(d)*phi(n/d) = Sum_{k=1..n} h(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). Similarly,
Sum_{d|n} phi(d)*h(d) = Sum_{k=1..n} h(n/gcd(n,k)) = Sum_{k=1..n} h(gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)).
More generally,
Sum_{d|n} h(d) = Sum_{k=1..n} h(gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))/phi(n/gcd(n,k)).
In particular, for sequences involving the Möbius transform:
Sum_{d|n} mu(d)*h(n/d) = Sum_{k=1..n} h(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)) = Sum_{k=1..n} h(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)), where mu = A008683.
Use of gcd(n,k)*lcm(n,k) = n*k and phi(gcd(n,k))*phi(lcm(n,k)) = phi(n)*phi(k) provide further variations. (End)
From Richard L. Ollerton, Nov 07 2021: (Start)
Formulas for products corresponding to the sums above may found using the substitution h(n) = log(f(n)) where f(n) > 0 (for example, cf. formulas for the sum A018804 and product A067911 of gcd(n,k)):
Product_{d|n} f(n/d)^phi(d) = Product_{k=1..n} f(gcd(n,k)) = Product_{d|n} f(d)^phi(n/d) = Product_{k=1..n} f(n/gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))),
Product_{d|n} f(d)^phi(d) = Product_{k=1..n} f(n/gcd(n,k)) = Product_{k=1..n} f(gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))),
Product_{d|n} f(d) = Product_{k=1..n} f(gcd(n,k))^(1/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(1/phi(n/gcd(n,k))),
Product_{d|n} f(n/d)^mu(d) = Product_{k=1..n} f(gcd(n,k))^(mu(n/gcd(n,k))/phi(n/gcd(n,k))) = Product_{k=1..n} f(n/gcd(n,k))^(mu(gcd(n,k))/phi(n/gcd(n,k))), where mu = A008683. (End)
a(n+1) is the number of binary words with exactly n distinct subsequences (when n > 0). - Radoslaw Zak, Nov 29 2021

Examples

			G.f. = x + x^2 + 2*x^3 + 2*x^4 + 4*x^5 + 2*x^6 + 6*x^7 + 4*x^8 + 6*x^9 + 4*x^10 + ...
a(8) = 4 with {1, 3, 5, 7} units modulo 8. a(10) = 4 with {1, 3, 7, 9} units modulo 10. - _Michael Somos_, Aug 27 2013
From _Eduard I. Vatutin_, Nov 01 2020: (Start)
The a(5)=4 cyclic Latin squares with the first row in ascending order are:
  0 1 2 3 4   0 1 2 3 4   0 1 2 3 4   0 1 2 3 4
  1 2 3 4 0   2 3 4 0 1   3 4 0 1 2   4 0 1 2 3
  2 3 4 0 1   4 0 1 2 3   1 2 3 4 0   3 4 0 1 2
  3 4 0 1 2   1 2 3 4 0   4 0 1 2 3   2 3 4 0 1
  4 0 1 2 3   3 4 0 1 2   2 3 4 0 1   1 2 3 4 0
(End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 840.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 24.
  • M. Baake and U. Grimm, Aperiodic Order Vol. 1: A Mathematical Invitation, Encyclopedia of Mathematics and its Applications 149, Cambridge University Press, 2013: see Tables 3.1 and 3.2.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 409.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 154-156.
  • C. W. Curtis, Pioneers of Representation Theory ..., Amer. Math. Soc., 1999; see p. 3.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, Paris, 2004, Problème 529, pp. 71-257.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, Chapter V.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 115-119.
  • Carl Friedrich Gauss, "Disquisitiones Arithmeticae", Yale University Press, 1965; see p. 21.
  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994, p. 137.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B36.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, th. 60, 62, 63, 288, 323, 328, 330.
  • Peter Hilton and Jean Pedersen, A Mathematical Tapestry, Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, pages 261-264, the Coach theorem.
  • Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21 pp. 281-294.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, New York, Heidelberg, Berlin, 2 vols., 1976, Vol. II, problem 71, p. 126.
  • Paulo Ribenboim, The New Book of Prime Number Records.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 28-33.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 162-167.

Crossrefs

Cf. A002088 (partial sums), A008683, A003434 (steps to reach 1), A007755, A049108, A002202 (values), A011755 (Sum k*phi(k)).
Cf. also A005277 (nontotient numbers). For inverse see A002181, A006511, A058277.
Jordan function J_k(n) is a generalization - see A059379 and A059380 (triangle of values of J_k(n)), this sequence (J_1), A007434 (J_2), A059376 (J_3), A059377 (J_4), A059378 (J_5).
Row sums of triangles A134540, A127448, A143239, A143353 and A143276.
Equals right and left borders of triangle A159937. - Gary W. Adamson, Apr 26 2009
Values for prime powers p^e: A006093 (e=1), A036689 (e=2), A135177 (e=3), A138403 (e=4), A138407 (e=5), A138412 (e=6).
Values for perfect powers n^e: A002618 (e=2), A053191 (e=3), A189393 (e=4), A238533 (e=5), A306411 (e=6), A239442 (e=7), A306412 (e=8), A239443 (e=9).
Cf. A076479.
Cf. A023900 (Dirichlet inverse of phi), A306633 (Dgf at s=3).

Programs

  • Axiom
    [eulerPhi(n) for n in 1..100]
    
  • Haskell
    a n = length (filter (==1) (map (gcd n) [1..n])) -- Allan C. Wechsler, Dec 29 2014
    
  • Julia
    # Computes the first N terms of the sequence.
    function A000010List(N)
        phi = [i for i in 1:N + 1]
        for i in 2:N + 1
            if phi[i] == i
                for j in i:i:N + 1
                    phi[j] -= div(phi[j], i)
        end end end
    return phi end
    println(A000010List(68))  # Peter Luschny, Sep 03 2023
  • Magma
    [ EulerPhi(n) : n in [1..100] ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
    
  • Maple
    with(numtheory): A000010 := phi; [ seq(phi(n), n=1..100) ]; # version 1
    with(numtheory): phi := proc(n) local i,t1,t2; t1 := ifactors(n)[2]; t2 := n*mul((1-1/t1[i][1]),i=1..nops(t1)); end; # version 2
    # Alternative without library function:
    A000010List := proc(N) local i, j, phi;
        phi := Array([seq(i, i = 1 .. N+1)]);
        for i from 2 to N + 1 do
            if phi[i] = i then
                for j from i by i to N + 1 do
                    phi[j] := phi[j] - iquo(phi[j], i) od
            fi od;
    return phi end:
    A000010List(68);  # Peter Luschny, Sep 03 2023
  • Mathematica
    Array[EulerPhi, 70]
  • Maxima
    makelist(totient(n),n,0,1000); /* Emanuele Munarini, Mar 26 2011 */
    
  • PARI
    {a(n) = if( n==0, 0, eulerphi(n))}; /* Michael Somos, Feb 05 2011 */
    
  • Python
    from sympy.ntheory import totient
    print([totient(i) for i in range(1, 70)])  # Indranil Ghosh, Mar 17 2017
    
  • Python
    # Note also the implementation in A365339.
    
