A001068 a(n) = floor(5*n/4), numbers that are congruent to {0, 1, 2, 3} mod 5.

0, 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 85, 86, 87, 88

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

From M. F. Hasler, Oct 21 2008: (Start)

Also, for n>0, the 4th term (after [0,n,3n]) in the continued fraction expansion of arctan(1/n). (Observation by V. Reshetnikov)

Proof:

arctan(1/n) = (1/n) / (1 + (1/n)^2/( 3 + (2/n)^2/( 5 + (3/n)^2/( 7 + ...)...)

= 1 / ( n + 1/( 3n + 4/( 5n + 9/( 7n + 25/(...)...)

= 1 / ( n + 1/( 3n + 1/( 5n/4 + (9/4)/( 7n + 25/(...)...),

and the term added to 5n/4, (9/4)/(7n+...) = (1/4)*9/(7n+...) is less than 1/4 for all n>=2. (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Paul Erdős, Some recent problems and results in graph theory, Discr. Math., Vol. 164, No. 1-3 (1997), pp. 81-85.
• Wikipedia, Continued fraction for arctangent.
• R. Witula, P. Lorenc, M. Rozanski and M. Szweda, Sums of the rational powers of roots of cubic polynomials, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, (2014), pp. 17-34.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Crossrefs

Cf. A001622, A002265, A030308, A047203, A110255, A110256, A110257, A110258, A110259, A110260, A177704 (first differences), A249547.

Programs

Magma

[Floor(5*n/4): n in [0..80]]; // Vincenzo Librandi, Nov 13 2011

Maple

A001068:=n->floor(5*n/4); seq(A001068(k), k=0..100); # Wesley Ivan Hurt, Nov 07 2013

Mathematica

Table[Floor[5*n/4],{n,0,120}] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
#+{0,1,2,3}&/@(5*Range[0,20])//Flatten (* or *) Complement[Range[0,103],5*Range[20]-1] (* Harvey P. Dale, Dec 03 2023 *)

PARI

a(n)=5*n\4 /* or, cf. comment: */
a(n)=contfrac(atan(1/n))[4] \\ M. F. Hasler, Oct 21 2008

Formula

contfrac( arctan( 1/n )) = 0 + 1/( n + 1/( 3n + 1/( a(n) + 1/(...)))). - M. F. Hasler, Oct 21 2008

a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=2 and b(k)=5*2^(k-2) for k>1. - Philippe Deléham, Oct 17 2011.

From Bruno Berselli, Oct 17 2011: (Start)

G.f.: x*(1+x+x^2+2*x^3)/((1+x)*(1-x)^2*(1+x^2)).

a(n) = (10*n+2*(-1)^((n-1)n/2)+(-1)^n-3)/8.

a(-n) = -A047203(n+1). (End)

From Wesley Ivan Hurt, Sep 17 2015: (Start)

a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.

a(n) = n + floor(n/4). (End)

a(n) = n + A002265(n). - Robert Israel, Sep 17 2015

E.g.f.: (sin(x) + cos(x) + (5*x - 2)*sinh(x) + (5*x - 1)*cosh(x))/4. - Ilya Gutkovskiy, May 06 2016

Sum_{n>=1} (-1)^(n+1)/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 + sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 10 2021

Extensions

More terms from James Sellers, Sep 19 2000