A047209 -id:A047209 - cp's OEIS frontend

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

1, 6, 1, 8, 0, 3, 3, 9, 8, 8, 7, 4, 9, 8, 9, 4, 8, 4, 8, 2, 0, 4, 5, 8, 6, 8, 3, 4, 3, 6, 5, 6, 3, 8, 1, 1, 7, 7, 2, 0, 3, 0, 9, 1, 7, 9, 8, 0, 5, 7, 6, 2, 8, 6, 2, 1, 3, 5, 4, 4, 8, 6, 2, 2, 7, 0, 5, 2, 6, 0, 4, 6, 2, 8, 1, 8, 9, 0, 2, 4, 4, 9, 7, 0, 7, 2, 0, 7, 2, 0, 4, 1, 8, 9, 3, 9, 1, 1, 3, 7, 4, 8, 4, 7, 5

Offset: 1

N. J. A. Sloane

Also decimal expansion of the positive root of (x+1)^n - x^(2n). (x+1)^n - x^(2n) = 0 has only two real roots x1 = -(sqrt(5)-1)/2 and x2 = (sqrt(5)+1)/2 for all n > 0. - Cino Hilliard, May 27 2004
The golden ratio phi is the most irrational among irrational numbers; its successive continued fraction convergents F(n+1)/F(n) are the slowest to approximate to its actual value (I. Stewart, in "Nature's Numbers", Basic Books, 1997). - Lekraj Beedassy, Jan 21 2005
Let t=golden ratio. The lesser sqrt(5)-contraction rectangle has shape t-1, and the greater sqrt(5)-contraction rectangle has shape t. For definitions of shape and contraction rectangles, see A188739. - Clark Kimberling, Apr 16 2011
The golden ratio (often denoted by phi or tau) is the shape (i.e., length/width) of the golden rectangle, which has the special property that removal of a square from one end leaves a rectangle of the same shape as the original rectangle. Analogously, removals of certain isosceles triangles characterize side-golden and angle-golden triangles. Repeated removals in these configurations result in infinite partitions of golden rectangles and triangles into squares or isosceles triangles so as to match the continued fraction, [1,1,1,1,1,...] of tau. For the special shape of rectangle which partitions into golden rectangles so as to match the continued fraction [tau, tau, tau, ...], see A188635. For other rectangular shapes which depend on tau, see A189970, A190177, A190179, A180182. For triangular shapes which depend on tau, see A152149 and A188594; for tetrahedral, see A178988. - Clark Kimberling, May 06 2011
Given a pentagon ABCDE, 1/(phi)^2 <= (A*C^2 + C*E^2 + E*B^2 + B*D^2 + D*A^2) / (A*B^2 + B*C^2 + C*D^2 + D*E^2 + E*A^2) <= (phi)^2. - Seiichi Kirikami, Aug 18 2011
If a triangle has sides whose lengths form a geometric progression in the ratio of 1:r:r^2 then the triangle inequality condition requires that r be in the range 1/phi < r < phi. - Frank M Jackson, Oct 12 2011
The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - Clark Kimberling, Oct 19 2011
Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - Michel Lagneau, Dec 02 2011
Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - Michel Lagneau, Apr 17 2012
This is the case n=5 of (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)): (1+sqrt(5))/2 = (Gamma(1/5)/Gamma(3/5))*(Gamma(4/5)/Gamma(2/5)). - Bruno Berselli, Dec 14 2012
Also decimal expansion of the only number x>1 such that (x^x)^(x^x) = (x^(x^x))^x = x^((x^x)^x). - Jaroslav Krizek, Feb 01 2014
For n >= 1, round(phi^prime(n)) == 1 (mod prime(n)) and, for n >= 3, round(phi^prime(n)) == 1 (mod 2*prime(n)). - Vladimir Shevelev, Mar 21 2014
The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - Giovanni Zedda, Jun 22 2019
Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - Diego Rattaggi, Apr 17 2021
Given any complex p such that real(p) > -1, phi is the only real solution of the equation z^p+z^(p+1)=z^(p+2), and the only attractor of the complex mapping z->M(z,p), where M(z,p)=(z^p+z^(p+1))^(1/(p+2)), convergent from any complex plane point. - Stanislav Sykora, Oct 14 2021
The only positive number such that its decimal part, its integral part and the number itself (x-[x], [x] and x) form a geometric progression is phi, with respectively (phi -1, 1, phi) and a ratio = phi. This is the answer to the 4th problem of the 7th Canadian Mathematical Olympiad in 1975 (see IMO link and Doob reference). - Bernard Schott, Dec 08 2021
The golden ratio is the unique number x such that f(n*x)*c(n/x) - f(n/x)*c(n*x) = n for all n >= 1, where f = floor and c = ceiling. - Clark Kimberling, Jan 04 2022
In The Second Scientific American Book Of Mathematical Puzzles and Diversions, Martin Gardner wrote that, by 1910, Mark Barr (1871-1950) gave phi as a symbol for the golden ratio. - Bernard Schott, May 01 2022
Phi is the length of the equal legs of an isosceles triangle with side c = phi^2, and internal angles (A,B) = 36 degrees, C = 108 degrees. - Gary W. Adamson, Jun 20 2022
The positive solution to x^2 - x - 1 = 0. - Michal Paulovic, Jan 16 2023
The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025
This is the largest real zero x of (x^4 + x^2 + 1)^2 = 2*(x^8 + x^4 + 1). - Thomas Ordowski, May 14 2025

			1.6180339887498948482045868343656381177203091798057628621...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 112, 123, 184, 190, 203.
Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993 - Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1975, pages 76-77, 1993.
Richard A. Dunlap, The Golden Ratio and Fibonacci Numbers, World Scientific, River Edge, NJ, 1997.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, Vol. 94, Cambridge University Press, 2003, Section 1.2.
Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.
Martin Gardner, Weird Water and Fuzzy Logic: More Notes of a Fringe Watcher, "The Cult of the Golden Ratio", Chapter 9, Prometheus Books, 1996, pages 90-97.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 287.
H. E. Huntley, The Divine Proportion, Dover, NY, 1970.
Mario Livio, The Golden Ratio, Broadway Books, NY, 2002. [see the review by G. Markowsky in the links field]
Gary B. Meisner, The Golden Ratio: The Divine Beauty of Mathematics, Race Point Publishing (The Quarto Group), 2018. German translation: Der Goldene Schnitt, Librero, 2023.
Scott Olsen, The Golden Section, Walker & Co., NY, 2006.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 137-139.
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).
Hans Walser, The Golden Section, Math. Assoc. of Amer. Washington DC 2001.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 36-40.
Claude-Jacques Willard, Le nombre d'or, Magnard, Paris, 1987.