  • Sage
    def A000010(n): return euler_phi(n) # Jaap Spies, Jan 07 2007
    
  • Sage
    [euler_phi(n) for n in range(1, 70)]  # Zerinvary Lajos, Jun 06 2009
    

Formula

phi(n) = n*Product_{distinct primes p dividing n} (1 - 1/p).
Sum_{d divides n} phi(d) = n.
phi(n) = Sum_{d divides n} mu(d)*n/d, i.e., the Moebius transform of the natural numbers; mu() = Moebius function A008683().
Dirichlet generating function Sum_{n>=1} phi(n)/n^s = zeta(s-1)/zeta(s). Also Sum_{n >= 1} phi(n)*x^n/(1 - x^n) = x/(1 - x)^2.
Multiplicative with a(p^e) = (p - 1)*p^(e-1). - David W. Wilson, Aug 01 2001
Sum_{n>=1} (phi(n)*log(1 - x^n)/n) = -x/(1 - x) for -1 < x < 1 (cf. A002088) - Henry Bottomley, Nov 16 2001
a(n) = binomial(n+1, 2) - Sum_{i=1..n-1} a(i)*floor(n/i) (see A000217 for inverse). - Jon Perry, Mar 02 2004
It is a classical result (certainly known to Landau, 1909) that lim inf n/phi(n) = 1 (taking n to be primes), lim sup n/(phi(n)*log(log(n))) = e^gamma, with gamma = Euler's constant (taking n to be products of consecutive primes starting from 2 and applying Mertens' theorem). See e.g. Ribenboim, pp. 319-320. - Pieter Moree, Sep 10 2004
a(n) = Sum_{i=1..n} |k(n, i)| where k(n, i) is the Kronecker symbol. Also a(n) = n - #{1 <= i <= n : k(n, i) = 0}. - Benoit Cloitre, Aug 06 2004 [Corrected by Jianing Song, Sep 25 2018]
Conjecture: Sum_{i>=2} (-1)^i/(i*phi(i)) exists and is approximately 0.558 (A335319). - Orges Leka (oleka(AT)students.uni-mainz.de), Dec 23 2004
From Enrique Pérez Herrero, Sep 07 2010: (Start)
a(n) = Sum_{i=1..n} floor(sigma_k(i*n)/sigma_k(i)*sigma_k(n)), where sigma_2 is A001157.
a(n) = Sum_{i=1..n} floor(tau_k(i*n)/tau_k(i)*tau_k(n)), where tau_3 is A007425.
a(n) = Sum_{i=1..n} floor(rad(i*n)/rad(i)*rad(n)), where rad is A007947. (End)
a(n) = A173557(n)*A003557(n). - R. J. Mathar, Mar 30 2011
a(n) = A096396(n) + A096397(n). - Reinhard Zumkeller, Mar 24 2012
phi(p*n) = phi(n)*(floor(((n + p - 1) mod p)/(p - 1)) + p - 1), for primes p. - Gary Detlefs, Apr 21 2012
For odd n, a(n) = 2*A135303((n-1)/2)*A003558((n-1)/2) or phi(n) = 2*c*k; the Coach theorem of Pedersen et al. Cf. A135303. - Gary W. Adamson, Aug 15 2012
G.f.: Sum_{n>=1} mu(n)*x^n/(1 - x^n)^2, where mu(n) = A008683(n). - Mamuka Jibladze, Apr 05 2015
a(n) = n - cototient(n) = n - A051953(n). - Omar E. Pol, May 14 2016
a(n) = lim_{s->1} n*zeta(s)*(Sum_{d divides n} A008683(d)/(e^(1/d))^(s-1)), for n > 1. - Mats Granvik, Jan 26 2017
Conjecture: a(n) = Sum_{a=1..n} Sum_{b=1..n} Sum_{c=1..n} 1 for n > 1. The sum is over a,b,c such that n*c - a*b = 1. - Benedict W. J. Irwin, Apr 03 2017
a(n) = Sum_{j=1..n} gcd(j, n) cos(2*Pi*j/n) = Sum_{j=1..n} gcd(j, n) exp(2*Pi*i*j/n) where i is the imaginary unit. Notice that the Ramanujan's sum c_n(k) := Sum_{j=1..n, gcd(j, n) = 1} exp(2*Pi*i*j*k/n) gives a(n) = Sum_{k|n} k*c_(n/k)(1) = Sum_{k|n} k*mu(n/k). - Michael Somos, May 13 2018
G.f.: x*d/dx(x*d/dx(log(Product_{k>=1} (1 - x^k)^(-mu(k)/k^2)))), where mu(n) = A008683(n). - Mamuka Jibladze, Sep 20 2018
a(n) = Sum_{d|n} A007431(d). - Steven Foster Clark, May 29 2019
G.f. A(x) satisfies: A(x) = x/(1 - x)^2 - Sum_{k>=2} A(x^k). - Ilya Gutkovskiy, Sep 06 2019
a(n) >= sqrt(n/2) (Nicolas). - Hugo Pfoertner, Jun 01 2020
a(n) > n/(exp(gamma)*log(log(n)) + 5/(2*log(log(n)))), except for n=223092870 (Rosser, Schoenfeld). - Hugo Pfoertner, Jun 02 2020
From Bernard Schott, Nov 28 2020: (Start)
Sum_{m=1..n} 1/a(m) = A028415(n)/A048049(n) -> oo when n->oo.
Sum_{n >= 1} 1/a(n)^2 = A109695.
Sum_{n >= 1} 1/a(n)^3 = A335818.
Sum_{n >= 1} 1/a(n)^k is convergent iff k > 1.
a(2n) = a(n) iff n is odd, and, a(2n) > a(n) iff n is even. (End) [Actually, a(2n) = 2*a(n) for even n. - Jianing Song, Sep 18 2022]
a(n) = 2*A023896(n)/n, n > 1. - Richard R. Forberg, Feb 03 2021
From Richard L. Ollerton, May 09 2021: (Start)
For n > 1, Sum_{k=1..n} phi^{(-1)}(n/gcd(n,k))*a(gcd(n,k))/a(n/gcd(n,k)) = 0, where phi^{(-1)} = A023900.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(gcd(n,k)))*rad(gcd(n,k))/gcd(n,k) = 0.
For n > 1, Sum_{k=1..n} a(gcd(n,k))*mu(rad(n/gcd(n,k)))*rad(n/gcd(n,k))*gcd(n,k) = 0.
Sum_{k=1..n} a(gcd(n,k))/a(n/gcd(n,k)) = n. (End)
a(n) = Sum_{d|n, e|n} gcd(d, e)*mobius(n/d)*mobius(n/e) (the sum is a multiplicative function of n by Tóth, and takes the value p^e - p^(e-1) for n = p^e, a prime power). - Peter Bala, Jan 22 2024
Sum_{n >= 1} phi(n)*x^n/(1 + x^n) = x + 3*x^3 + 5*x^5 + 7*x^7 + ... = Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(4*n-2)). For the first equality see Pólya and Szegő, problem 71, p. 126. - Peter Bala, Feb 29 2024
Conjecture: a(n) = lim_{k->oo} (n^(k + 1))/A000203(n^k). - Velin Yanev, Dec 04 2024 [A000010(p) = p-1, A000203(p^k) = (p^(k+1)-1)/(p-1), so the conjecture is true if n is prime. - Vaclav Kotesovec, Dec 19 2024]

