Geoff Ahiakwo - cp's OEIS frontend

A255875 a(n) = Fibonacci(n+2) + n - 2.

1, 3, 6, 10, 16, 25, 39, 61, 96, 152, 242, 387, 621, 999, 1610, 2598, 4196, 6781, 10963, 17729, 28676, 46388, 75046, 121415, 196441, 317835, 514254, 832066, 1346296, 2178337, 3524607, 5702917, 9227496, 14930384, 24157850, 39088203, 63246021, 102334191, 165580178, 267914334

Offset: 1

Geoff Ahiakwo, Mar 08 2015

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

Array[Fibonacci[# + 2] + # - 2 &, 30] (* Michael De Vlieger, Jul 10 2015 *)

main(size)={return( vector(size,n,fibonacci(n+2)+n-2));} /* Anders Hellström, Jul 11 2015 */

G.f.: x*(-1+x^2+x^3) / ( (x^2+x-1)*(x-1)^2 ). - R. J. Mathar, Jul 10 2015

a(n) = A232896(n+1)-2. - R. J. Mathar, Jul 10 2015

Terms a(31) and beyond from Andrew Howroyd, Dec 23 2019

A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.

Original entry on oeis.org

1983, 1984, 1986, 1989, 1994, 2002, 2015, 2036, 2070, 2125, 2214, 2358, 2591, 2968, 3578, 4565, 6162, 8746, 12927, 19692, 30638, 48349, 77006, 123374, 198399, 319792, 516210, 834021, 1348250, 2180290, 3526559, 5704868, 9229446, 14932333, 24159798

Offset: 1

Geoff Ahiakwo, Oct 22 2009

Starting at some a(1)=s and creating further terms with the recurrence a(n)=a(n-1)+A000045(n) defines a family of sequences with recurrences a(n)= 2*a(n-1) -a(n-3).
The generating functions are x*( s+(1-s)*x+(1-s)*x^2 )/((1-x) * (1-x-x^2)).
The terms are a(n) = A000045(n+2)+s-2 = s + A001911(n-1) = (2*s+1+k)/2 where k=A166863(n-1), n>=1.
Examples: Up to offsets, s=1 yields A000071, s=2 yields A000045 shifted left thrice, s=3 yields A001611 shifted left thrice, s=4 yields A018910.
I appreciate the editing by R. J. Mathar. However I would like further analysis of the following formula. The sequence which I call GAP can have any integer as its first term, not just 1983. Thus a(1) can be 0, 1, 2, 3,... Then a(2) is always a(1)+ 1, while a(3) is a(1) + k(n)/2; where k(n) = k(n-2)+ k(n-1)+4 (This is a separate sequence submitted for consideration). [Geoff Ahiakwo, Nov 19 2009]

			For s=1983, n=3, we have k= A166863(2)= 5, a(3) = (2s+1+k)/2 = (2*1983+1+5)/2 = 1986.
For n=3, a(3)= a(1)+ k(3)/2 = a(1)+ [K(3-2)+ k(3-1)]/2 + 2 = a(1)+ 1 + 2 thus if a(1)is 0, a(3)= 3; if a(1)= 5, a(3)= 8; if a(1)=1983, a(3)= 1986, etc. [_Geoff Ahiakwo_, Nov 19 2009]

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

Cf. A001911, A000071

LinearRecurrence[{2, 0, -1}, {1983, 1984, 1986}, 100] (* G. C. Greubel, May 27 2016 *)

a(n) = 2*a(n-1) - a(n-3).
G.f.: x*(-1983 + 1982*x + 1982*x^2)/((1-x)*(x^2+x-1)).
Let a(n)= a(1)+ k(n)/2, then G.f.: k(n)= k(n-2)+ k(n-1) + 4. - Geoff Ahiakwo, Nov 19 2009

Definition and comments edited by R. J. Mathar, Oct 26 2009

A166863 a(1)= 1; a(2)= 5; thereafter a(n)= a(n-1) + a(n-2) + 5.

Original entry on oeis.org

1, 5, 11, 21, 37, 63, 105, 173, 283, 461, 749, 1215, 1969, 3189, 5163, 8357, 13525, 21887, 35417, 57309, 92731, 150045, 242781, 392831, 635617, 1028453, 1664075, 2692533, 4356613, 7049151, 11405769, 18454925, 29860699, 48315629, 78176333, 126491967

Offset: 1

Geoff Ahiakwo, Oct 22 2009

			a(3) = 5 + 1 + 5 = 11.

