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

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

Author

Paul Barry, Apr 04 2003

Keywords

• easy
• nonn

Comments

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

Links

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

Crossrefs

First row of A082110.

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

Programs

Mathematica

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 *)

PARI

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

Formula

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

Extensions

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