

A005408


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


839



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 ).
Also continued fraction for coth(1) (A073747 is decimal expansion).  Rick L. Shepherd, Aug 07 2002
a(1) = 1; a(n) = 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
Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010).  Lekraj Beedassy, Aug 27 2004
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) = L(n,2)*(1)^n, where L is defined as in A108299.  Reinhard Zumkeller, Jun 01 2005
For n > 1, numbers having 2 as an antidivisor.  Alexandre Wajnberg, Oct 02 2005
a(n) = shortest side a of all integersided triangles with sides a <= b <= c and inradius n >= 1.
First differences of squares (A000290).  Lekraj Beedassy, Jul 15 2006
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 > 0: number of divisors of (n1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n1)); a(n) = A000005(A000302(n1)) = A000005(A001019(n1)) = A000005(A009969(n1)) = A000005(A087752(n1)).  Reinhard Zumkeller, Mar 04 2007
For n > 2, a(n1) is the least integer not the sum of < n ngonal numbers (0 allowed).  Jonathan Sondow, Jul 01 2007
A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454.  Reinhard Zumkeller, Oct 27 2007
Numbers n such that sigma(2n) = 3*sigma(n).  Farideh Firoozbakht, Feb 26 2008
a(n) = A139391(A016825(n)) = A006370(A016825(n)).  Reinhard Zumkeller, Apr 17 2008
Number of divisors of 4^(n1) for n > 0.  J. Lowell, Aug 30 2008
Equals INVERT transform of A078050 (signed  cf. comments); and row sums of triangle A144106.  Gary W. Adamson, Sep 11 2008
Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1).  Pierre CAMI, Sep 27 2008
A000035(a(n))=1, A059841(a(n))=0.  Reinhard Zumkeller, Sep 29 2008
Multiplicative closure of A065091.  Reinhard Zumkeller, Oct 14 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
Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0).  Philippe Deléham, Sep 17 2009
Also the 3rough numbers: positive integers that have no prime factors less than 3.  Michael B. Porter, Oct 08 2009
Or n without 2 as prime factor.  JuriStepan Gerasimov, Nov 19 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
A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)).  Reinhard Zumkeller, Apr 13 2010
For n >= 1, a(n1) = numbers k such that arithmetic mean of the first k positive integers is integer. A040001(a(n1)) = 1. See A145051 and A040001.  Jaroslav Krizek, May 28 2010
Union of A179084 and A179085.  Reinhard Zumkeller, Jun 28 2010
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)^nh)/4 (h and n in A000027), therefore ((2*h*n+(h4)*(1)^nh)/4)^21==0 (mod h); in this case, a(n)^21==0 (mod 4). Also a(n)^21=0 (mod 8).  Bruno Berselli, Nov 17 2010
A004767 = a(a(n)).  Reinhard Zumkeller, Jun 27 2011
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
A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0.  Reinhard Zumkeller, Jan 21 2012
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
A007814(a(n)) = 0; A037227(a(n)) = 1.  Reinhard Zumkeller, Jun 30 2012
1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,inf} 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 n such that (n^n+1)/(n+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


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

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.
Hongwei Chen, Evaluations of Some Variant Euler Sums, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.3.
K. Ciesielski and Z. Pogoda, On ordering the natural numbers, or the Sharkovski theorem, Amer. Math. Monthly, 115 (No. 2, 2008), 158165.
Mark W. Coffey, Bernoulli identities, zeta relations, determinant expressions, Mellin transforms, and representation of the Hurwitz numbers, arXiv:1601.01673 [math.NT], 2016. See p. 35.
T.X. He, L. W. Shapiro, FussCatalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 2541, theorem 2.5, k=4.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 935
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
M. Somos, Rational Function Multiplicative Coefficients
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N))
William A. Stein, The modular forms database
Eric Weisstein's World of Mathematics, Odd Number, DavenportSchinzel Sequence, Gnomonic Number, Pythagorean Triple, Inverse Tangent, Inverse Cotangent, Inverse Hyperbolic Cotangent, Inverse Hyperbolic Tangent, Nexus Number.
Index entries for "core" sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).


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]
From Hieronymus Fischer, May 25 2007: (Start)
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(0) = 1, a(1) = 3, a(n) = 2*a(n1)  a(n2).  Jaume Oliver Lafont, May 07 2008
A005408(n)=A000330(A016777(n))/A000217(A016777(n)).  Pierre CAMI, Sep 27 2008
a(n) = A034856(n+1)  A000217(n) = A005843(n) + A000124(n)  A000217(n) = A005843(n) + 1.  Jaroslav Krizek, Sep 05 2009
a(n) = (n  1) + n (sum of two sequential integers).  Dominick Cancilla, Aug 09 2010
a(n) = 4*A000217(n)+1  2*Sum_{i=1..n1} a(i) for n > 1.  Bruno Berselli, Nov 17 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) = floor(sqrt(2*A000384(n+1))).  Ivan N. Ianakiev, Jun 17 2013
a(n) = 3*A000330(n)/A000217(n), n > 0.  Ivan N. Ianakiev, Jul 12 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
a(n) = A000225(n+1)  A005803(n+1).  Miquel Cerda, Nov 25 2016


EXAMPLE

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


MAPLE

A005408 := n>2*n+1;
A005408:=(1+z)/(z1)^2; # Simon Plouffe in his 1992 dissertation


MATHEMATICA

Table[2n  1, {n, 1, 50}] (* Stefan Steinerberger, Apr 01 2006 *)
Range[1, 131, 2] (* Harvey P. Dale, Apr 26 2011 *)


PROG

(MAGMA) [ 2*n+1 : n in [0..100]];
(PARI) {a(n) = 2*n + 1}
(Haskell)
a005408 n = (+ 1) . (* 2)
a005408_list = [1, 3 ..]  Reinhard Zumkeller, Feb 11 2012, Jun 28 2011
(Maxima) makelist(2*n+1, n, 0, 30); /* Martin Ettl, Dec 11 2012 */
(Python) a=lambda n: 2*n+1 # Indranil Ghosh, Jan 04 2017


CROSSREFS

Cf. A000027, A005843, A065091.
See A120062 for sequences related to integersided triangles with integer inradius n.
Cf. A128200, A000290, A078050, A144106, A109613, A167875.
Cf. A001651 (n=1 or 2 mod 3), A047209 (n=1 or 4 mod 5).
Cf. A003558, A216371, A179480 (relating to the Coach theorem).
Cf. A000754 (boustrophedon transform).
Cf. A240876.
Sequence in context: A247328 A004273 * A176271 A144396 A060747 A089684
Adjacent sequences: A005405 A005406 A005407 * A005409 A005410 A005411


KEYWORD

nonn,core,nice,easy,changed


AUTHOR

N. J. A. Sloane


EXTENSIONS

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


STATUS

approved



