A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

1, 1, 3, 5, 10, 17, 30, 50, 84, 138, 227, 370, 603, 979, 1589, 2575, 4172, 6755, 10936, 17700, 28646, 46356, 75013, 121380, 196405, 317797, 514215, 832025, 1346254, 2178293, 3524562, 5702870, 9227448, 14930334, 24157799, 39088150, 63245967, 102334135, 165580121

Offset

0,3

Comments

If the first two terms are {0,1}, we get A020956 except for the first term.

If the first two terms are {1,2}, we get A281362.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Jean-Luc Baril, Sergey Kirgizov, and Armen Petrossian, Dyck Paths with catastrophes modulo the positions of a given pattern, Australasian J. Comb. (2022) Vol. 84, No. 2, 398-418.

Nathan Fox, Proof of formula for a(n).

Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Formula

From Colin Barker, Sep 16 2015: (Start)

a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5) for n>4.

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

a(n) = Fibonacci(n+3) - floor((n+3)/2). - Nathan Fox, Jan 27 2017

a(n) = (-3/4 + (-1)^n/4 + (2^(-n)*((1-t)^n*(-2+t) + (1+t)^n*(2+t)))/t + (-1-n)/2) where t=sqrt(5). - Colin Barker, Feb 09 2017

From G. C. Greubel, Apr 05 2024: (Start)

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

E.g.f.: 2*exp(x/2)*( cosh(sqrt(5)*x/2) + (2/sqrt(5))*sinh(sqrt(5)*x/2) ) - (1/2)*( (x+2)*cosh(x) + (x+3)*sinh(x) ). (End)

Mathematica

Table[((-1)^n - 2 n + 8 Fibonacci[n] + 4 LucasL[n] - 5)/4, {n, 0, 20}] (* Vladimir Reshetnikov, May 18 2016 *)
RecurrenceTable[{a[0]==a[1]==1, a[n]==a[n-1]+a[n-2]+Floor[n/2]}, a, {n, 40}] (* or *) LinearRecurrence[{2, 1, -3, 0, 1}, {1, 1, 3, 5, 10}, 40] (* Harvey P. Dale, Jul 11 2020 *)

Prog

(Python)
prpr = prev = 1
for n in range(2, 100):
    print(prpr, end=', ')
    curr = prpr+prev + n//2
    prpr = prev
    prev = curr
(PARI) Vec(-(x^3-x+1)/((x-1)^2*(x+1)*(x^2+x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2015
(PARI) a(n)=([0, 1, 0, 0, 0; 0, 0, 1, 0, 0; 0, 0, 0, 1, 0; 0, 0, 0, 0, 1; 1, 0, -3, 1, 2]^n* [1; 1; 3; 5; 10])[1, 1] \\ Charles R Greathouse IV, Jan 16 2017
(Magma) [Fibonacci(n+3)-(2*n+5-(-1)^n)/4: n in [0..40]]; // _G. C. Greubel, Feb 01 2018
(SageMath) [fibonacci(n+3) -(n+2+(n%2))//2 for n in range(41)] # G. C. Greubel, Apr 05 2024

Crossrefs

Cf. A020956, except for first term: same formula, seed {0,1}.

Cf. A000045, A281362.

Sequence in context: A270415 A192757 A079934 * A005403 A018072 A090170

Adjacent sequences: A215001 A215002 A215003 * A215005 A215006 A215007

Keyword

nonn,easy

Author

Alex Ratushnyak, Jul 31 2012

Status

approved