A101220 -id:A101220 - cp's OEIS frontend

A071568 Smallest k>n such that n^3+1 divides k*n^2+1.

1, 3, 11, 31, 69, 131, 223, 351, 521, 739, 1011, 1343, 1741, 2211, 2759, 3391, 4113, 4931, 5851, 6879, 8021, 9283, 10671, 12191, 13849, 15651, 17603, 19711, 21981, 24419, 27031, 29823, 32801, 35971, 39339, 42911, 46693, 50691, 54911, 59359

Offset: 0

Benoit Cloitre, May 31 2002

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A101220.

sk[n_]:=Module[{k=n+1,n2=n^2,n3=n^3+1},While[!Divisible[k*n2+1,n3], k++]; k]; Array[sk,40] (* Harvey P. Dale, Jun 13 2013 *)

for(n=1,50,s=n+1; while((s*n^2+1)%(n^3+1)>0,s++); print1(s,","))

a(n) = n^3+n+1.
a(n+1) = A101220(n, n+1, 4).
G.f.: (1 - x + 5*x^2 + x^3)/(1 - x)^4. - Philippe Deléham, Jun 06 2015
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Wesley Ivan Hurt, May 04 2021
E.g.f.: exp(x)*(1 + 2*x + 3*x^2 + x^3). - Stefano Spezia, Jul 21 2025

a(0) from Philippe Deléham, Jun 06 2015

A007502 Les Marvin sequence: a(n) = F(n) + (n-1)*F(n-1), F() = Fibonacci numbers.

Original entry on oeis.org

1, 2, 4, 9, 17, 33, 61, 112, 202, 361, 639, 1123, 1961, 3406, 5888, 10137, 17389, 29733, 50693, 86204, 146246, 247577, 418299, 705479, 1187857, 1997018, 3352636, 5621097, 9412937, 15744681, 26307469, 43912648

Offset: 1

N. J. A. Sloane, Robert G. Wilson v

Denominators of convergents of the continued fraction with the n partial quotients: [1;1,1,...(n-1 1's)...,1,n], starting with [1], [1;2], [1;1,3], [1;1,1,4], ... Numerators are A088209(n-1). - Paul D. Hanna, Sep 23 2003

			a(7) = F(7) + 6*F(6) = 13 + 6*8 = 61.

Les Marvin, Problem, J. Rec. Math., Vol. 10 (No. 3, 1976-1977), p. 213.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..500
Ignas Gasparavičius, Andrius Grigutis, and Juozas Petkelis, Picturesque convolution-like recurrences and partial sums' generation, arXiv:2507.23619 [math.NT], 2025. See p. 27.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A045925, A088209 (numerators), A101220, A109754.

a007502 n = a007502_list !! (n-1)
a007502_list = zipWith (+) a045925_list $ tail a000045_list
-- Reinhard Zumkeller, Oct 01 2012, Mar 04 2012

# The function 'fibrec' is defined in A354044.
function A007502(n)
    n == 0 && return BigInt(1)
    a, b = fibrec(n-1)
    (n-1)*a + b
end
println([A007502(n) for n in 1:32]) # Peter Luschny, May 18 2022

A007502:= func< n | Fibonacci(n) +(n-1)*Fibonacci(n-1) >;
[A007502(n): n in [1..40]]; // G. C. Greubel, Aug 26 2025

Table[Fibonacci[n]+(n-1)*Fibonacci[n-1], {n,40}] (* or *) LinearRecurrence[ {2,1,-2,-1}, {1,2,4,9}, 40](* Harvey P. Dale, Jul 13 2011 *)
f[n_] := Denominator@  FromContinuedFraction@ Join[ Table[1, {n}], {n + 1}]; Array[f, 30, 0] (* Robert G. Wilson v, Mar 04 2012 *)

Vec((1-x^2+x^3)/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 04 2012

def A007502(n): return fibonacci(n) +(n-1)*fibonacci(n-1)
print([A007502(n) for n in range(1,41)]) # G. C. Greubel, Aug 26 2025

G.f.: (1-x^2+x^3)/(1-x-x^2)^2. - Paul D. Hanna, Sep 23 2003
a(n+1) = A109754(n, n+1) = A101220(n, 0, n+1). - N. J. A. Sloane, May 19 2006
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) for n>3, a(0)=1, a(1)=2, a(2)=4, a(3)=9. - Harvey P. Dale, Jul 13 2011
E.g.f.: exp(x/2)*( ((3 + 2*x)/sqrt(5))*sinh(sqrt(5)*x/2) - cosh(sqrt(5)*x/2) ) + 1. - G. C. Greubel, Aug 26 2025

A022099 Fibonacci sequence beginning 1, 9.

