A023537 a(n) = Lucas(n+4) - (3*n+7).

1, 5, 13, 28, 54, 98, 171, 291, 487, 806, 1324, 2164, 3525, 5729, 9297, 15072, 24418, 39542, 64015, 103615, 167691, 271370, 439128, 710568, 1149769, 1860413, 3010261, 4870756, 7881102, 12751946, 20633139, 33385179, 54018415, 87403694, 141422212, 228826012

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Define a triangle with T(n, 1) = n*(n-1) + 1 and T(n, n) = n for n = 1, 2, 3, ... The interior terms T(r, c) = T(r - 1, c) + T(r - 2, c - 1); this triangle will give the sum of terms in row(n) = a(n). The rows begin 1; 3 2; 7 3 3; 13 6 5 4; 21 13 8 7 5 having row(n) sums 1, 5, 13, 28, 54. - J. M. Bergot, Feb 17 2013

References

• Wolfdieter Lang in "Applications of Fibonacci Numbers", Vol. 7, p. 235, eds.: G. E. Bergum et al, Kluwer, 1998.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Crossrefs

Cf. A000032, A000045, A027960, A027961, A101220.

Programs

GAP

List([1..40], n-> Lucas(1, -1, n+4)[2] -(3*n+7) ); # G. C. Greubel, Jun 01 2019

Magma

[Lucas(n+4) -(3*n+7): n in [1..40]]; // Vincenzo Librandi, Apr 16 2011

Maple

with(combinat): L:=n->fibonacci(n+2)-fibonacci(n-2): seq(L(n),n=0..12): seq(L(n+4)-3*n-7,n=1..40); # Emeric Deutsch, Aug 08 2005

Mathematica

Table[LucasL[n + 4] - (3n + 7), {n, 40}] (* Alonso del Arte, Feb 17 2013 *)

PARI

Vec(x*(1+2*x)/((1-x-x^2)*(1-x)^2) + O(x^40)) \\ Colin Barker, Mar 11 2017

Sage

[lucas_number2(n+4, 1, -1) -(3*n+7) for n in (1..40)] # G. C. Greubel, Jun 01 2019

Scala

def lucas(n: BigInt): BigInt = {
  val zero = BigInt(0)
  def fibTail(n: BigInt, a: BigInt, b: BigInt): BigInt = n match {
    case `zero` => a
    case _ => fibTail(n - 1, b, a + b)
  }
  fibTail(n, 2, 1)
}
(1 to 40).map(n => lucas(n + 4) - (3 * n + 7)) // Alonso del Arte, Oct 20 2019

Formula

Convolution of natural numbers with Lucas numbers A000204.

a(n) = A027960(n+1, n+3).

From Wolfdieter Lang: (Start)

a(n) = 7*(F(n+1) - 1) + 4*F(n) - 3*n; F(n) = A000045 (Fibonacci);

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

a(n) - a(n-1) = A101220(3, 1, n). - Ross La Haye, May 31 2006

a(n+1) - a(n) = A027961(n+1). - R. J. Mathar, Feb 21 2013

From Colin Barker, Mar 11 2017: (Start)

a(n) = -4 + (2^(-1 - n)*((1 - sqrt(5))^n*(-15 + 7*sqrt(5)) + (1 + sqrt(5))^n*(15 + 7*sqrt(5)))) / sqrt(5) - 3*(1+n).

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n > 4. (End)

From G. C. Greubel, Jun 08 2025: (Start)

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

a(n) = Sum_{i=0..n-1} Sum_{j=0..i} A027960(i,j).

E.g.f.: exp(x/2)*( 7*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2) ) - (3*x+7)*exp(x). (End)

Extensions

More terms from Emeric Deutsch, Aug 08 2005