A249547 -id:A249547 - cp's OEIS frontend

A047215 Numbers that are congruent to {0, 2} mod 5.

0, 2, 5, 7, 10, 12, 15, 17, 20, 22, 25, 27, 30, 32, 35, 37, 40, 42, 45, 47, 50, 52, 55, 57, 60, 62, 65, 67, 70, 72, 75, 77, 80, 82, 85, 87, 90, 92, 95, 97, 100, 102, 105, 107, 110, 112, 115, 117, 120, 122, 125, 127, 130, 132, 135, 137, 140, 142, 145, 147, 150, 152, 155, 157

Offset: 0

N. J. A. Sloane

Number of partitions of 5n into exactly 2 parts. - Colin Barker, Mar 23 2015

Numbers k such that k^2/5 + k*(k + 1)/5 = k*(2*k + 1)/5 is a nonnegative integer. - Bruno Berselli, Feb 14 2017

David Lovler, Table of n, a(n) for n = 0..1000
Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See p. 4.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Different from A038126.
Cf. A001622, A010693, A030308, A084215
Cf. A249547 (partial sums), A010693 (1st differences).

[(5*n - (n mod 2))/2: n in [1..100]]; // G. C. Greubel, Jun 23 2024

Table[Floor[5 n/2], {n, 0, 100}] (* or *) LinearRecurrence[{1, 1, -1}, {0, 2, 5},101] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)

a(n)=5*n\2 \\ Charles R Greathouse IV, Oct 17 2011

[int(5*n//2) for n in range(1,101)] # G. C. Greubel, Jun 23 2024

a(n) = floor(5*n/2).
From R. J. Mathar, Sep 23 2008: (Start)
G.f.: x*(2 + 3*x)/((1 + x)*(1 - x)^2).
a(n) = 5*n/2 + ((-1)^n-1)/4.
a(n+1) - a(n) = A010693(n+1). (End)
a(n) = 5*n - a(n-1) - 8 with a(1)=0. - Vincenzo Librandi, Aug 05 2010
a(n+1) = Sum_{k>=0} A030308(n,k)*A084215(k+1). - Philippe Deléham, Oct 17 2011
a(n) = 2*n + floor(n/2). - Arkadiusz Wesolowski, Sep 19 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = log(5)/4 - sqrt(5)*log(phi)/10 + sqrt(1-2/sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021
E.g.f.: (5*x*exp(x) - sinh(x))/2. - David Lovler, Aug 22 2022

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

Original entry on oeis.org

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

N. J. A. Sloane

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)

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).

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

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

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

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 *)

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

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

More terms from James Sellers, Sep 19 2000

A082111 a(n) = n^2 + 5*n + 1.

Original entry on oeis.org

1, 7, 15, 25, 37, 51, 67, 85, 105, 127, 151, 177, 205, 235, 267, 301, 337, 375, 415, 457, 501, 547, 595, 645, 697, 751, 807, 865, 925, 987, 1051, 1117, 1185, 1255, 1327, 1401, 1477, 1555, 1635, 1717, 1801, 1887, 1975, 2065, 2157, 2251, 2347, 2445, 2545, 2647

Offset: 0

Paul Barry, Apr 04 2003

From Gary W. Adamson, Jul 29 2009: (Start)
Let (a,b) = roots to x^2 - 5*x + 1 = 0 = 4.79128... and 0.208712...
Then a(n) = (n + a) * (n + b). Example: a(5) = 51 = (5 + 4.79128...) * (5 + 0.208712...) (End)
For n > 0: a(n) = A176271(n+2,n). - Reinhard Zumkeller, Apr 13 2010
a(n-2) = n*(n+1) - 5, n >= 0, with a(-2) = -5 and a(-1) = -3, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 21 for b = 2*n + 1. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
Numbers m > 0 such that 4m+21 is a square. - Bruce J. Nicholson, Jul 19 2017
Numbers represented as 151 in number base B. If 'digits' from B upwards are allowed then 151(2)=15, 151(3)=25, 151(4)=37, 151(5)=51 also. - Ron Knott, Nov 14 2017
If A and B are sequences satisfying the recurrence t(n) = 5*t(n-1) - t(n-2) with initial values A(0) = 1, A(1) = n+5 and B(0) = -1, B(1) = n, then a(n) = A(i)^2 - A(i-1)*A(i+1) = B(j)^2 - B(j-1)*B(j+1) for i, j > 0. - Klaus Purath, Oct 18 2020
The prime terms in this sequence are listed in A089376. The prime factors are given in A038893. With the exception of 3 and 7, each prime factor p divides exactly 2 out of any p consecutive terms. If a(i) and a(k) form such a pair that are divisible by p, then i + k == -5 (mod p). - Klaus Purath, Nov 24 2020

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

First row of A082110.

Cf. A002522, A014209, A028387, A028872, A338041.

Table[n^2 + 5*n + 1,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011 *)
LinearRecurrence[{3,-3,1},{1,7,15},80] (* Harvey P. Dale, Apr 22 2012 *)

a(n)=n^2+5*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 2*n + a(n-1) + 4 (with a(0)=1). - Vincenzo Librandi, Aug 08 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=1, a(1)=7, a(2)=15. - Harvey P. Dale, Apr 22 2012
Sum_{n>=0} 1/a(n) = 8/15 + Pi*tan(sqrt(21)*Pi/2)/sqrt(21) = 1.424563592286456286... . - Vaclav Kotesovec, Apr 10 2016
From G. C. Greubel, Jul 19 2017: (Start)
G.f.: (1 + 4*x - 3*x^2)/(1 - x)^3.
E.g.f.: (x^2 + 6*x + 1)*exp(x). (End)
a(n) = A014209(n+1) - 2 = A338041(2*n+1). - Hugo Pfoertner, Oct 08 2020
a(n) = A249547(n+1) - A024206(n-4), n >= 5. - Klaus Purath, Nov 24 2020

New title (using given formula) from Hugo Pfoertner, Oct 08 2020

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A047215 Numbers that are congruent to {0, 2} mod 5.

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

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A082111 a(n) = n^2 + 5*n + 1.

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