A000196 Integer part of square root of n. Or, number of positive squares <= n. Or, n appears 2n+1 times.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Also the integer part of the geometric mean of the divisors of n. - Amarnath Murthy, Dec 19 2001
Number of numbers k (<= n) with an odd number of divisors. - Benoit Cloitre, Sep 07 2002
Also, for n > 0, the number of digits when writing n in base where place values are squares, cf. A007961; A190321(n) <= a(n). - Reinhard Zumkeller, May 08 2011
The least monotonic left inverse of squares, A000290. That is, the lexicographically least nondecreasing sequence a(n) such that a(A000290(n)) = n. - Antti Karttunen, Oct 06 2017

Examples

			G.f. = x + x^2 + x^3 + 2*x^4 + 2*x^5 + 2*x^6 + 2*x^7 + 2*x^8 + 3*x^9 + ...
		

References

  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, p. 73, problem 23.
  • Lionel Levine, Fractal sequences and restricted Nim, Ars Combin. 80 (2006), 113-127.
  • Paul J. McCarthy, Introduction to Arithmetical Functions, Springer Verlag, 1986, p. 28.
  • N. J. A. Sloane and Allan Wilks, On sequences of Recaman type, paper in preparation, 2006.

Crossrefs

Programs

  • Haskell
    import Data.Bits (shiftL, shiftR)
    a000196 :: Integer -> Integer
    a000196 0 = 0
    a000196 n = newton n (findx0 n 1) where
       -- find x0 == 2^(a+1), such that 4^a <= n < 4^(a+1).
       findx0 0 b = b
       findx0 a b = findx0 (a `shiftR` 2) (b `shiftL` 1)
       newton n x = if x' < x then newton n x' else x
                    where x' = (x + n `div` x) `div` 2
    a000196_list = concat $ zipWith replicate [1,3..] [0..]
    -- Reinhard Zumkeller, Apr 12 2012, Oct 23 2010
    
  • Julia
    a(n) = isqrt(n) # Paul Muljadi, Jun 03 2024
  • Magma
    [Isqrt(n) : n in [0..100]];
    
  • Maple
    Digits := 100; A000196 := n->floor(evalf(sqrt(n)));
  • Mathematica
    Table[n, {n, 0, 20}, {2n + 1}] //Flatten (* Zak Seidov Mar 19 2011 *)
    IntegerPart[Sqrt[Range[0, 110]]] (* Harvey P. Dale, May 23 2012 *)
    Floor[Sqrt[Range[0, 99]]] (* Alonso del Arte, Dec 31 2013 *)
    a[ n_] := SeriesCoefficient[ (EllipticTheta[ 3, 0, x]  - 1) / (2 (1 - x)), {x, 0, n}]; (* Michael Somos, May 28 2014 *)
  • PARI
    {a(n) = if( n<0, 0, sqrtint(n))};
    
  • Python
    # from http://code.activestate.com/recipes/577821-integer-square-root-function/
    def A000196(n):
      if n < 0:
        raise ValueError('only defined for nonnegative n')
      if n == 0:
        return 0
      a, b = divmod(n.bit_length(), 2)
      j = 2**(a+b)
      while True:
        k = (j + n//j)//2
        if k >= j:
          return j
        j = k
    print([A000196(n)for n in range(102)])
    # Jason Kimberley, Nov 09 2016
    
  • Python
    from math import isqrt
    def a(n): return isqrt(n)
    print([a(n) for n in range(102)]) # Michael S. Branicky, Feb 15 2023
    
  • Scheme
    ;; The following implementation uses higher order function LEFTINV-LEASTMONO-NC2NC from my IntSeq-library. It returns the least monotonic left inverse of any strictly growing function (see the comment-section for the definition) and although it does not converge as fast to the result as many specialized integer square root algorithms, at least it does not involve any floating point arithmetic. Thus with correctly implemented bignums it will produce correct results even with very large arguments, in contrast to just taking the floor of (sqrt n).
    ;; Source of LEFTINV-LEASTMONO-NC2NC can be found under https://github.com/karttu/IntSeq/blob/master/src/Transforms/transforms-core.ss and the definition of A000290 is given under that entry.
    (define A000196 (LEFTINV-LEASTMONO-NC2NC 0 0 A000290)) ;; Antti Karttunen, Oct 06 2017
    

