

A005408


The odd numbers: a(n) = 2*n + 1.
(Formerly M2400)


1139



1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131
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OFFSET

0,2


COMMENTS

Leibniz's series: Pi/4 = Sum_{n>=0} (1)^n/(2n+1) (cf. A072172).
Beginning of the ordering of the natural numbers used in Sharkovski's theorem  see the CielsielskiPogoda paper.
The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).
a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n1.  Amarnath Murthy, Jul 14 2003
Smallest number greater than n, not a multiple of n, but containing it in binary representation.  Reinhard Zumkeller, Oct 06 2003
Pi*sqrt(2)/4 = Sum_{n>=0} (1)^floor(n/2)/(2n+1) = 1 + 1/3  1/5  1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over Pi < x < Pi = 2(sin(x)/1  sin(2x)/2 + sin(3x)/3  ...) using x = Pi/4 (Maor)].  Gerald McGarvey, Feb 04 2005
a(n) = shortest side a of all integersided triangles with sides a <= b <= c and inradius n >= 1.
The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1.  Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006
2n5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123.  David Hoek (david.hok(AT)telia.com), Feb 28 2007
For n > 2, a(n1) is the least integer not the sum of < n ngonal numbers (0 allowed).  Jonathan Sondow, Jul 01 2007
Number of divisors of 4^(n1) for n > 0.  J. Lowell, Aug 30 2008
Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1).  Pierre CAMI, Sep 27 2008
a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc.  Carmine Suriano, Jun 08 2009
Also the 3rough numbers: positive integers that have no prime factors less than 3.  Michael B. Porter, Oct 08 2009
Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices.  K.V.Iyer, Dec 19 2009
For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15.  Gary W. Adamson, Jul 15 2010
Numbers that are the sum of two sequential integers.  Dominick Cancilla, Aug 09 2010
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h4)*(1)^n  h)/4 (h and n in A000027), therefore ((2*h*n + (h4)*(1)^n  h)/4)^2  1 == 0 (mod h); in this case, a(n)^2  1 == 0 (mod 4). Also a(n)^2  1 == 0 (mod 8).  Bruno Berselli, Nov 17 2010
For n >= 3 they are the numbers for which the product of their proper divisors divides the product of their antidivisors.  Paolo P. Lava, Jul 07 2011
a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads2n+1 tosses) = 1/2.  Dennis P. Walsh, Apr 04 2012
1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/(exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2)  1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc.  Gary W. Adamson, Jul 01 2012
Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19  1) = (2^1 * 9), (19  9) = (2^1 * 5), (19  5) = (2^1  7), (19  7) = (2^2 * 3), (19  3) = (2^4 * 1).  Gary W. Adamson, Sep 09 2012
A005408 is the numerator 2n1 of the term (1/m^2  1/n^2) = (2n1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogenlike elements.  Freimut Marschner, Aug 10 2013
This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.)  Franklin T. AdamsWatters, Sep 28 2013
These are also numbers k such that (k^k+1)/(k+1) is an integer.  Derek Orr, May 22 2014
a(n1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5.  Derek Orr, Nov 22 2014
The number of partitions of 4*n into at most 2 parts.  Colin Barker, Mar 31 2015
a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591).  Martin Renner, Mar 14 2016
Unique solution a( ) of the complementary equation a(n) = a(n1)^2  a(n2)*b(n1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences.  Clark Kimberling, Nov 21 2017
Also the number of maximal and maximum cliques in the ncentipede graph.  Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.)  Ivan Neretin, Dec 21 2017
Maximum number of nonintersecting line segments between vertices of a convex (n+2)gon.  Christoph B. Kassir, Oct 21 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231.  Lara Pudwell, Apr 10 2023


REFERENCES

T. M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, 1976, page 2.
T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203205.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

D. Applegate and J. C. Lagarias, The 3x+1 semigroup, Journal of Number Theory, Vol. 177, Issue 1, March 2006, pp. 146159; see also the arXiv version, arXiv:math/0411140 [math.NT], 20042005.