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

Cf. A001595, A154691.

a166863 n = a166863_list !! (n-1)
a166863_list = 1 : zipWith (+) a166863_list (drop 3 $ map (* 2) a000045_list)
-- Reinhard Zumkeller, Nov 17 2013

2 * Fibonacci[Range[4,4! ]] - 5 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *)
RecurrenceTable[{a[1]==1,a[2]==5,a[n]==a[n-1]+a[n-2]+5},a,{n,40}] (* or *) LinearRecurrence[{2,0,-1},{1,5,11},40] (* Harvey P. Dale, Jan 29 2021 *)

a(n) = A154691(n) - 2 = 2*A000045(n+3) - 5. - R. J. Mathar, Oct 26 2009
From R. J. Mathar, Oct 26 2009: (Start)
a(n) = 2*a(n-1) - a(n-3).
G.f: x*(1+3*x+x^2)/((x-1)* (x^2+x-1)). (End)
a(n+1) = a(n) + 2*A000045(n+2). - Reinhard Zumkeller, Nov 17 2013

Missing value for a(29) inserted by Reinhard Zumkeller, Nov 17 2013

A169622 a(n) = a(n-1) + Fibonacci(n), a(1)=5.

Original entry on oeis.org

5, 6, 8, 11, 16, 24, 37, 58, 92, 147, 236, 380, 613, 990, 1600, 2587, 4184, 6768, 10949, 17714, 28660, 46371, 75028, 121396, 196421, 317814, 514232, 832043, 1346272, 2178312, 3524581, 5702890, 9227468, 14930355, 24157820, 39088172, 63245989, 102334158

Offset: 1

Geoff Ahiakwo, Dec 03 2009

			n=2: a(1)+Fibonacci(2) = 5+1 = 6.
n=3: a(2)+Fibonacci(3) = 6+2 = 8.

Vincenzo Librandi, Table of n, a(n) for n = 1..250
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A166876, A168193, A000045.

[ n eq 1 select 5 else Self(n-1)+Fibonacci(n): n in [1..40] ];  // Klaus Brockhaus, Jan 31 2011

RecurrenceTable[{a[1]==5,a[n]==a[n-1]+Fibonacci[n]},a[n],{n,40}] (* or *) LinearRecurrence[{2,0,-1},{5,6,8},40] (* Harvey P. Dale, Jul 20 2011 *)

Vec((5 - 4*x - 4*x^2) / ((1 - x)*(1 - x - x^2)) + O(x^40)) \\ Colin Barker, Apr 20 2017

a(n) = 5 + A168193(n)/2.
a(n) = 2*a(n-1) - a(n-3) = 3 + A000045(n+2). - R. J. Mathar Dec 04 2009
G.f.: x*(-5+4*x+4*x^2) / ((1-x)*(x^2+x-1)). - R. J. Mathar Dec 04 2009
a(n) = 3 + (2^(-1-n)*((1-sqrt(5))^n*(-3+sqrt(5)) + (1+sqrt(5))^n*(3+sqrt(5)))) / sqrt(5). - Colin Barker, Apr 20 2017

A168193 a(n) = a(n-1) + a(n-2) + 4, with a(0)=0, a(1)=2.

Original entry on oeis.org

0, 2, 6, 12, 22, 38, 64, 106, 174, 284, 462, 750, 1216, 1970, 3190, 5164, 8358, 13526, 21888, 35418, 57310, 92732, 150046, 242782, 392832, 635618, 1028454, 1664076, 2692534, 4356614, 7049152, 11405770, 18454926, 29860700, 48315630, 78176334, 126491968

Offset: 0

Geoff Ahiakwo, Nov 19 2009

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

I:=[0,2,6]; [n le 3 select I[n] else Self(n-1)+Self(n-2)+4: n in [1..40]]; // Vincenzo Librandi, Jul 16 2016

Fibonacci[Range[3,4! ]]*2-4 (* Vladimir Joseph Stephan Orlovsky, Mar 19 2010 *)
LinearRecurrence[{2, 0, -1}, {0, 2, 6}, 50] (* G. C. Greubel, Jul 15 2016 *)

From R. J. Mathar, Nov 22 2009: (Start)
a(n)= 2*a(n-1) - a(n-3) = 2*A001911(n).
G.f.: 2*x*(1+x)/((x-1)*(x^2+x-1)). (End)
a(n) = a(n-1) + 2*Fibonacci(n+1), with a(0)=0. - Taras Goy, Mar 24 2019
E.g.f.: 4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 4*exp(x). - Stefano Spezia, Oct 14 2022
a(n) = A019274(n+1)+A019274(n+2). - R. J. Mathar, Jul 07 2023

Definition replaced by recurrence from R. J. Mathar, Nov 23 2009

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User: Geoff Ahiakwo

A255875 a(n) = Fibonacci(n+2) + n - 2.

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A166876 a(n) = a(n-1) + Fibonacci(n), a(1)=1983.

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A166863 a(1)= 1; a(2)= 5; thereafter a(n)= a(n-1) + a(n-2) + 5.

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A169622 a(n) = a(n-1) + Fibonacci(n), a(1)=5.

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A168193 a(n) = a(n-1) + a(n-2) + 4, with a(0)=0, a(1)=2.

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