Robert G. Wilson v, Table of n, a(n) for n = 1..100000
Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; Solution, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.
John Baez, This week's finds in mathematical physics, Week 203.
John Baez, The Rankin Lectures 2008, My Favorite Numbers: 5. [video]
Murray Berg, Phi, the golden ratio (to 4599 decimal places) and Fibonacci numbers, Fib. Quart., Vol. 4, No. 2 (1961), pp. 157-162.
Ömür Deveci, Zafer Adıgüzel and Taha Doğan, On the Generalized Fibonacci-circulant-Hurwitz numbers, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 1, 179-190.
T. Eveilleau, Le nombre d'or (in French).
Abdul Gaffar, Anand B. Joshi, Sonali Singh, and Keerti Srivastava, A high capacity multi-image steganography technique based on golden ratio and non-subsampled contourlet transform, Multimedia Tools and Applications (2022).
Gutenberg Project, The golden ratio to 20000 places.
ICON Project, The golden ratio to 50000 places.
The IMO Compendium, Problem 4, 7th Canadian Mathematical Olympiad 1975.
L. B. W. Jolley, Summation of Series, Dover, 1961
Franklin H. J. Kenter, It's good to be phi: a solution to a problem of Gosper and Knuth, arXiv:1712.04856 [math.HO], 2017.
Clark Kimberling, Two kinds of golden triangles, generalized to match continued fractions, Journal for Geometry and Graphics, Vol. 11 (2007), pp. 165-171.
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
Ron Knott, Fibonacci numbers and the golden section.
Wolfdieter Lang, A list of representative simple difference sets of the Singer type for small orders m, Karlsruher Institut für Technologie (Karlsruhe, Germany 2020).
Wolfdieter Lang, Cantor's List of Real Algebraic Numbers of Heights 1 to 7, arXiv:2307.10645 [math.NT], 2023.
Simon Litsyn and Vladimir Shevelev, Irrational Factors Satisfying the Little Fermat Theorem, International Journal of Number Theory, Vol. 1, No. 4 (2005), pp. 499-512.
Gary B. Meisner, Phi, The Golden Number.
George Markowsky, Misconceptions About the Golden Ratio, College Mathematics Journal, 23:1 (January 1992), 2-19.
George Markowsky, Book review: The Golden Ratio, Notices of the AMS, 52:3 (March 2005), 344-347.
R. S. Melham and A. G. Shannon, Inverse Trigonometric Hyperbolic Summation Formulas Involving Generalized Fibonacci Numbers, The Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 32-40.
Jean-Christophe Michel, Le nombre d'or.
J. J. O'Connor and E. F. Robertson, The Golden ratio.
Hugo Pfoertner, 1 million digits of phi, Computed using A. J. Yee's y-cruncher.
Simon Plouffe, Plouffe's Inverter, The golden ratio to 10 million digits. [Only announcement, file truncated]
Simon Plouffe, The golden ratio:(1+sqrt(5))/2 to 20000 places.
Fred Richman, Fibonacci sequence with multiprecision Java, Successive approximations to phi from ratios of consecutive Fibonacci numbers.
Herman P. Robinson, The CSR Function, Popular Computing (Calabasas, CA), Vol. 4, No. 35 (Feb 1976), pages PC35-3 to PC35-4. Annotated and scanned copy.
E. F. Schubert, The Fibonacci series.
Vladimir Shevelev, A property of n-bonacci constant, Seqfan (Mar 23 2014).
Jonathan Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, Diophantine Analysis and Related Fields 2011 - AIP Conference Proceedings, Vol. 1385, pp. 97-100; arXiv:1106.4246 [math.NT], 2011.
Matthew R. Watkins, The "Golden Mean" in number theory.
Eric Weisstein's World of Mathematics, Golden Ratio.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
Eric Weisstein's World of Mathematics, Pisot Number.
Eric Weisstein's World of Mathematics, Silver Ratio.
Wikipedia, Mark Barr.
Wikipedia, Golden ratio.
Wikipedia, Kronecker-Weber theorem.
Wikipedia, Metallic mean.
Alexander J. Yee, y-cruncher - A Multi-Threaded Pi-Program.
Index entries for algebraic numbers, degree 2.
Index to sequences related to Olympiads.

Cf. A000012 (continued fraction coefficients), A000032, A000045, A006497, A080039, A104457, A188635, A192222, A192223, A145996, A139339, A197762, A002163, A094874, A134973.
Cf. A102208, A102769, A131595.
Cf. A302973, A303069, A304022.

Digits:=1000; evalf((1+sqrt(5))/2); # Wesley Ivan Hurt, Nov 01 2013

RealDigits[(1 + Sqrt[5])/2, 10, 130] (* Stefan Steinerberger, Apr 02 2006 *)
RealDigits[ Exp[ ArcSinh[1/2]], 10, 111][[1]] (* Robert G. Wilson v, Mar 01 2008 *)
RealDigits[GoldenRatio,10,120][[1]] (* Harvey P. Dale, Oct 28 2015 *)

default(realprecision, 20080); x=(1+sqrt(5))/2; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b001622.txt", n, " ", d));  \\ Harry J. Smith, Apr 19 2009

/* Digit-by-digit method: write it as 0.5+sqrt(1.25) and start at hundredths digit */
r=11; x=400; print(1); print(6);
for(dig=1, 110, {d=0; while((20*r+d)*d <= x, d++);
d--; /* while loop overshoots correct digit */
print(d); x=100*(x-(20*r+d)*d); r=10*r+d})
\\ Michael B. Porter, Oct 24 2009

a(n) = floor(10^(n-1)*(quadgen(5))%10);
alist(len) = digits(floor(quadgen(5)*10^(len-1))); \\ Chittaranjan Pardeshi, Jun 22 2022

from sympy import S
def alst(n): # truncate extra last digit to avoid rounding
  return list(map(int, str(S.GoldenRatio.n(n+1)).replace(".", "")))[:-1]
print(alst(105)) # Michael S. Branicky, Jan 06 2021