Formula

a(n) = Card(k, 0 < k <= n such that k is relatively prime to core(k)) where core(x) is the squarefree part of x. - Benoit Cloitre, May 02 2002
a(n) = a(n-1) + floor(n/(a(n-1)+1)^2), a(0) = 0. - Reinhard Zumkeller, Apr 12 2004
From Hieronymus Fischer, May 26 2007: (Start)
a(n) = Sum_{k=1..n} A010052(k).
G.f.: g(x) = (1/(1-x))*Sum_{j>=1} x^(j^2) = (theta_3(0, x) - 1)/(2*(1-x)) where theta_3 is a Jacobi theta function. (End)
a(n) = floor(A000267(n)/2). - Reinhard Zumkeller, Jun 27 2011
a(n) = floor(sqrt(n)). - Arkadiusz Wesolowski, Jan 09 2013
Sum_{n>0} 1/a(n)^s = 2*zeta(s-1) + zeta(s), where zeta is the Riemann zeta function. - Enrique Pérez Herrero, Oct 15 2013
From Wesley Ivan Hurt, Dec 31 2013: (Start)
a(n) = Sum_{i=1..n} (A000005(i) mod 2), n > 0.
a(n) = (1/2)*Sum_{i=1..n} (1 - (-1)^A000005(i)), n > 0. (End)
a(n) = sqrt(A048760(n)), n >= 0. - Wolfdieter Lang, Mar 24 2015
a(n) = Sum_{k=1..n} floor(n/k)*lambda(k) = Sum_{m=1..n} Sum_{d|m} lambda(d) where lambda(j) is Liouville lambda function, A008836. - Geoffrey Critzer, Apr 01 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) (A002162). - Amiram Eldar, May 02 2023

A000201 Lower Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi), where phi = (1+sqrt(5))/2 = A001622.

Original entry on oeis.org

1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 27, 29, 30, 32, 33, 35, 37, 38, 40, 42, 43, 45, 46, 48, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 66, 67, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 92, 93, 95, 97, 98, 100, 101, 103, 105, 106, 108, 110
Offset: 1

Views

Author

Keywords

Comments

This is the unique sequence a satisfying a'(n)=a(a(n))+1 for all n in the set N of natural numbers, where a' denotes the ordered complement (in N) of a. - Clark Kimberling, Feb 17 2003
This sequence and A001950 may be defined as follows. Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
These are the numbers whose lazy Fibonacci representation (see A095791) includes 1; the complementary sequence (the upper Wythoff sequence, A001950) are the numbers whose lazy Fibonacci representation includes 2 but not 1.
a(n) is the unique monotonic sequence satisfying a(1)=1 and the condition "if n is in the sequence then n+(rank of n) is not in the sequence" (e.g. a(4)=6 so 6+4=10 and 10 is not in the sequence) - Benoit Cloitre, Mar 31 2006
Write A for A000201 and B for A001950 (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB,...,BBB,... appear in many complementary equations having solution A000201 (or equivalently, A001950). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Cumulative sum of A001468 terms. - Eric Angelini, Aug 19 2008
The lower Wythoff sequence also can be constructed by playing the so-called Mancala-game: n piles of total d(n) chips are standing in a row. The piles are numbered from left to right by 1, 2, 3, ... . The number of chips in a pile at the beginning of the game is equal to the number of the pile. One step of the game is described as follows: Distribute the pile on the very left one by one to the piles right of it. If chips are remaining, build piles out of one chip subsequently to the right. After f(n) steps the game ends in a constant row of piles. The lower Wythoff sequence is also given by n -> f(n). - Roland Schroeder (florola(AT)gmx.de), Jun 19 2010
With the exception of the first term, a(n) gives the number of iterations required to reverse the list {1,2,3,...,n} when using the mapping defined as follows: remove the first term of the list, z(1), and add 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list. See A183110 where this mapping is used and other references given. This appears to be essentially the Mancala-type game interpretation given by R. Schroeder above. - John W. Layman, Feb 03 2011
Also row numbers of A213676 starting with an even number of zeros. - Reinhard Zumkeller, Mar 10 2013
From Jianing Song, Aug 18 2022: (Start)
Numbers k such that {k*phi} > phi^(-2), where {} denotes the fractional part.
Proof: Write m = floor(k*phi).
If {k*phi} > phi^(-2), take s = m-k+1. From m < k*phi < m+1 we have k < (m-k+1)*phi < k + phi, so floor(s*phi) = k or k+1. If floor(s*phi) = k+1, then (see A003622) floor((k+1)*phi) = floor(floor(s*phi)*phi) = floor(s*phi^2)-1 = s+floor(s*phi)-1 = m+1, but actually we have (k+1)*phi > m+phi+phi^(-2) = m+2, a contradiction. Hence floor(s*phi) = k.
If floor(s*phi) = k, suppose otherwise that k*phi - m <= phi^(-2), then m < (k+1)*phi <= m+2, so floor((k+1)*phi) = m+1. Suppose that A035513(p,q) = k for p,q >= 1, then A035513(p,q+1) = floor((k+1)*phi) - 1 = m = A035513(s,1). But it is impossible for one number (m) to occur twice in A035513. (End)
The formula from Jianing Song above is a direct consequence of an old result by Carlitz et al. (1972). Their Theorem 11 states that (a(n)) consists of the numbers k such that {k*phi^(-2)} < phi^(-1). One has {k*phi^(-2)} = {k*(2-phi)} = {-k*phi}. Using that 1-phi^(-1) = phi^(-2), the Jianing Song formula follows. - Michel Dekking, Oct 14 2023
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,1)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 28 2025

Examples

			From Roland Schroeder (florola(AT)gmx.de), Jul 13 2010: (Start)
Example for n = 5; a(5) = 8;
(Start: [1,2,3,4,5]; 8 steps until [5,4,3,2,1]):
[1,2,3,4,5]; [3,3,4,5]; [4,5,6]; [6,7,1,1]; [8,2,2,1,1,1]: [3,3,2,2,2,1,1,1]; [4,3,3,2,1,1,1]; [4,4,3,2,1,1]; [5,4,3,2,1]. (End)
		

References

  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, date?
  • M. Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • 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).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Crossrefs

a(n) = least k such that s(k) = n, where s = A026242. Complement of A001950. See also A058066.
The permutation A002251 maps between this sequence and A001950, in that A002251(a(n)) = A001950(n), A002251(A001950(n)) = a(n).
First differences give A014675. a(n) = A022342(n) + 1 = A005206(n) + n + 1. a(2n)-a(n)=A007067(n). a(a(a(n)))-a(n) = A026274(n-1). - Benoit Cloitre, Mar 08 2003
A185615 gives values n such that n divides A000201(n)^m for some integer m>0.
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021
Bisections: A276854, A342279.

