A122195 -id:A122195 - cp's OEIS frontend

A022308 a(n) = a(n-1) + a(n-2) + 1 for n>1, a(0)=0, a(1)=3.

0, 3, 4, 8, 13, 22, 36, 59, 96, 156, 253, 410, 664, 1075, 1740, 2816, 4557, 7374, 11932, 19307, 31240, 50548, 81789, 132338, 214128, 346467, 560596, 907064, 1467661, 2374726, 3842388, 6217115, 10059504, 16276620, 26336125, 42612746, 68948872, 111561619

Offset: 0

N. J. A. Sloane

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000032, A000045, A000285, A122195.

with(combinat): seq(fibonacci(n)+fibonacci(n+5)-1, n=-2..30); # Zerinvary Lajos, Feb 01 2008

Table[(3 LucasL[n] - Fibonacci[n] - 2)/2, {n, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)
LinearRecurrence[{2, 0, -1}, {0, 3, 4}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *)
CoefficientList[Series[x (3 - 2 x)/(x^3 - 2 x + 1), {x, 0, 20}], x] (* Eric W. Weisstein, Mar 26 2018 *)

concat(0, Vec(x*(3-2*x)/(x^3-2*x+1) + O(x^50))) \\ Colin Barker, Feb 20 2017

a(n) = if(n==0, 0, if(n==1, 3, a(n-1)+a(n-2)+1)) \\ Felix Fröhlich, Mar 26 2018

G.f.: x*(3-2*x) / (x^3-2*x+1).
a(n) = 2*A000045(n) + A000045(n+2) - 1 = A000285(n)-1.
a(n) = 2*a(n-1) - a(n-3) for n>=3. - Ron Knott, Aug 25 2006
a(n) = (3*A000032(n) - A000045(n) - 2)/2. - Vladimir Joseph Stephan Orlovsky, Feb 02 2012
a(n) = 4*F(n) + F(n-1) - 1, where F = A000045. - Bruno Berselli, Feb 20 2017
a(n) = (-10 + (5-7*sqrt(5))*((1-sqrt(5))/2)^n + ((1+sqrt(5))/2)^n*(5+7*sqrt(5))) / 10. - Colin Barker, Feb 20 2017

A022318 a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 4.

Original entry on oeis.org

1, 4, 6, 11, 18, 30, 49, 80, 130, 211, 342, 554, 897, 1452, 2350, 3803, 6154, 9958, 16113, 26072, 42186, 68259, 110446, 178706, 289153, 467860, 757014, 1224875, 1981890, 3206766, 5188657, 8395424

Offset: 0

N. J. A. Sloane

G. C. Greubel, Table of n, a(n) for n = 0..1000
Diana Savin and Elif Tan, On Companion sequences associated with Leonardo quaternions: Applications over finite fields, arXiv:2403.01592 [math.CO], 2024. See p. 2.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A122195.

LinearRecurrence[{2,0,-1}, {1,4,6}, 50] (* G. C. Greubel, Aug 25 2017 *)

x='x+O('x^50); Vec((1+2*x-2*x^2)/((1-x)*(1-x-x^2))) \\ G. C. Greubel, Aug 25 2017

From Ron Knott, Aug 25 2006: (Start)
a(n) = 2*A000045(n+2) + A000045(n) - 1.
G.f.: (1+2*x-2*x^2)/((1-x)*(1-x-x^2)).
a(0)=1, a(1)=4, a(2)=6, a(n) = 2*a(n-1) - a(n-3). (End)
a(n) - a(n-1) = A013655(n-1). - R. J. Mathar, May 06 2014

A022406 a(0)=3, a(1)=7; thereafter a(n) = a(n-1) + a(n-2) + 1.

Original entry on oeis.org

3, 7, 11, 19, 31, 51, 83, 135, 219, 355, 575, 931, 1507, 2439, 3947, 6387, 10335, 16723, 27059, 43783, 70843, 114627, 185471, 300099, 485571, 785671, 1271243, 2056915, 3328159, 5385075, 8713235, 14098311, 22811547, 36909859, 59721407, 96631267, 156352675

Offset: 0

N. J. A. Sloane

a(n) is the minimum number of nodes required for a full binary AVL tree of height n+1 whose root node has a balance factor of 0. - Sumukh Patel, Jun 24 2022

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000045, A022087, A122195. See A022403 for a very similar sequence.