Equals Sum_{n>=2} 1/A064170(n) = 1/1 + 1/2 + 1/(2*5) + 1/(5*13) + 1/(13*34) + ... - Gary W. Adamson, Dec 15 2007
Equals Hypergeometric2F1([1/5, 4/5], [1/2], 3/4) = 2*cos((3/5)*arcsin(sqrt(3/4))). - Artur Jasinski, Oct 26 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
The fractional part of phi^n equals phi^(-n), if n is odd. For even n, the fractional part of phi^n is equal to 1-phi^(-n).
General formula: Provided x>1 satisfies x-x^(-1)=floor(x), where x=phi for this sequence, then:
for odd n: x^n - x^(-n) = floor(x^n), hence fract(x^n) = x^(-n),
for even n: x^n + x^(-n) = ceiling(x^n), hence fract(x^n) = 1 - x^(-n),
for all n>0: x^n + (-x)^(-n) = round(x^n).
x=phi is the minimal solution to x - x^(-1) = floor(x) (where floor(x)=1 in this case).
Other examples of constants x satisfying the relation x - x^(-1) = floor(x) include A014176 (the silver ratio: where floor(x)=2) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
Equals 2*cos(Pi/5) = e^(i*Pi/5) + e^(-i*Pi/5). - Eric Desbiaux, Mar 19 2010
The solutions to x-x^(-1)=floor(x) are determined by x=(1/2)*(m+sqrt(m^2+4)), m>=1; x=phi for m=1. In terms of continued fractions the solutions can be described by x=[m;m,m,m,...], where m=1 for x=phi, and m=2 for the silver ratio A014176, and m=3 for the bronze ratio A098316. - Hieronymus Fischer, Oct 20 2010
Sum_{n>=1} x^n/n^2 = Pi^2/10 - (log(2)*sin(Pi/10))^2 where x = 2*sin(Pi/10) = this constant here. [Jolley, eq 360d]
phi = 1 + Sum_{k>=1} (-1)^(k-1)/(F(k)*F(k+1)), where F(n) is the n-th Fibonacci number (A000045). Proof. By Catalan's identity, F^2(n) - F(n-1)*F(n+1) = (-1)^(n-1). Therefore,(-1)^(n-1)/(F(n)*F(n+1)) = F(n)/F(n+1) - F(n-1)/F(n). Thus Sum_{k=1..n} (-1)^(k-1)/(F(k)*F(k+1)) = F(n)/F(n+1). If n goes to infinity, this tends to 1/phi = phi - 1. - Vladimir Shevelev, Feb 22 2013
phi^n = (A000032(n) + A000045(n)*sqrt(5)) / 2. - Thomas Ordowski, Jun 09 2013
Let P(q) = Product_{k>=1} (1 + q^(2*k-1)) (the g.f. of A000700), then A001622 = exp(Pi/6) * P(exp(-5*Pi)) / P(exp(-Pi)). - Stephen Beathard, Oct 06 2013
phi = i^(2/5) + i^(-2/5) = ((i^(4/5))+1) / (i^(2/5)) = 2*(i^(2/5) - (sin(Pi/5))i) = 2*(i^(-2/5) + (sin(Pi/5))i). - Jaroslav Krizek, Feb 03 2014
phi = sqrt(2/(3 - sqrt(5))) = sqrt(2)/A094883. This follows from the fact that ((1 + sqrt(5))^2)*(3 - sqrt(5)) = 8, so that ((1 + sqrt(5))/2)^2 = 2/(3 - sqrt(5)). - Geoffrey Caveney, Apr 19 2014
exp(arcsinh(cos(Pi/2-log(phi)*i))) = exp(arcsinh(sin(log(phi)*i))) = (sqrt(3) + i) / 2. - Geoffrey Caveney, Apr 23 2014
exp(arcsinh(cos(Pi/3))) = phi. - Geoffrey Caveney, Apr 23 2014
cos(Pi/3) + sqrt(1 + cos(Pi/3)^2). - Geoffrey Caveney, Apr 23 2014
2*phi = z^0 + z^1 - z^2 - z^3 + z^4, where z = exp(2*Pi*i/5). See the Wikipedia Kronecker-Weber theorem link. - Jonathan Sondow, Apr 24 2014
phi = 1/2 + sqrt(1 + (1/2)^2). - Geoffrey Caveney, Apr 25 2014
Phi is the limiting value of the iteration of x -> sqrt(1+x) on initial value a >= -1. - Chayim Lowen, Aug 30 2015
From Isaac Saffold, Feb 28 2018: (Start)
1 = Sum_{k=0..n} binomial(n, k) / phi^(n+k) for all nonnegative integers n.
1 = Sum_{n>=1} 1 / phi^(2n-1).
1 = Sum_{n>=2} 1 / phi^n.
phi = Sum_{n>=1} 1/phi^n. (End)
From Christian Katzmann, Mar 19 2018: (Start)
phi = Sum_{n>=0} (15*(2*n)! + 8*n!^2)/(2*n!^2*3^(2*n+2)).
phi = 1/2 + Sum_{n>=0} 5*(2*n)!/(2*n!^2*3^(2*n+1)). (End)
phi = Product_{k>=1} (1 + 2/(-1 + 2^k*(sqrt(4+(1-2/2^k)^2) + sqrt(4+(1-1/2^k)^2)))). - Gleb Koloskov, Jul 14 2021
Equals Product_{k>=1} (Fibonacci(3*k)^2 + (-1)^(k+1))/(Fibonacci(3*k)^2 + (-1)^k) (Melham and Shannon, 1995). - Amiram Eldar, Jan 15 2022
From Michal Paulovic, Jan 16 2023: (Start)
Equals the real part of 2 * e^(i * Pi / 5).
Equals 2 * sin(3 * Pi / 10) = 2*A019863.
Equals -2 * sin(37 * Pi / 10).
Equals 1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / (1 + 1 / ...)))).
Equals (2 + 3 * (2 + 3 * (2 + 3 * ...)^(1/4))^(1/4))^(1/4).
Equals (1 + 2 * (1 + 2 * (1 + 2 * ...)^(1/3))^(1/3))^(1/3).
Equals (1 + phi + (1 + phi + (1 + phi + ...)^(1/3))^(1/3))^(1/3).
Equals 13/8 + Sum_{k=0..oo} (-1)^(k+1)*(2*k+1)!/((k+2)!*k!*4^(2*k+3)).
(End)
phi^n = phi * A000045(n) + A000045(n-1). - Gary W. Adamson, Sep 09 2023
The previous formula holds for integer n, with F(-n) = (-1)^(n+1)*F(n), for n >= 0, with F(n) = A000045(n), for n >= 0. phi^n are integers in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Sep 16 2023
Equals Product_{k>=0} ((5*k + 2)*(5*k + 3))/((5*k + 1)*(5*k + 4)). - Antonio Graciá Llorente, Feb 24 2024
From Antonio Graciá Llorente, Apr 21 2024: (Start)
Equals Product_{k>=1} phi^(-2^k) + 1, with phi = A001622.
Equals Product_{k>=0} ((5^(k+1) + 1)*(5^(k-1/2) + 1))/((5^k + 1)*(5^(k+1/2) + 1)).
Equals Product_{k>=1} 1 - (4*(-1)^k)/(10*k - 5 + (-1)^k) = Product_{k>=1} A047221(k)/A047209(k).
Equals Product_{k>=0} ((5*k + 7)*(5*k + 1 + (-1)^k))/((5*k + 1)*(5*k + 7 + (-1)^k)).
Equals Product_{k>=0} ((10*k + 3)*(10*k + 5)*(10*k + 8)^2)/((10*k + 2)*(10*k + 4)*(10*k + 9)^2).
Equals Product_{k>=5} 1 + 1/(Fibonacci(k) - (-1)^k).
Equals Product_{k>=2} 1 + 1/Fibonacci(2*k).
Equals Product_{k>=2} (Lucas(k)^2 + (-1)^k)/(Lucas(k)^2 - 4*(-1)^k). (End)