Programs

  • Haskell
    a000201 n = a000201_list !! (n-1)
    a000201_list = f [1..] [1..] where
       f (x:xs) (y:ys) = y : f xs (delete (x + y) ys)
    -- Reinhard Zumkeller, Jul 02 2015, Mar 10 2013
    
  • Maple
    Digits := 100; t := evalf((1+sqrt(5))/2); A000201 := n->floor(t*n);
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])/2]], {n, 1, 75}]
    Array[ Floor[ #*GoldenRatio] &, 68] (* Robert G. Wilson v, Apr 17 2010 *)
  • Maxima
    makelist(floor(n*(1+sqrt(5))/2),n,1,60); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    a(n)=floor(n*(sqrt(5)+1)/2)
    
  • PARI
    a(n)=(n+sqrtint(5*n^2))\2 \\ Charles R Greathouse IV, Feb 07 2013
    
  • Python
    def aupton(terms):
      alst, aset = [None, 1], {1}
      for n in range(1, terms):
        an = alst[n] + (1 if n not in aset else 2)
        alst.append(an); aset.add(an)
      return alst[1:]
    print(aupton(68)) # Michael S. Branicky, May 14 2021
    
  • Python
    from math import isqrt
    def A000201(n): return (n+isqrt(5*n**2))//2 # Chai Wah Wu, Jan 11 2022

Formula

Zeckendorf expansion of n (cf. A035517) ends with an even number of 0's.
Other properties: a(1)=1; for n>1, a(n) is taken to be the smallest integer greater than a(n-1) which is consistent with the condition "n is in the sequence if and only if a(n)+1 is not in the sequence".
a(1) = 1; for n>0, a(n+1) = a(n)+1 if n is not in the sequence, a(n+1) = a(n)+2 if n is in the sequence.
a(a(n)) = floor(n*phi^2) - 1 = A003622(n).
{a(k)} union {a(k)+1} = {1, 2, 3, 4, ...}. Hence a(1) = 1; for n>1, a(a(n)) = a(a(n)-1)+2, a(a(n)+1) = a(a(n))+1. - Benoit Cloitre, Mar 08 2003
{a(n)} is a solution to the recurrence a(a(n)+n) = 2*a(n)+n, a(1)=1 (see Barbeau et al.).
a(n) = A001950(n) - n. - Philippe Deléham, May 02 2004
a(0) = 0; a(n) = n + Max_{k : a(k) < n}. - Vladeta Jovovic, Jun 11 2004
a(Fibonacci(r-1)+j) = Fibonacci(r)+a(j) for 0 < j <= Fibonacci(r-2); 2 < r. - Paul Weisenhorn, Aug 18 2012
With 1 < k and A001950(k-1) < n <= A001950(k): a(n) = 2*n-k; A001950(n) = 3*n-k. - Paul Weisenhorn, Aug 21 2012

A051953 Cototient(n) := n - phi(n).

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 1, 4, 3, 6, 1, 8, 1, 8, 7, 8, 1, 12, 1, 12, 9, 12, 1, 16, 5, 14, 9, 16, 1, 22, 1, 16, 13, 18, 11, 24, 1, 20, 15, 24, 1, 30, 1, 24, 21, 24, 1, 32, 7, 30, 19, 28, 1, 36, 15, 32, 21, 30, 1, 44, 1, 32, 27, 32, 17, 46, 1, 36, 25, 46, 1, 48, 1, 38, 35, 40, 17, 54, 1, 48, 27
Offset: 1

Views

Author

Labos Elemer, Dec 21 1999

Keywords

Comments

Unlike totients, cototient(n+1) = cototient(n) never holds -- except 2-phi(2) = 3 - phi(3) = 1 -- because cototient(n) is congruent to n modulo 2. - Labos Elemer, Aug 08 2001
Theorem (L. Redei): b^a(n) == b^n (mod n) for every integer b. - Thomas Ordowski and Robert Israel, Mar 11 2016
Let S be the sum of the cototients of the divisors of n (A001065). S < n iff n is deficient, S = n iff n is perfect, and S > n iff n is abundant. - Ivan N. Ianakiev, Oct 06 2023

Examples

			n = 12, phi(12) = 4 = |{1, 5, 7, 11}|, a(12) = 12 - phi(12) = 8, numbers not exceeding 12 and not coprime to 12: {2, 3, 4, 6, 8, 9, 10, 12}.
		

Crossrefs

Cf. A000010, A001065 (inverse Möbius transform), A005278, A001274, A083254, A098006, A049586, A051612, A053579, A054525, A062790 (Möbius transform), A063985 (partial sums), A063986, A290087.
Records: A065385, A065386.
Number of zeros in the n-th row of triangle A054521. - Omar E. Pol, May 13 2016
Cf. A063740 (number of k such that cototient(k) = n). - M. F. Hasler, Jan 11 2018

Programs

  • Haskell
    a051953 n = n - a000010 n  -- Reinhard Zumkeller, Jan 21 2014
    
  • Maple
    with(numtheory); A051953 := n->n-phi(n);
  • Mathematica
    Table[n - EulerPhi[n], {n, 1, 80}] (* Carl Najafi, Aug 16 2011 *)
  • PARI
    A051953(n) = n - eulerphi(n); \\ Michael B. Porter, Jan 28 2010
    
  • Python
    from sympy.ntheory import totient
    print([i - totient(i) for i in range(1, 101)]) # Indranil Ghosh, Mar 17 2017