Original entry on oeis.org

1, 9, 10, 19, 29, 48, 77, 125, 202, 327, 529, 856, 1385, 2241, 3626, 5867, 9493, 15360, 24853, 40213, 65066, 105279, 170345, 275624, 445969, 721593, 1167562, 1889155, 3056717, 4945872, 8002589, 12948461, 20951050, 33899511, 54850561, 88750072, 143600633, 232350705

Offset: 0

N. J. A. Sloane, Jun 14 1998

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(9;n-1-k,k) with n>=1, a(-1)=8. These are the SW-NE diagonals in P(9;n,k), the (9,1) Pascal triangle A093644. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have: b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths:  1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012
No, it is not the same as in A001175. The Pisano periods are different for moduli 71 and 142, where they are 35 and 105 instead of 70 and 210. Otherwise they coincide with those of the Fibonacci sequence. - Klaus Purath, Jun 26 2022

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A001175, A093644, A101220, A109754.

a0:=1; a1:=9; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 9}, 36] (* Robert G. Wilson v, Apr 11 2014 *)

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=9. a(-1):=8.
G.f.: (1+8*x)/(1-x-x^2).
a(n) = A109754(8, n+1) = A101220(8, 0, n+1).
a(n+1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5))+ 4*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) =  8*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 10*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+3) + Fibonacci(n-4). - Greg Dresden and Mary Beth Pittman, Mar 12 2022

A022100 Fibonacci sequence beginning 1, 10.

Original entry on oeis.org

1, 10, 11, 21, 32, 53, 85, 138, 223, 361, 584, 945, 1529, 2474, 4003, 6477, 10480, 16957, 27437, 44394, 71831, 116225, 188056, 304281, 492337, 796618, 1288955, 2085573, 3374528, 5460101, 8834629, 14294730, 23129359, 37424089, 60553448, 97977537, 158530985

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(10; n-1-k, k), n >= 1, with a(-1)=9. These are the SW-NE diagonals in P(10; n, k), the (10,1) Pascal triangle A093645. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1, 1).

Cf. A000045.

a(n) = A109754(9, n+1) = A101220(9, 0, n+1).

a0:=1; a1:=10; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1,1},{1,10},40] (* Harvey P. Dale, May 17 2017 *)

a(n) = a(n-1) + a(n-2) for n >= 2, a(0)=1, a(1)=10, a(-1):=9.
G.f.: (1 + 9*x)/(1 - x - x^2).
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*(9*binomial(1, k) - 8*binomial(0, k)). - Paul Barry, May 05 2005
a(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)) + (9/2)*((1+sqrt(5))^(n-1) - (1-sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
From Bruno Berselli, Feb 20 2017: (Start)
a(n) =  9*A000045(n) + A000045(n+1).
a(n) = 11*A000045(n) - A000045(n-2). (End)

A027973 a(n) = T(n,n) + T(n,n+1) + ... + T(n,2n), T given by A027960.

Original entry on oeis.org

1, 4, 9, 21, 46, 99, 209, 436, 901, 1849, 3774, 7671, 15541, 31404, 63329, 127501, 256366, 514939, 1033449, 2072676, 4154701, 8324529, 16673534, 33386671, 66837421, 133778524, 267724809, 535721061, 1071881326, 2144473299, 4290096449, 8582053396, 17167117141

Offset: 0

Clark Kimberling

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2).

Cf. A000032, A000045, A027960.

List([0..40], n-> 2^(n+2) - Lucas(1,-1,n+2)[2]); # G. C. Greubel, Sep 26 2019

[2^n-Lucas(n): n in [2..40]]; // Vincenzo Librandi, May 05 2017

with(combinat): a[0]:=1: for n from 1 to 30 do a[n]:=2*a[n-1]+fibonacci(n+1)-fibonacci(n-3) od: seq(a[n],n=0..30); # Emeric Deutsch, Nov 29 2006

Table[2^n - LucasL[n], {n, 2, 50}] (* Vincenzo Librandi, May 05 2017 *)

vector(40, n, f=fibonacci; 2^(n+1) - f(n+2) - f(n) ) \\ G. C. Greubel, Sep 26 2019

[2^(n+2) - lucas_number2(n+2,1,-1) for n in (0..40)] # G. C. Greubel, Sep 26 2019