Additional links contributed by Lekraj Beedassy, Dec 23 2003
More terms from Gabriel Cunningham (gcasey(AT)mit.edu), Oct 24 2004
More terms from Stefan Steinerberger, Apr 02 2006
Broken URL to Project Gutenberg replaced by Georg Fischer, Jan 03 2009
Edited by M. F. Hasler, Feb 24 2014

A005408 The odd numbers: a(n) = 2*n + 1.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

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 Cielsielski-Pogoda 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) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - 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
For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
a(n) = shortest side a of all integer-sided 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
2n-5 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 (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007
For n > 2, a(n-1) is the least integer not the sum of < n n-gonal 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^(n-1) 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 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009
Or n without 2 as prime factor. - Juri-Stepan 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(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 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 + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-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
A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 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 heads|2n+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,...,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 4q-1 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 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(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 hydrogen-like 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. Adams-Watters, Sep 28 2013
These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014
a(n-1) 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(n-1)^2 - a(n-2)*b(n-1), 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 n-centipede 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 non-intersecting 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

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

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology, p. 264.
D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203-205.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Ayomikun Adeniran and Lara Pudwell, Pattern avoidance in parking functions, Enumer. Comb. Appl. 3:3 (2023), Article S2R17.
D. Applegate and J. C. Lagarias, The 3x+1 semigroup, Journal of Number Theory, Vol. 177, Issue 1, March 2006, pp. 146-159; see also the arXiv version, arXiv:math/0411140 [math.NT], 2004-2005.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
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), 158-165.
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 and L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, theorem 2.5, k=4.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 935
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Bridges.
Tanya Khovanova, Recursive Sequences
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Franck Ramaharo, Enumerating the states of the twist knot, arXiv:1712.06543 [math.CO], 2017.
Michael Somos, Rational Function Multiplicative Coefficients
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N))
William A. Stein, The modular forms database
Leo Tavares, Illustration: Triangular Sides
Eric Weisstein's World of Mathematics, Centipede Graph
Eric Weisstein's World of Mathematics, Davenport-Schinzel Sequence
Eric Weisstein's World of Mathematics, Gnomonic Number
Eric Weisstein's World of Mathematics, Inverse Cotangent,
Eric Weisstein's World of Mathematics, Inverse Hyperbolic Cotangent
Eric Weisstein's World of Mathematics, Inverse Hyperbolic Tangent
Eric Weisstein's World of Mathematics, Inverse Tangent
Eric Weisstein's World of Mathematics, Maximal Clique
Eric Weisstein's World of Mathematics, Maximum Clique
Eric Weisstein's World of Mathematics, Nexus Number
Eric Weisstein's World of Mathematics, Odd Number
Eric Weisstein's World of Mathematics, Pythagorean Triple
Chai Wah Wu, Can machine learning identify interesting mathematics? An exploration using empirically observed laws, arXiv:1805.07431 [cs.LG], 2018.
Index entries for "core" sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000027, A005843, A065091.
See A120062 for sequences related to integer-sided 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).

List([0..100],n->2*n+1); # Muniru A Asiru, Oct 16 2018

a005408 n = (+ 1) . (* 2)
a005408_list = [1, 3 ..]  -- Reinhard Zumkeller, Feb 11 2012, Jun 28 2011

Magma
```
[ 2*n+1 : n in [0..100]];
```

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

Table[2 n - 1, {n, 1, 50}] (* Stefan Steinerberger, Apr 01 2006 *)
Range[1, 131, 2] (* Harvey P. Dale, Apr 26 2011 *)
2 Range[0, 20] + 1 (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{2, -1}, {1, 3}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + x)/(-1 + x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

makelist(2*n+1, n, 0, 30); /* Martin Ettl, Dec 11 2012 */

PARI
```
{a(n) = 2*n + 1}
```

first(n) = Vec((1 + x)/(1 - x)^2 + O(x^n)) \\ Iain Fox, Dec 29 2017

a=lambda n: 2*n+1 # Indranil Ghosh, Jan 04 2017

[2*n+1 for n in range(100)] # G. C. Greubel, Nov 28 2018

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)/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(-x))/2. - Geoffrey Critzer, Aug 25 2012
a(n) = L(n,-2)*(-1)^n, where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
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(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(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..n-1} 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(i-j+1))A113311(n);%20for%20n%20%3C%200%20we%20have%20f(n)=0.%20-%20_Mircea%20Merca">{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
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
Sum_{n>=0} 1/a(n)^2 = Pi^2/8 = A111003. - Bernard Schott, Dec 10 2020
Sum_{n >= 1} (-1)^n/(a(n)*a(n+1)) = Pi/4 - 1/2 = 1/(3 + (1*3)/(4 + (3*5)/(4 + ... + (4*n^2 - 1)/(4 + ... )))). Cf. A016754. - Peter Bala, Mar 28 2024
a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024
a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024
a(n) = A000217(a(n))/n for n > 0. - Stefano Spezia, Feb 15 2025

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

Peripheral comments deleted by N. J. A. Sloane, May 09 2022

A007310 Numbers congruent to 1 or 5 mod 6.