Formula

a(n) = n - A000010(n).
Equals Mobius transform (A054525) of A001065. - Gary W. Adamson, Jul 11 2008
a(A006881(n)) = sopf(A006881(n)) - 1; a(A000040(n)) = 1. - Wesley Ivan Hurt, May 18 2013
G.f.: sum(n>=1, A000010(n)*x^(2*n)/(1-x^n) ). - Mircea Merca, Feb 23 2014
From Ilya Gutkovskiy, Apr 13 2017: (Start)
G.f.: -Sum_{k>=2} mu(k)*x^k/(1 - x^k)^2.
Dirichlet g.f.: zeta(s-1)*(1 - 1/zeta(s)). (End)
From Antti Karttunen, Sep 05 2018 & Apr 29 2022: (Start)
Dirichlet convolution square of A317846/A046644 gives this sequence + A063524.
a(n) = A003557(n) * A318305(n).
a(n) = A000010(n) - A083254(n).
a(n) = A318325(n) - A318326(n).
a(n) = Sum_{d|n} A062790(d) = Sum_{d|n, dA007431(d)*(A000005(n/d)-1).
a(n) = A048675(A318834(n)) = A276085(A353564(n)). [These follow from the formula below]
a(n) = Sum_{d|n, dA000010(d).
a(n) = A051612(n) - A001065(n).
(End)

A001950 Upper Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi = (1+sqrt(5))/2.

Original entry on oeis.org

2, 5, 7, 10, 13, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 41, 44, 47, 49, 52, 54, 57, 60, 62, 65, 68, 70, 73, 75, 78, 81, 83, 86, 89, 91, 94, 96, 99, 102, 104, 107, 109, 112, 115, 117, 120, 123, 125, 128, 130, 133, 136, 138, 141, 143, 146, 149, 151, 154, 157
Offset: 1

Views

Author

Keywords

Comments

Indices at which blocks (1;0) occur in infinite Fibonacci word; i.e., n such that A005614(n-2) = 0 and A005614(n-1) = 1. - Benoit Cloitre, Nov 15 2003
A000201 and this sequence may be defined as follows: Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004
a(n) = n-th integer which is not equal to the floor of any multiple of phi, where phi = (1+sqrt(5))/2 = golden number. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), May 09 2007
Write A for A000201 and B for the present sequence (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB, ..., BBB, ... appear in many complementary equations having solution A000201 (or equivalently, the present sequence). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007
Apart from the initial 0 in A090909, is this the same as that sequence? - Alec Mihailovs (alec(AT)mihailovs.com), Jul 23 2007
If we define a base-phi integer as a positive number whose representation in the golden ratio base consists only of nonnegative powers of phi, and if these base-phi integers are ordered in increasing order (beginning 1, phi, ...), then it appears that the difference between the n-th and (n-1)-th base-phi integer is phi-1 if and only if n belongs to this sequence, and the difference is 1 otherwise. Further, if each base-phi integer is written in linear form as a + b*phi (for example, phi^2 is written as 1 + phi), then it appears that there are exactly two base-phi integers with b=n if and only if n belongs to this sequence, and exactly three base-phi integers with b=n otherwise. - Geoffrey Caveney, Apr 17 2014
Numbers with an odd number of trailing zeros in their Zeckendorf representation (A014417). - Amiram Eldar, Feb 26 2021
Numbers missing from A066096. - Philippe Deléham, Jan 19 2023

Examples

			From _Paul Weisenhorn_, Aug 18 2012 and Aug 21 2012: (Start)
a(14) = floor(14*phi^2) = 36; a'(14) = floor(14*phi)=22;
with r=9 and j=1: a(13+1) = 34 + 2 = 36;
with r=8 and j=1: a'(13+1) = 21 + 1 = 22.
k=6 and a(5)=13 < n <= a(6)=15
a(14) = 3*14 - 6 = 36; a'(14) = 2*14 - 6 = 22;
a(15) = 3*15 - 6 = 39; a'(15) = 2*15 - 6 = 24. (End)
		

References

  • Claude Berge, Graphs and Hypergraphs, North-Holland, 1973; p. 324, Problem 2.
  • Eric Friedman, Scott M. Garrabrant, Ilona K. Phipps-Morgan, A. S. Landsberg and Urban Larsson, Geometric analysis of a generalized Wythoff game, in Games of no Chance 5, MSRI publ. Cambridge University Press, 2019.
  • Martin Gardner, Penrose Tiles to Trapdoor Ciphers, W. H. Freeman, 1989; see p. 107.
  • 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).
  • I. M. Yaglom, Two games with matchsticks, pp. 1-7 of Qvant Selecta: Combinatorics I, Amer Math. Soc., 2001.

Crossrefs

a(n) = greatest k such that s(k) = n, where s = A026242.
Complement of A000201 or A066096.
A002251 maps between A000201 and A001950, in that A002251(A000201(n)) = A001950(n), A002251(A001950(n)) = A000201(n).
Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864.
First differences give (essentially) A076662.
Bisections: A001962, A001966.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a001950 n = a000201 n + n  -- Reinhard Zumkeller, Mar 10 2013
    
  • Magma
    [Floor(n*((1+Sqrt(5))/2)^2): n in [1..80]]; // Vincenzo Librandi, Nov 19 2016
    
  • Maple
    A001950 := proc(n)
        floor(n*(3+sqrt(5))/2) ;
    end proc:
    seq(A001950(n),n=0..40) ; # R. J. Mathar, Jul 16 2024
  • Mathematica
    Table[Floor[N[n*(1+Sqrt[5])^2/4]], {n, 1, 75}]
    Array[ Floor[ #*GoldenRatio^2] &, 60] (* Robert G. Wilson v, Apr 17 2010 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+3)/2)
    
  • PARI
    A001950(n)=(sqrtint(n^2*5)+n*3)\2 \\ M. F. Hasler, Sep 17 2014
    
  • Python
    from math import isqrt
    def A001950(n): return (n+isqrt(5*n**2)>>1)+n # Chai Wah Wu, Aug 10 2022

Formula

a(n) = n + floor(n*phi). In general, floor(n*phi^m) = Fibonacci(m-1)*n + floor(Fibonacci(m)*n*phi). - Benoit Cloitre, Mar 18 2003
a(n) = n + floor(n*phi) = n + A000201(n). - Paul Weisenhorn and Philippe Deléham
Append a 0 to the Zeckendorf expansion (cf. A035517) of n-th term of A000201.
a(n) = A003622(n) + 1. - Philippe Deléham, Apr 30 2004
a(n) = Min(m: A134409(m) = A006336(n)). - Reinhard Zumkeller, Oct 24 2007
If a'=A000201 is the ordered complement (in N) of {a(n)}, then a(Fib(r-2) + j) = Fib(r) + a(j) for 0 < j <= Fib(r-2), 3 < r; and a'(Fib(r-1) + j) = Fib(r) + a'(j) for 0 < j <= Fib(r-2), 2 < r. - Paul Weisenhorn, Aug 18 2012
With a(1)=2, a(2)=5, a'(1)=1, a'(2)=3 and 1 < k and a(k-1) < n <= a(k) one gets a(n)=3*n-k, a'(n)=2*n-k. - Paul Weisenhorn, Aug 21 2012

Extensions

Corrected by Michael Somos, Jun 07 2000

A003849 The infinite Fibonacci word (start with 0, apply 0->01, 1->0, take limit).

