

A016825


Positive integers congruent to 2 mod 4: a(n) = 4n+2, for n >= 0.


121



2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 190, 194, 198, 202, 206, 210, 214, 218, 222, 226, 230, 234
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OFFSET

0,1


COMMENTS

Twice the odd numbers, also called singly even numbers.
Numbers having equal numbers of odd and even divisors: A001227(a(n))=A000005(2*a(n)).  Reinhard Zumkeller, Dec 28 2003
Continued fraction for coth(1/2) = (e+1)/(e1). The continued fraction for tanh(1/2) = (e1)/(e+1) would be a(0) = 0, a(n) = A016825(n1), n >= 1.
No solutions to a(n)=b^2c^2.  Henry Bottomley, Jan 13 2001
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 70 ).
Sequence gives n such that 8 is the largest power of 2 dividing A003629(k)^n1 for any k.  Benoit Cloitre, Apr 05 2002
n such that sum(dn,(1)^d)=A048272(n)=0.  Benoit Cloitre, Apr 15 2002
Also n such that sum(dn,phi(d)*mu(n/d))=A007431(n)=0.  Benoit Cloitre, Apr 15 2002
Also n such that sum(dn,(d/A00005(d))*mu(n/d))=0, n such that sum(dn,(A00005(d)/d)*mu(n/d))=0.  Benoit Cloitre, Apr 19 2002
Solutions to phi(x)=phi(x/2); primorial numbers are here.  Labos Elemer, Dec 16 2002
Together with 1, numbers that are not the leg of a primitive Pythagorean triangle.  Lekraj Beedassy, Nov 25 2003
Maximum number of electrons in an atomic subshell with orbital quantum number l is 2(2l+1) since the magnetic quantum number m goes from l to +l and the electron spin is either 1/2 or +1/2 for each m.
For n>0: complement of A107750 and A023416(a(n)1)=A023416(a(n))<>A023416(a(n)+1).  Reinhard Zumkeller, May 23 2005
Also the minimal value of sum([p(i)p(i+1)]^2, i=1..n+2), where p(n+3)=p(1), as p ranges over all permutations of {1,2,...,n+2} (see the Mihai reference). Example: a(2)=10 because the values of the sum for the permutations of {1,2,3,4}are 10 (8 times), 12 (8 times) and 18 (8 times).  Emeric Deutsch, Jul 30 2005
Except for a(n)=2, numbers having 4 as an antidivisor.  Alexandre Wajnberg, Oct 02 2005
This is also the number of polyacenes in carbon nanotubes. See page 413 equation 12 of the paper by I. Lukovits and D. Janezic.  Parthasarathy Nambi, Aug 22 2006
A139391(a(n)) = A006370(a(n)) = A005408(n).  Reinhard Zumkeller, Apr 17 2008
Also a(n) = (n1) + n + (n+1) + (n+2), so a(n) and a(n) are all the integers that are sums of four consecutive integers.  Rick L. Shepherd, Mar 21 2009
(e1)/(e+1) = tanh(1/2).  Harry J. Smith, May 09 2009
The denominator in Pi/8 = (1/2)(1/6)+(1/10)(1/14)+(1/18)(1/22)+....  Mohammad K. Azarian, Oct 13 2011
A080736(a(n)) = 0.  Reinhard Zumkeller, Jun 11 2012
A007814(a(n)) = 1; A037227(a(n)) = 3.  Reinhard Zumkeller, Jun 30 2012
A214546(a(n)) = 0.  Reinhard Zumkeller, Jul 20 2012
Also, for a(n)>=6, the orders of the dihedral groups D_{2n+1} which are Frobenius groups. See A178498.  Bernard Schott, Dec 21 2012
Let D0 = {d0(n,i)}, i = 1..p, denote the set of the p even divisors of n and D1 = {d1(n,j)}, j = 1..q the set of the q odd divisors of n; then a(n) are the numbers such that sum_{i=1..p} 1/phi(d0(i)) = sum_{j=1..q} 1/phi(d1(j)).  Michel Lagneau, Dec 26 2014
This sequence gives the positive zeros of i^x = 0, x real, because i^x = exp(i*x*Pi/2).  Ilya Gutkovskiy, Aug 08 2015


REFERENCES

Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183185.
H. Bass, Mathematics, Mathematicians and Mathematics Education, Bull. Amer. Math. Soc. (N.S.) 42 (2004), no. 4, 417430.
A. Beiser, Concepts of Modern Physics, 2nd Ed., McGrawHill, 1973.
J. R. Goldman, The Queen of Mathematics, 1998, p. 70.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGrawHill Book Company, New York (1968).


LINKS

Harry J. Smith, Table of n, a(n) for n = 0..20000
Daniel Forgues, Number of electrons per filled orbital
Tanya Khovanova, Recursive Sequences
I. Lukovits and D. Janezic, Enumeration of conjugated circuits in nanotubes, J. Chem. Inf. Comput. Sci. 44 (2004), 410414.
Vasile Mihai and Michael Woltermann, Problem 10725: The Smoothest and Roughest Permutations, Amer. Math. Monthly, 108 (March 2001), pp. 272273.
William A. Stein, Dimensions of the spaces S_k^{new}(Gamma_0(N))
William A. Stein, The modular forms database
Eric Weisstein's World of Mathematics, Singly Even Number
Eric Weisstein's World of Mathematics, Square Number
G. Xiao, Contfrac
Index entries for continued fractions for constants
Index entries for linear recurrences with constant coefficients, signature (2,1).


FORMULA

a(n) = 4*n + 2, for n >= 0.
a(n) = 2*A005408(n).  Lekraj Beedassy, Nov 28 2003
a(n) = A118413(n+1,2) for n>1.  Reinhard Zumkeller, Apr 27 2006
G.f.: 2* (1+x)/(1x)^2. E.g.f.: 2*(1+2*x)*exp(x). a(n)= a(n1) + 4. a(1n)= a(n).  Michael Somos, Apr 11 2007
a(n) = 8*na(n1) (with a(0)=2).  Vincenzo Librandi, Nov 20 2010


EXAMPLE

0.4621171572600097585023184... = 0 + 1/(2 + 1/(6 + 1/(10 + 1/(14 + ...)))).
2.1639534137386528487700040... = 2 + 1/(6 + 1/(10 + 1/(14 + 1/(18 + ...)))), i.e., CF for coth(1/2).


MATHEMATICA

Range[2, 500, 4] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)


PROG

(MAGMA) [4*n+2 : n in [0..100] ];
(PARI) a(n)= 4*n+2
(PARI) allocatemem(932245000); default(realprecision, 180000); x=contfrac(tanh(1/2)); for (n=2, 20002, write("b016825.txt", n2, " ", x[n])); \\ Harry J. Smith, May 09 2009
(Haskell)
a016825 = (+ 2) . (* 4)
a016825_list = [2, 6 ..]  Reinhard Zumkeller, Feb 14 2012


CROSSREFS

Cf. A107687. First differences of A001105.
Cf. A160327 = Decimal expansion.
Subsequence of A042963.
Sequence in context: A111284 A130824 * A161718 A122905 A132417 A103747
Adjacent sequences: A016822 A016823 A016824 * A016826 A016827 A016828


KEYWORD

nonn,easy,nice,cofr,changed


AUTHOR

N. J. A. Sloane


STATUS

approved