Original entry on oeis.org

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175

Offset: 1

C. Christofferson (Magpie56(AT)aol.com)

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1.  Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...

K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016-2018.
B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers.
L. B. W. Jolley, Summation of Series, Dover, 1961
Cedric A. B. Smith, Prime factors and recurring duodecimals, Math. Gaz. 59 (408) (1975) 106-109.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N).
Eric Weisstein's World of Mathematics, Rough Number.
Eric Weisstein's World of Mathematics, Pi Formulas. [_Jaume Oliver Lafont_, Oct 23 2009]
Index entries for sequences related to smooth numbers [_Michael B. Porter_, Oct 09 2009]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Filtered([1..150],n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018

a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016

seq(seq(6*i+j,j=[1,5]),i=0..100); # Robert Israel, Sep 08 2014

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *)
a[n_] := (6 n + (-1)^n - 3)/2; a[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)
Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)
Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)

isA007310(n) = gcd(n,6)==1 \\ Michael B. Porter, Oct 09 2009

A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014

\\ given an element from the sequence, find the next term in the sequence.
nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016

def A007310(n): return (n+(n>>1)<<1)-1 # Chai Wah Wu, Oct 10 2023

[i for i in range(150) if gcd(6,i) == 1] # Zerinvary Lajos, Apr 21 2009

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)

A003114 Number of partitions of n into parts 5k+1 or 5k+4.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 9, 10, 12, 14, 17, 19, 23, 26, 31, 35, 41, 46, 54, 61, 70, 79, 91, 102, 117, 131, 149, 167, 189, 211, 239, 266, 299, 333, 374, 415, 465, 515, 575, 637, 709, 783, 871, 961, 1065, 1174, 1299, 1429, 1579, 1735, 1913, 2100, 2311, 2533, 2785

Offset: 0

N. J. A. Sloane and Herman P. Robinson

Expansion of Rogers-Ramanujan function G(x) in powers of x.
Same as number of partitions into distinct parts where the difference between successive parts is >= 2.
As a formal power series, the limit of polynomials S(n,x): S(n,x)=sum(T(i,x),0<=i<=n); T(i,x)=S(i-2,x).x^i; T(0,x)=1,T(1,x)=x; S(n,1)=A000045(n+1), the Fibonacci sequence. - Claude Lenormand (claude.lenormand(AT)free.fr), Feb 04 2001
The Rogers-Ramanujan identity is 1 + Sum_{n >= 1} t^(n^2)/((1-t)*(1-t^2)*...*(1-t^n)) = Product_{n >= 1} 1/((1-t^(5*n-1))*(1-t^(5*n-4))).
Coefficients in expansion of permanent of infinite tridiagonal matrix:
  1 1 0 0 0 0 0 0 ...
  x 1 1 0 0 0 0 0 ...
  0 x^2 1 1 0 0 0 ...
  0 0 x^3 1 1 0 0 ...
  0 0 0 x^4 1 1 0 ...
  ................... - Vladeta Jovovic, Jul 17 2004
Also number of partitions of n such that the smallest part is greater than or equal to number of parts. - Vladeta Jovovic, Jul 17 2004
Also number of partitions of n such that if k is the largest part, then each of {1, 2, ..., k-1} occur at least twice. Example: a(9)=5 because we have [3, 2, 2, 1, 1], [2, 2, 2, 1, 1, 1], [2, 2, 1, 1, 1, 1, 1], [2, 1, 1, 1, 1, 1, 1, 1] and [1, 1, 1, 1, 1, 1, 1, 1, 1]. - Emeric Deutsch, Feb 27 2006
Also number of partitions of n such that if k is the largest part, then k occurs at least k times. Example: a(9)=5 because we have [3, 3, 3], [2, 2, 2, 2, 1], [2, 2, 2, 1, 1, 1], [2, 2, 1, 1, 1, 1, 1] and [1, 1, 1, 1, 1, 1, 1, 1, 1]. - Emeric Deutsch, Apr 16 2006
a(n) = number of NW partitions of n, for n >= 1; see A237981.
For more about the generalized Rogers-Ramanujan series G[i](x) see the Andrews-Baxter and Lepowsky-Zhu papers. The present series is G[1](x). - N. J. A. Sloane, Nov 22 2015
Convolution of A109700 and A109697. - Vaclav Kotesovec, Jan 21 2017

			G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 4*x^8 + 5*x^9 + ...
G.f. = 1/q + q^59 + q^119 + q^179 + 2*q^239 + 2*q^299 + 3*q^359 + 3*q^419 + ...
From _Joerg Arndt_, Dec 27 2012: (Start)
The a(16)=17 partitions of 16 where all parts are 1 or 4 (mod 5) are
  [ 1]  [ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ]
  [ 2]  [ 4 1 1 1 1 1 1 1 1 1 1 1 1 ]
  [ 3]  [ 4 4 1 1 1 1 1 1 1 1 ]
  [ 4]  [ 4 4 4 1 1 1 1 ]
  [ 5]  [ 4 4 4 4 ]
  [ 6]  [ 6 1 1 1 1 1 1 1 1 1 1 ]
  [ 7]  [ 6 4 1 1 1 1 1 1 ]
  [ 8]  [ 6 4 4 1 1 ]
  [ 9]  [ 6 6 1 1 1 1 ]
  [10]  [ 6 6 4 ]
  [11]  [ 9 1 1 1 1 1 1 1 ]
  [12]  [ 9 4 1 1 1 ]
  [13]  [ 9 6 1 ]
  [14]  [ 11 1 1 1 1 1 ]
  [15]  [ 11 4 1 ]
  [16]  [ 14 1 1 ]
  [17]  [ 16 ]
The a(16)=17 partitions of 16 where successive parts differ by at least 2 are
  [ 1]  [ 7 5 3 1 ]
  [ 2]  [ 8 5 3 ]
  [ 3]  [ 8 6 2 ]
  [ 4]  [ 9 5 2 ]
  [ 5]  [ 9 6 1 ]
  [ 6]  [ 9 7 ]
  [ 7]  [ 10 4 2 ]
  [ 8]  [ 10 5 1 ]
  [ 9]  [ 10 6 ]
  [10]  [ 11 4 1 ]
  [11]  [ 11 5 ]
  [12]  [ 12 3 1 ]
  [13]  [ 12 4 ]
  [14]  [ 13 3 ]
  [15]  [ 14 2 ]
  [16]  [ 15 1 ]
  [17]  [ 16 ]
(End)