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1
Offset: 0

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Keywords

Comments

A Sturmian word.
Define strings S(0)=0, S(1)=01, S(n)=S(n-1)S(n-2); iterate; sequence is S(infinity). If the initial 0 is omitted from S(n) for n>0, we obtain A288582(n+1).
The 0's occur at positions in A022342 (i.e., A000201 - 1), the 1's at positions in A003622.
Replace each run (1;1) with (1;0) in the infinite Fibonacci word A005614 (and add 0 as prefix) A005614 begins: 1,0,1,1,0,1,0,1,1,0,1,1,... changing runs (1,1) with (1,0) produces 1,0,0,1,0,1,0,0,1,0,0,1,... - Benoit Cloitre, Nov 10 2003
Characteristic function of A003622. - Philippe Deléham, May 03 2004
The fraction of 0's in the first n terms approaches 1/phi (see for example Allouche and Shallit). - N. J. A. Sloane, Sep 24 2007
The limiting mean and variance of the first n terms are 2-phi and 2*phi-3, respectively. - Clark Kimberling, Mar 12 2014, Aug 16 2018
Let S(n) be defined as above. Then this sequence is S(1) + Sum_{n=0..} S(n), where the addition of strings represents concatenation. - Isaac Saffold, May 03 2019
The word is a concatenation of three runs: 0, 1, and 00. The limiting proportions of these are respectively 1 - phi/2, 1/2, and (phi - 1)/2. The mean runlength is (phi + 1)/2. - Clark Kimberling, Dec 26 2010
From Amiram Eldar, Mar 10 2021: (Start)
a(n) is the number of the trailing 0's in the dual Zeckendorf representation of (n+1) (A104326).
The asymptotic density of the occurrences of k (0 or 1) is 1/phi^(k+1), where phi is the golden ratio (A001622).
The asymptotic mean of this sequence is 1/phi^2 (A132338). (End)

Examples

			The word is 010010100100101001010010010100...
Over the alphabet {a,b} this is a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, a, b, a, b, a, a, b, a, a, b, a, b, a, ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • Jean Berstel, Fibonacci words—a survey, In The book of L, pp. 13-27. Springer Berlin Heidelberg, 1986.
  • J. C. Lagarias, Number Theory and Dynamical Systems, pp. 35-72 of S. A. Burr, ed., The Unreasonable Effectiveness of Number Theory, Proc. Sympos. Appl. Math., 46 (1992). Amer. Math. Soc. - see p. 64.
  • Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337. [See A317208 for a link.]
  • G. Melançon, Factorizing infinite words using Maple, MapleTech journal, vol. 4, no. 1, 1997, pp. 34-42, esp. p. 36.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

There are several versions of this sequence in the OEIS. This one and A003842 are probably the most important. See also A008352, A076662, A288581, A288582.
Positions of 1's gives A003622.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003849 n = a003849_list !! n
    a003849_list = tail $ concat fws where
       fws = [1] : [0] : (zipWith (++) fws $ tail fws)
    -- Reinhard Zumkeller, Nov 01 2013, Apr 07 2012
    
  • Magma
    t1:=[ n le 2 select ["0","0,1"][n] else Self(n-1) cat "," cat Self(n-2) : n in [1..12]]; t1[12];
    
  • Maple
    z := proc(m) option remember; if m=0 then [0] elif m=1 then [0,1] else [op(z(m-1)),op(z(m-2))]; fi; end; z(12);
    M:=19; S[0]:=`0`; S[1]:=`01`; for n from 2 to M do S[n]:=cat(S[n-1], S[n-2]); od:
    t0:=S[M]: l:=length(t0); for i from 1 to l do lprint(i-1,substring(t0,i..i)); od: # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 10] (* Robert G. Wilson v, Mar 05 2005 *)
    Flatten[Nest[{#, #[[1]]} &, {0, 1}, 9]] (* IWABUCHI Yu(u)ki, Oct 23 2013 *)
    Table[Floor[(n + 2) #] - Floor[(n + 1) #], {n, 0, 120}] &[2 - GoldenRatio] (* Michael De Vlieger, Aug 15 2016 *)
    SubstitutionSystem[{0->{0,1},1->{0}},{0},{10}][[1]] (* Harvey P. Dale, Dec 20 2021 *)
  • PARI
    a(n)=my(k=2);while(fibonacci(k)<=n,k++);while(n>1,while(fibonacci(k--)>n,); n-=fibonacci(k)); n==1 \\ Charles R Greathouse IV, Feb 03 2014
    
  • PARI
    M3849=[2,2,1,0]/*L(k),S(k),L(k-1),S(k-1)*/; A003849(n)={while(n>M3849[1],M3849=vecextract(M3849,[1,2,1,2])+[M3849[3],M3849[4]<M. F. Hasler, Apr 07 2021
    
  • Python
    def fib(n):
        """Return the concatenation of A003849(0..F-1) where F is the smallest
           Fibonacci number > n, so that the result contains a(n) at index n."""
        a, b = '10'
        while len(b)<=n:
            a, b = b, b + a
        return b # Robert FERREOL, Apr 15 2016, edited by M. F. Hasler, Apr 07 2021
    
  • Python
    from math import isqrt
    def A003849(n): return 2-(n+2+isqrt(m:=5*(n+2)**2)>>1)+(n+1+isqrt(m-10*n-15)>>1) # Chai Wah Wu, Aug 25 2022

