A108195 a(n) = n^2 + 5*n - 1.

5, 13, 23, 35, 49, 65, 83, 103, 125, 149, 175, 203, 233, 265, 299, 335, 373, 413, 455, 499, 545, 593, 643, 695, 749, 805, 863, 923, 985, 1049, 1115, 1183, 1253, 1325, 1399, 1475, 1553, 1633, 1715, 1799, 1885, 1973, 2063, 2155, 2249, 2345, 2443, 2543, 2645, 2749

Offset: 1

Author

Reinhard Zumkeller, Jun 15 2005

Keywords

• nonn
• easy

Comments

a(n-2) = n*(n + 1) - 7, n >= 0, with a(-2) = -7, a(-1) = -5 and a(0) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 29 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 16 2013

Numbers m such that 4*m + 29 is an odd square, starting with 7^2 = A016754(3). - Bruce J. Nicholson, Jul 11 2017

Examples

			The Cross of Lorraine having n=2 crossbeams consists of a(2)=13 squares
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+---+---+---+---+---+
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Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, Gaullist Cross.
• Eric Weisstein's World of Mathematics, Greek Cross.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A002522, A016754, A176271.

Programs

Magma

[n^2+5*n-1: n in [1..40]]; // Vincenzo Librandi, Jun 11 2014

Maple

with (combinat):seq(fibonacci(3, n)+n-8, n=3..51); # Zerinvary Lajos, Jun 07 2008

Mathematica

CoefficientList[Series[(5 + 3 x - x^2 - 5 x)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 11 2014 *)
Array[#^2 + 5 # - 1 &, 49] (* Michael De Vlieger, Jul 12 2017 *)

PARI

a(n)=n^2+5*n-1 \\ Charles R Greathouse IV, Oct 07 2015

Formula

For n > 1: a(n) = A176271(n+2,n-1). - Reinhard Zumkeller, Apr 13 2010

a(n) = 2*n + a(n-1) + 4, with n > 1, a(1)=5. - Vincenzo Librandi, Nov 13 2010

G.f.: x*(5 - 2*x - x^2)/(1 - x)^3. - Vincenzo Librandi, Jun 11 2014

From Elmo R. Oliveira, Nov 01 2024: (Start)

E.g.f.: exp(x)*(x^2 + 6*x - 1) + 1.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

Sum_{n>=1} 1/a(n) = 47/35 + tan(sqrt(29)*Pi/2)*Pi/sqrt(29). - Amiram Eldar, May 12 2025