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 109, 238.
G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1999; Exercise 6(e), p. 591.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 669.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 107.
G. H. Hardy, Ramanujan, AMS Chelsea Publ., Providence, RI, 2002, pp. 90-92.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth ed., Clarendon Press, Oxford, 2003, pp. 290-291.
H. P. Robinson, Letter to N. J. A. Sloane, Jan 04 1974.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
George E. Andrews, Three aspects of partitions, Séminaire Lotharingien de Combinatoire, B25f (1990), 1 p.
George E. Andrews, Euler's "De Partitio Numerorum", Bull. Amer. Math. Soc., 44 (No. 4, 2007), 561-573.
George E. Andrews; R. J. Baxter, A motivated proof of the Rogers-Ramanujan identities, Amer. Math. Monthly 96 (1989), no. 5, 401-409.
M. Archibald, A. Blecher, S. Elizalde, and A. Knopfmacher, Subdiagonal and superdiagonal partitions, Afr. Mat. 36, 77 (2025). See p. 14.
R. K. Guy, Letter to N. J. A. Sloane, Sep 25 1986.
R. K. Guy, Letter to N. J. A. Sloane, 1987
R. K. Guy, Letter to N. J. A. Sloane, 1988-04-12 (annotated scanned copy)
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712.
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
P. Jacob and P. Mathieu, Parafermionic derivation of Andrews-type multiple-sums, arXiv:hep-th/0505097, 2005.
James Lepowsky and Minxian Zhu, A motivated proof of Gordon's identities, The Ramanujan Journal 29.1-3 (2012): 199-211.
I. Martinjak, D. Svrtan, New Identities for the Polarized Partitions and Partitions with d-Distant Parts, J. Int. Seq. 17 (2014) # 14.11.4.
Herman P. Robinson, Letter to N. J. A. Sloane, Jan 1974.
A. V. Sills, Finite Rogers-Ramanujan type identities, Electron. J. Combin. 10 (2003), Research Paper 13, 122 pp.
Michael Somos, Introduction to Ramanujan theta functions
Eric Weisstein's World of Mathematics, Rogers-Ramanujan Identities.
Mingjia Yang, Doron Zeilberger, Systematic Counting of Restricted Partitions, arXiv:1910.08989 [math.CO], 2019.

Cf. A003106, A003116, A127836, A003113, A006141, A039899, A039900.
Cf. A188216 (least part k occurs at least k times).
Cf. A047209, A203776, A237981.
For the generalized Rogers-Ramanujan series G[1], G[2], G[3], G[4], G[5], G[6], G[7], G[8] see A003114, A003106, A006141, A264591, A264592, A264593, A264594, A264595. G[0] = G[1]+G[2] is given by A003113.
Row sums of A268187.

a003114 = p a047209_list where
   p _      0 = 1
   p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
-- Reinhard Zumkeller, Jan 05 2011

a003114 = p 1 where
   p _ 0 = 1
   p k m = if k > m then 0 else p (k + 2) (m - k) + p (k + 1) m
-- Reinhard Zumkeller, Feb 19 2013

g:=sum(x^(k^2)/product(1-x^j,j=1..k),k=0..10): gser:=series(g,x=0,65): seq(coeff(gser,x,n),n=0..60); # Emeric Deutsch, Feb 27 2006

CoefficientList[ Series[Sum[x^k^2/Product[1 - x^j, {j, 1, k}], {k, 0, 10}], {x, 0, 65}], x][[1 ;; 61]] (* Jean-François Alcover, Apr 08 2011, after Emeric Deutsch *)
Table[Count[IntegerPartitions[n], p_ /; Min[p] >= Length[p]], {n, 0, 24}] (* Clark Kimberling, Feb 13 2014 *)
a[ n_] := SeriesCoefficient[ 1 / (QPochhammer[ x^1, x^5] QPochhammer[ x^4, x^5]), {x, 0, n}]; (* Michael Somos, May 17 2015 *)
a[ n_] := SeriesCoefficient[ Product[ (1 - x^k)^{-1, 0, 0, -1, 0}[[Mod[k, 5, 1]]], {k, n}], {x, 0, n}]; (* Michael Somos, May 17 2015 *)
nmax = 60; kmax = nmax/5;
s = Flatten[{Range[0, kmax]*5 + 1}~Join~{Range[0, kmax]*5 + 4}];
Table[Count[IntegerPartitions@n, x_ /; SubsetQ[s, x]], {n, 0, nmax}] (* Robert Price, Aug 02 2020 *)

{a(n) = my(t); if( n<0, 0, t = 1 + x * O(x^n); polcoeff( sum(k=1, sqrtint(n), t *= x^(2*k - 1) / (1 - x^k) * (1 + x * O(x^(n - k^2))), 1), n))}; /* Michael Somos, Oct 15 2008 */

G.f.: Sum_{k>=0} x^(k^2)/(Product_{i=1..k} 1-x^i).
The g.f. above is the special case D=2 of sum(n>=0, x^(D*n*(n+1)/2 - (D-1)*n) / prod(k=1..n, 1-x^k) ), the g.f. for partitions into distinct part where the difference between successive parts is >= D. - Joerg Arndt, Mar 31 2014
G.f.: 1 + sum(i=1, oo, x^(5i+1)/prod(j=1 or 4 mod 5 and j<=5i+1, 1-x^j) + x^(5i+4)/prod(j=1 or 4 mod 5 and j<=5i+4, 1-x^j)). - Jon Perry, Jul 06 2004
G.f.: (Product_{k>0} 1 + x^(2*k)) * (Sum_{k>=0} x^(k^2) / (Product_{i=1..k} 1 - x^(4*i))). - Michael Somos, Oct 19 2006
Euler transform of period 5 sequence [ 1, 0, 0, 1, 0, ...]. - Michael Somos, Oct 15 2008
Expansion of f(-x^5) / f(-x^1, -x^4) in powers of x where f(,) is the Ramanujan general theta function. - Michael Somos, May 17 2015
Expansion of f(-x^2, -x^3) / f(-x) in powers of x where f(,) is the Ramanujan general theta function. - Michael Somos, Jun 13 2015
a(n) ~ phi^(1/2) * exp(2*Pi*sqrt(n/15)) / (2 * 3^(1/4) * 5^(1/2) * n^(3/4)) * (1 - (3*sqrt(15)/(16*Pi) + Pi/(60*sqrt(15))) / sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Aug 23 2015, extended Jan 24 2017
a(n) = (1/n)*Sum_{k=1..n} A284150(k)*a(n-k), a(0) = 1. - Seiichi Manyama, Mar 21 2017

A042965 Nonnegative integers not congruent to 2 mod 4.