Formula

a(n) = floor((n+2)*r) - floor((n+1)*r) where r=phi/(1+2*phi) and phi is the Golden Ratio. - Benoit Cloitre, Nov 10 2003
a(n) = A003714(n) mod 2 = A014417(n) mod 2. - Philippe Deléham, Jan 04 2004
The first formula by Cloitre is just one of an infinite family of formulas. Using phi^2=1+phi, it follows that r=phi/(1+2*phi)=2-phi. Then from floor(-x)=-floor(x)-1 for non-integer x, it follows that a(n)=2-A014675(n)=2-(floor((n+2)* phi)-floor((n+1)*phi)). - Michel Dekking, Aug 27 2016
a(n) = 1 - A096270(n+1), i.e., A096270 is the complement of this sequence. - A.H.M. Smeets, Mar 31 2024

Extensions

Revised by N. J. A. Sloane, Jul 03 2012

A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19, 21, 22, 24, 25, 26, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 42, 43, 45, 46, 48, 49, 50, 52, 53, 55, 56, 57, 59, 60, 62, 63, 65, 66, 67, 69, 70, 72, 73, 74, 76, 77, 79, 80, 82, 83, 84, 86, 87, 89, 90, 91, 93, 94, 96, 97, 98, 100
Offset: 0

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Comments

Earliest monotonic sequence greater than 0 satisfying the condition: "a(n) + 2n is not in the sequence". - Benoit Cloitre, Mar 25 2004
Also the integer part of the hypotenuse of isosceles right triangles. The real part of these numbers is irrational. For proof see Jones and Jones.
First differences are 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, ... (A006337 with a 1 in front). - Philippe Deléham, May 29 2006
It appears that the distance between the a(n)-th triangular number and the nearest square is not greater than floor(a(n)/2). - Ralf Stephan, Sep 14 2013
These are the nonnegative integers m satisfying sin(m*Pi/r)*sin((m+1)*Pi/r) <= 0, where r = sqrt(2). In general, the Beatty sequence of an irrational number r > 1 consists of the numbers m satisfying sin(m*x)*sin((m+1)*x) <= 0, where x = Pi/r. Thus the numbers m satisfying sin(m*x)*sin((m+1)*x) > 0 form the Beatty sequence of r/(1-r). - Clark Kimberling, Aug 21 2014
For n > 0: A080764(a(n)) = 1. - Reinhard Zumkeller, Jul 03 2015
From Clark Kimberling, Oct 17 2016: (Start)
We can generate A001951 and A001952 without using sqrt(2).
First write the even positive integers in a row:
2 4 6 8 10 12 14 . . .
Then put 1 under 2 and add:
2 4 6 8 10 12 14 . . .
1
3
Next, under 4, put the least positive integer that is not yet in rows 2 and 3;
it is 2; and add:
2 4 6 8 10 12 14 . . .
1 2
3 6
Next, under the 6 in row 1, put the least positive integer not yet in rows 2 and 3;
it is 4, and add:
2 4 6 8 10 12 14 . . .
1 2 4
3 6 10
Continue in this manner. (End)
This sequence contains an infinite number of powers of 2 (proof in Crux Mathematicorum link). See A103341. - Bernard Schott, Mar 08 2019
The terms of this sequence generate the multiplicative group of positive rational numbers (observation by Stephen M. Gagola, Jr.; see References). - Allen Stenger, Aug 05 2023
a(n) is also the number of distinct straight cylinders with integer radius and height having the same surface as a sphere with radius n. - Felix Huber, Sep 20 2024
Let P(x,y) be the condition x^2 + y^2 <= n^2, then 4*a(n) is the number of integer points (x,y) such that P(x,y) is true and at least one of P(x+1,y), P(x-1,y), P(x,y+1), P(x,y-1) is false. See LINKS for examples. - Bob de Boisvilliers, May 14 2025

References

  • Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
  • Stephen M. Gagola Jr., Solution of Problem 12282, Am. Math. Monthly, 130 (2023), pp. 682-683.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 77.
  • Gareth A. Jones and J. Mary Jones, Elementary Number Theory, Springer, 1998; pp. 221-222.
  • 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).
  • Roland Sprague, Recreations in Mathematics, Blackie and Son, (1963).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, Revised edition (1997), Entry sqrt(2), p. 18.

Crossrefs

Complement of A001952. Equals A001952(n) - 2*n for n>0.
Equals A003151(n) - n; a bisection of A094077.
Bisections: A022842, A342281.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021
Partial sums: A194102.

Programs

  • Haskell
    a001951 = floor . (* sqrt 2) . fromIntegral
    -- Reinhard Zumkeller, Sep 14 2014
    
  • Magma
    [Floor(n*Sqrt(2)): n in [0..60]]; // Vincenzo Librandi, Oct 22 2011
    
  • Magma
    [Isqrt(2*n^2):n in[0..60]]; // Jason Kimberley, Oct 28 2016
    
  • Maple
    a:=n->floor(n*sqrt(2)): seq(a(n),n=0..80); # Muniru A Asiru, Mar 09 2019
  • Mathematica
    Floor[Range[0, 72] Sqrt[2]] (* Robert G. Wilson v, Oct 17 2012 *)
  • Maxima
    makelist(floor(n*sqrt(2)), n, 0, 100); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    f(n) = for(j=1,n,print1(floor(sqrt(2*j^2))","))
    
  • PARI
    a(n)=sqrtint(2*n^2) \\ Charles R Greathouse IV, Oct 19 2016
    
  • Python
    from sympy import integer_nthroot
    def A001951(n): return integer_nthroot(2*n**2,2)[0] # Chai Wah Wu, Mar 16 2021

Formula

a(n) = A000196(A001105(n)). - Jason Kimberley, Oct 26 2016
a(n) = floor(csc(1/(sqrt(2)*n))) for n > 0, since sqrt(2)*n < csc(1/(sqrt(2)*n)) < sqrt(2)*n + 1/(3*sqrt(2)*n) < floor(sqrt(2)*n) + 1 for n > 0. - Jianing Song, Sep 07 2021
a(n) = A194102(n) - A194102(n-1) for n > 0. - M. F. Hasler, Apr 23 2022

Extensions

More terms from David W. Wilson, Sep 20 2000
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