FORMULA

a(n) = 2*n + 1. a(1  n) = a(n). a(n+1) = a(n) + 2.
G.f.: (1 + x) / (1  x)^2.
E.g.f.: (1 + 2*x) * exp(x).
G.f. with interpolated zeros: (x^3+x)/((1x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(x))/2.  Geoffrey Critzer, Aug 25 2012
Euler transform of length 2 sequence [3, 1].  Michael Somos, Mar 30 2007
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = v * (1 + 2*u) * (1  2*u + 16*v)  (u  4*v)^2 * (1 + 2*u + 2*u^2).  Michael Somos, Mar 30 2007
a(n) = b(2*n + 1) where b(n) = n if n is odd is multiplicative. [This seems to say that A000027 is multiplicative?  R. J. Mathar, Sep 23 2011]
a(n) = (n+1)^2  n^2.
G.f. g(x) = Sum_{k>=0} x^floor(sqrt(k)) = Sum_{k>=0} x^A000196(k). (End)
a(n) = (n  1) + n (sum of two sequential integers).  Dominick Cancilla, Aug 09 2010
n*a(2n+1)^2+1 = (n+1)*a(2n)^2; e.g., 3*15^2+1 = 4*13^2.  Charlie Marion, Dec 31 2010
arctanh(x) = Sum_{n>=0} x^(2n+1)/a(n).  R. J. Mathar, Sep 23 2011
a(n) = det(f(ij+1))_{1<=i,j<=n}, where f(n) = A113311(n); for n < 0 we have f(n)=0.  Mircea Merca, Jun 23 2012
G.f.: Q(0), where Q(k) = 1 + 2*(k+1)*x/( 1  1/(1 + 2*(k+1)/Q(k+1))); (continued fraction).  Sergei N. Gladkovskii, May 11 2013
a(n) = Product_{k=1..2*n} 2*sin(Pi*k/(2*n+1)) = Product_{k=1..n} (2*sin(Pi*k/(2*n+1)))^2, n >= 0 (undefined product = 1). See an Oct 09 2013 formula contribution in A000027 with a reference.  Wolfdieter Lang, Oct 10 2013
Noting that as n > infinity, sqrt(n^2 + n) > n + 1/2, let f(n) = n + 1/2  sqrt(n^2 + n). Then for n > 0, a(n) = round(1/f(n))/4.  Richard R. Forberg, Feb 16 2014
a(n) = Sum_{k=0..n+1} binomial(2*n+1,2*k)*4^(k)*bernoulli(2*k).  Vladimir Kruchinin, Feb 24 2015
a(n) = Sum_{k=0..n} binomial(6*n+3, 6*k)*Bernoulli(6*k).  Michel Marcus, Jan 11 2016
O.g.f.: Sum_{n >= 1} phi(2*n1)*x^(n1)/(1  x^(2*n1)), where phi(n) is the Euler totient function A000010.  Peter Bala, Mar 22 2019


EXAMPLE

G.f. = q + 3*q^3 + 5*q^5 + 7*q^7 + 9*q^9 + 11*q^11 + 13*q^13 + 15*q^15 + ...


MAPLE



MATHEMATICA

CoefficientList[Series[(1 + x)/(1 + x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)


PROG

(Magma) [ 2*n+1 : n in [0..100]];
(PARI) {a(n) = 2*n + 1}
(PARI) first(n) = Vec((1 + x)/(1  x)^2 + O(x^n)) \\ Iain Fox, Dec 29 2017
(Haskell)
a005408 n = (+ 1) . (* 2)
(Maxima) makelist(2*n+1, n, 0, 30); /* Martin Ettl, Dec 11 2012 */


CROSSREFS

See A120062 for sequences related to integersided triangles with integer inradius n.
Cf. A000754 (boustrophedon transform).


KEYWORD

nonn,core,nice,easy


AUTHOR



EXTENSIONS

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


STATUS

approved