Original entry on oeis.org

0, 1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 23, 24, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 72, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 88, 89, 91, 92

Offset: 1

N. J. A. Sloane

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence (starting at 3) gives values of AUB, sorted and duplicates removed. Values of AUBUC give same sequence. - David W. Wilson
These are the nonnegative integers that can be written as a difference of two squares, i.e., n = x^2 - y^2 for integers x,y. - Sharon Sela (sharonsela(AT)hotmail.com), Jan 25 2002. Equivalently, nonnegative numbers represented by the quadratic form x^2-y^2 of discriminant 4. The primes in this sequence are all the odd primes. - N. J. A. Sloane, May 30 2014
Numbers n such that Kronecker(4,n) == mu(gcd(4,n)). - Jon Perry, Sep 17 2002
Count, sieving out numbers of the form 2*(2*n+1) (A016825, "nombres pair-impairs"). A generalized Chebyshev transform of the Jacobsthal numbers: apply the transform g(x) -> (1/(1+x^2)) g(x/(1+x^2)) to the g.f. of A001045(n+2). Partial sums of 1,2,1,1,2,1,.... - Paul Barry, Apr 26 2005
For n>1, equals union of A020883 and A020884. - Lekraj Beedassy, Sep 28 2004
The sequence 1,1,3,4,5,... is the image of A001045(n+1) under the mapping g(x) -> g(x/(1+x^2)). - Paul Barry, Jan 16 2005
With offset 0 starting (1, 3, 4,...) = INVERT transform of A009531 starting (1, 2, -1, -4, 1, 6,...) with offset 0.
Apparently these are the regular numbers modulo 4 [Haukkanan & Toth]. - R. J. Mathar, Oct 07 2011
Numbers of the form x*y in nonnegative integers x,y with x+y even. - Michael Somos, May 18 2013
Convolution of A106510 with A000027. - L. Edson Jeffery, Jan 24 2015
Numbers that are the sum of zero or more consecutive odd positive numbers. - Gionata Neri, Sep 01 2015
Numbers that are congruent to {0, 1, 3} mod 4. - Wesley Ivan Hurt, Jun 10 2016
Nonnegative integers of the form (2+(3*m-2)/4^j)/3, j,m >= 0. - L. Edson Jeffery, Jan 02 2017
This is { x^2 - y^2; x >= y >= 0 }; with the restriction x > y one gets the same set without zero; with the restriction x > 0 (i.e., differences of two nonzero squares) one gets the set without 1. An odd number 2n-1 = n^2 - (n-1)^2, a number 4n = (n+1)^2 - (n-1)^2. - M. F. Hasler, May 08 2018

			G.f. = x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 7*x^6 + 8*x^7 + 9*x^8 + 11*x^9 + 12*x^10 + ...

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section D9.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 83.

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Pentti Haukkanen and László Tóth, An analogue of Ramanujan's sum with respect to regular integers (mod r), Ramanujan J., Vol. 27, No. 1 (2012), pp. 71-88.
Ron Knott, Pythagorean Triples and Online Calculators.
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references).
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A001045, A006165, A009531, A020883, A020884, A047209, A047222, A214546, A143978, A042968.
Essentially the complement of A016825.
See A267958 for these numbers multiplied by 4.

a042965 =  (`div` 3) . (subtract 3) . (* 4)
a042965_list = 0 : 1 : 3 : map (+ 4) a042965_list
-- Reinhard Zumkeller, Nov 09 2012

[n: n in [0..100] | not n mod 4 in [2]]; // Vincenzo Librandi, Sep 03 2015

a_list := proc(len) local rec; rec := proc(n) option remember;
ifelse(n <= 4, [0, 1, 3, 4][n], rec(n-1) + rec(n-3) - rec(n-4)) end:
seq(rec(n), n=1..len) end: a_list(76); # Peter Luschny, Aug 06 2022

nn=100; Complement[Range[0,nn], Range[2,nn,4]] (* Harvey P. Dale, May 21 2011 *)
f[n_]:=Floor[(4*n-3)/3]; Array[f,70] (* Robert G. Wilson v, Jun 26 2012 *)
LinearRecurrence[{1, 0, 1, -1}, {0, 1, 3, 4}, 70] (* L. Edson Jeffery, Jan 21 2015 *)
Select[Range[0, 100], ! MemberQ[{2}, Mod[#, 4]] &] (* Vincenzo Librandi, Sep 03 2015 *)

a(n)=(4*n-3)\3 \\ Charles R Greathouse IV, Jul 25 2011

def A042965(n): return (n<<2)//3-1 # Chai Wah Wu, Feb 10 2025

Recurrence: a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = n - 1 + (3n-3-sqrt(3)*(1-2*cos(2*Pi*(n-1)/3))*sin(2*Pi*(n-1)/3))/9. Partial sums of the period-3 sequence 0, 1, 1, 2, 1, 1, 2, 1, 1, 2, ... (A101825). - Ralf Stephan, May 19 2013
G.f.: A(x) = x^2*(1+x)^2/((1-x)^2*(1+x+x^2)); a(n)=Sum{k=0..floor(n/2)}, binomial(n-k-1, k)*A001045(n-2*k), n>0. - Paul Barry, Jan 16 2005, R. J. Mathar, Dec 09 2009
a(n) = floor((4*n-3)/3). - Gary Detlefs, May 14 2011
A214546(a(n)) != 0. - Reinhard Zumkeller, Jul 20 2012
From Michael Somos, May 18 2013: (Start)
Euler transform of length 3 sequence [3, -2, 1].
a(2-n) = -a(n). (End)
From Wesley Ivan Hurt, Jun 10 2016: (Start)
a(n) = (12*n-12+3*cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/9.
a(3k) = 4k-1, a(3k-1) = 4k-3, a(3k-2) = 4k-4. (End)
a(n) = round((4*n-4)/3). - Mats Granvik, Sep 24 2016
The g.f. A(x) satisfies (A(x)/x)^2 + A(x)/x = x*B(x)^2, where B(x) is the o.g.f. of A042968. - Peter Bala, Apr 12 2017
Sum_{n>=2} (-1)^n/a(n) = log(sqrt(2)+2)/sqrt(2) - (sqrt(2)-1)*log(2)/4. - Amiram Eldar, Dec 05 2021
From Peter Bala, Aug 03 2022: (Start)
a(n) = a(floor(n/2)) + a(1 + ceiling(n/2)) for n >= 2, with a(2) = 1 and a(3) = 3.
a(2*n) = a(n) + a(n+1); a(2*n+1) = a(n) + a(n+2). Cf. A047222 and A006165. (End)
E.g.f.: (9 + 12*exp(x)*(x - 1) + exp(-x/2)*(3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/9. - Stefano Spezia, Apr 05 2023