[4*Fibonacci(n+2) - 1: n in [0..40]]; // G. C. Greubel, Mar 01 2018

Table[4*Fibonacci[n + 2] - 1, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
CoefficientList[Series[(3+x-3*x^2)/((1-x)*(1-x-x^2)), {x, 0, 50}], x] (* G. C. Greubel, Mar 01 2018 *)

a(n) = 4*fibonacci(n+2)-1; \\ G. C. Greubel, Mar 01 2018

a(n) = 4*A000045(n+2) - 1. - Ron Knott, Aug 25 2006
From R. J. Mathar, May 28 2008: (Start)
a(n) = A022403(n+1).
O.g.f.: (3+x-3*x^2)/((1-x)*(1-x-x^2)).
a(n+1) - a(n) = A022087(n+1). (End)
a(n) = (2^(-n)*(-5*2^n + (10-6*sqrt(5))*(1-sqrt(5))^n + 2*(1+sqrt(5))^n*(5+3*sqrt(5)))) / 5. - Colin Barker, Mar 02 2018
E.g.f.: 4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x). - Stefano Spezia, Feb 01 2025

A122194 Numbers that are the sum of exactly two sets of Fibonacci numbers.

Original entry on oeis.org

3, 5, 6, 9, 10, 15, 17, 25, 28, 41, 46, 67, 75, 109, 122, 177, 198, 287, 321, 465, 520, 753, 842, 1219, 1363, 1973, 2206, 3193, 3570, 5167, 5777, 8361, 9348, 13529, 15126, 21891, 24475, 35421, 39602, 57313, 64078, 92735, 103681, 150049, 167760, 242785

Offset: 1

Ron Knott, Aug 25 2006

			a(1)=3 as 3 is the sum of just 2 Fibonacci sets {3=Fibonacci(4)} and {1=Fibonacci(2), 2=Fibonacci(3)};
a(2)=5 as 5 is sum of Fibonacci sets {5} and {2,3} only.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
J. Berstel, An Exercise on Fibonacci Representations, RAIRO/Informatique Theorique, Vol. 35, No 6, 2001, pp. 491-498, in the issue dedicated to Aldo De Luca on the occasion of his 60th anniversary.
M. Bicknell-Johnson & D. C. Fielder, The number of Representations of N Using Distinct Fibonacci Numbers, Counted by Recursive Formulas, Fibonacci Quart. 37.1 (1999) pp. 47 ff.
Ron Knott Sumthing about Fibonacci Numbers
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1).

Cf. A000032, A000045, A000071, A000119, A013583, A122195.

a:= function(n)
    if n mod 2=0 then return 2*Fibonacci(Int((n+6)/2)) -1;
    else return Lucas(1,-1, Int((n+5)/2))[2] -1;
    fi;
  end;
List([1..50], n-> a(n) ); # G. C. Greubel, Jul 13 2019

f:=Floor; [(n mod 2) eq 0 select 2*Fibonacci(f((n+6)/2))-1 else Lucas(f((n+5)/2))-1: n in [1..50]]; // G. C. Greubel, Jul 13 2019

fib:= combinat[fibonacci]:
lucas:=n->fib(n-1)+fib(n+1):
a:=n -> if n mod 2 = 0 then 2 *fib(n/2+3) -1 else lucas((n+1)/2+2)-1 fi:
seq(a(n), n=1..50);

LinearRecurrence[{1, 1, -1, 1, -1}, {3, 5, 6, 9, 10, 15}, 40] (* Vincenzo Librandi, Jul 25 2017 *)
Table[If[Mod[n,2]==0, 2*Fibonacci[(n+6)/2]-1, LucasL[(n+5)/2]-1], {n,50}] (* G. C. Greubel, Jul 13 2019 *)

vector(50, n, f=fibonacci; if(n%2==0, 2*f((n+6)/2)-1, f((n+7)/2) + f((n+3)/2)-1)) \\ G. C. Greubel, Jul 13 2019

def a(n):
    if (mod(n,2)==0): return 2*fibonacci((n+6)/2) - 1
    else: return lucas_number2((n+5)/2, 1,-1) -1
[a(n) for n in (1..50)] # G. C. Greubel, Jul 13 2019

a(2n-1) = A000032(n+2) - 1,
a(2n) = 2*A000045(n+3) - 1.
a(2n-1) = A001610(n+2), a(2n) = A001595(n+2).
a(1)=3, a(2)=5, a(3)=6, a(4)=9, a(n) = a(n-2) + a(n-4) + 1, n > 4.
G.f.: (3 + 2*x - 2*x^2 + x^3 - 3*x^4)/(1-x-x^2+x^3-x^4+x^5).
a(n) = A272632(n)-1. - R. J. Mathar, Jan 13 2023

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A022308 a(n) = a(n-1) + a(n-2) + 1 for n>1, a(0)=0, a(1)=3.

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A022318 a(n) = a(n-1) + a(n-2) + 1, with a(0) = 1 and a(1) = 4.

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A022406 a(0)=3, a(1)=7; thereafter a(n) = a(n-1) + a(n-2) + 1.

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A122194 Numbers that are the sum of exactly two sets of Fibonacci numbers.

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