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Peter Pein and Ralf Stephan, Jun 17 2007

Typos fixed in Gary Detlefs's formula and in PARI program by Reinhard Zumkeller, Nov 09 2012

For a background discussion of dispersions and their fractal sequences, see A191426.  For dispersions of congruence sequences mod 3, mod 4, or mod 5, see A191655, A191663, A191667, A191702.
...
Suppose that {2,3,4,5,6} is partitioned as {x1, x2} and {x3,x4,x5}.  Let S be the increasing sequence of numbers >1 and congruent to x1 or x2 mod 5, and let T be the increasing sequence of numbers >1 and congruent to x3 or x4 or x5 mod 5.  There are 10 sequences in S, each matched by a (nearly) complementary sequence in T.  Each of the 20 sequences generates a dispersion, as listed here:
...
A191722=dispersion of A008851 (0, 1 mod 5 and >1)
A191723=dispersion of A047215 (0, 2 mod 5 and >1)
A191724=dispersion of A047218 (0, 3 mod 5 and >1)
A191725=dispersion of A047208 (0, 4 mod 5 and >1)
A191726=dispersion of A047216 (1, 2 mod 5 and >1)
A191727=dispersion of A047219 (1, 3 mod 5 and >1)
A191728=dispersion of A047209 (1, 4 mod 5 and >1)
A191729=dispersion of A047221 (2, 3 mod 5 and >1)
A191730=dispersion of A047211 (2, 4 mod 5 and >1)
A191731=dispersion of A047204 (3, 4 mod 5 and >1)
...
A191732=dispersion of A047202 (2,3,4 mod 5 and >1)
A191733=dispersion of A047206 (1,3,4 mod 5 and >1)
A191734=dispersion of A032793 (1,2,4 mod 5 and >1)
A191735=dispersion of A047223 (1,2,3 mod 5 and >1)
A191736=dispersion of A047205 (0,3,4 mod 5 and >1)
A191737=dispersion of A047212 (0,2,4 mod 5 and >1)
A191738=dispersion of A047222 (0,2,3 mod 5 and >1)
A191739=dispersion of A008854 (0,1,4 mod 5 and >1)
A191740=dispersion of A047220 (0,1,3 mod 5 and >1)
A191741=dispersion of A047217 (0,1,2 mod 5 and >1)
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For further information about these 20 dispersions, see A191722.
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Regarding the dispersions A191722-A191741, there are general formulas for sequences of the type "(a or b mod m)" and "(a or b or c mod m)" used in the relevant Mathematica programs.

Rational primes that decompose in the field Q(sqrt(5)). - N. J. A. Sloane, Dec 26 2017
These are also primes p that divide Fibonacci(p-1). - Jud McCranie
Primes ending in 1 or 9. - Lekraj Beedassy, Oct 27 2003
Also primes p such that p divides 5^(p-1)/2 - 4^(p-1)/2. - Cino Hilliard, Sep 06 2004
Primes p such that the polynomial x^2-x-1 mod p has 2 distinct zeros. - T. D. Noe, May 02 2005
Same as A038872, apart from the term 5. - R. J. Mathar, Oct 18 2008
Appears to be the primes p such that p^6 mod 210 = 1. - Gary Detlefs, Dec 29 2011
Primes in A047209, also in A090771. - Reinhard Zumkeller, Jan 07 2012
Primes p such that p does not divide Sum_{i=1..p} Fibonacci(i)^2. The sum is A001654(p). - Arkadiusz Wesolowski, Jul 23 2012
Primes congruent to {1, 9} mod 10. Legendre symbol (5, a(n)) = +1. For prime 5 this symbol (5, 5) is set to 0, and (5, prime) = -1 for prime == {3, 7} (mod 10), given in A003631. - Wolfdieter Lang, Mar 05 2021

Conjecture: n such that the characteristic polynomial of M(n) is irreducible over the rationals where M(n) is an n X n matrix with ones on the skew diagonal and below it and the skew line two positions above it and otherwise zeros; see example for one such matrix. Tested up to n=177. - Joerg Arndt, Aug 10 2011

Also n such that Kronecker(2,n) = mu(gcd(2,n)). - Jon Perry and T. D. Noe, Jun 13 2003
Also n such that x^2 == 2 (mod n) has a solution. The primes are given in sequence A001132. - T. D. Noe, Jun 13 2003
As indicated in the formula, a(n) is related to the even triangular numbers. - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
A089911(3*a(n)) = 2. - Reinhard Zumkeller, Jul 05 2013
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2)))  = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (For T see A053120.) - Wolfdieter Lang, Jun 04 2023

Or, numbers k such that k^2 == 1 (mod 9).

Or, numbers k such that the iterative cycle j -> sum of digits of j^2 when started at k contains a 1. E.g., 8 -> 6+4 = 10 -> 1+0+0 = 1 and 17 -> 2+8+9 = 19 -> 3+6+1 = 10 -> 1+0+0 = 1. - Asher Auel, May 17 2001

cp's OEIS Frontend

A191728 Dispersion of A047209, (numbers >1 and congruent to 1 or 4 mod 5), by antidiagonals.

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Mathematica

A001622 Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.

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Maple

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PARI

PARI

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A005408 The odd numbers: a(n) = 2*n + 1.

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Maxima

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A007310 Numbers congruent to 1 or 5 mod 6.

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A003114 Number of partitions of n into parts 5k+1 or 5k+4.

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