With a different offset: recurrence: a(-1)=a(0)=1 a(n+2) = a(n+1) + a(n) + 2^n; formula: a(n-2) = floor(2^n - phi^n) - (1-(-1)^n)/2. - Benoit Cloitre, Sep 02 2002
a(n) = A101220(4, 2, n+1) - A101220(4, 2, n). - Ross La Haye, Aug 05 2005
a(n) = 2*a(n-1) + Fibonacci(n+1) - Fibonacci(n-3) for n>=1; a(0)=1. - Emeric Deutsch, Nov 29 2006
O.g.f.: 4/(1-2*x) - (x+3)/(1-x-x^2). - R. J. Mathar, Nov 23 2007
a(n) = 2^(n+2) + F(n) - F(n+4) with F(n)=A000045(n). - Johannes W. Meijer, Aug 15 2010
Eigensequence of an infinite lower triangular matrix with the Lucas series (1, 3, 4, 7, ...) as the left border and the rest ones. - Gary W. Adamson, Jan 30 2012
a(n) = 2^(n+2) - Lucas(n+2). - Vincenzo Librandi, May 05 2017, corrected by Greg Dresden, Sep 13 2021

A027974 a(n) = Sum_{k=1..n+1} A027960(n+1, n+1+k).

Original entry on oeis.org

1, 5, 14, 35, 81, 180, 389, 825, 1726, 3575, 7349, 15020, 30561, 61965, 125294, 252795, 509161, 1024100, 2057549, 4130225, 8284926, 16609455, 33282989, 66669660, 133507081, 267285605, 535010414, 1070731475, 2142612801, 4287086100, 8577182549, 17159235945

Offset: 0

Clark Kimberling

The former name, a(n) = Sum_{i=0..n} Sum_{j=0..i} T(i,j), T given by A027960, was in error with the data given. [This double summation gives A023537(n+1), or A027960(n+2, n+4) for n >= 0]. - G. C. Greubel, Jun 08 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
Philipp Emanuel Weidmann, The Sequencer OEIS Survey
Susanne Wienand, Suggestion for a proof of Philipp Emanuel Weidman's conjecture concerning A027983
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2).

Cf. A000032, A000045, A027960, A101220.

List([0..30], n-> 2^(n+3) - Lucas(1,-1,n+4)[2]); # G. C. Greubel, Sep 26 2019

[2^(n+3) - Lucas(n+4): n in [0..30]]; // G. C. Greubel, Sep 26 2019

with(combinat); f:=fibonacci; seq(2^(n+3) - f(n+5) - f(n+3), n=0..30); # G. C. Greubel, Sep 26 2019

Table[2^(n+3) - LucasL[n+4], {n,0,30}] (* G. C. Greubel, Sep 26 2019 *)

vector(31, n, f=fibonacci; 2^(n+2) - f(n+4) - f(n+2)) \\ G. C. Greubel, Sep 26 2019

def A027974(n): return 2**(n+3) - lucas_number2(n+4,1,-1)
[A027974(n) for n in range(31)] # G. C. Greubel, Sep 26 2019; Jun 08 2025

a(n) = 2^(n+3) - Fibonacci(n+5) - Fibonacci(n+3).
a(n) = A101220(4, 2, n+1).
G.f.: (1+2*x)/((1-2*x)*(1-x-x^2)). - R. J. Mathar, Sep 22 2008
a(n) = 2*a(n-1) + A000032(n+1). - David A. Corneth, Apr 16 2015
a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3). - Colin Barker, Feb 17 2016
From G. C. Greubel, Jun 08 2025: (Start)
a(n) = 2^(n+3) - A000032(n+4).
E.g.f.: 8*exp(2*x) - exp(x/2)*( 7*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2) ). (End)

A088209 Numerators of convergents of the continued fraction with the n+1 partial quotients: [1;1,1,...(n 1's)...,1,n+1], starting with [1], [1;2], [1;1,3], [1;1,1,4], ...

Original entry on oeis.org

1, 3, 7, 14, 28, 53, 99, 181, 327, 584, 1034, 1817, 3173, 5511, 9527, 16402, 28136, 48109, 82023, 139481, 236631, 400588, 676822, 1141489, 1921993, 3231243, 5424679, 9095126, 15230452, 25475429, 42566379, 71052157, 118489383

Offset: 0

Paul D. Hanna, Sep 23 2003

Denominators form the Les Marvin sequence: A007502(n+1).

			a(3)/A007502(4) = [1;1,1,4] = 14/9.

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

a(n) = A109754(n, n+2) = A101220(n, 0, n+2).

Cf. A007502 (the denominators), A000045, A045925.

a088209 n = a088209_list !! n
a088209_list = zipWith (+) a000045_list $ tail a045925_list
-- Reinhard Zumkeller, Oct 01 2012, Mar 04 2012

# The function 'fibrec' is defined in A354044.
function A088209(n)
    n == 0 && return BigInt(1)
    a, b = fibrec(n)
    a + (n + 1)*b
end
println([A088209(n) for n in 0:32]) # Peter Luschny, May 18 2022

f[n_] := Numerator@  FromContinuedFraction@ Join[ Table[1, {n}], {n + 1}]; Array[f, 30, 0] (* Robert G. Wilson v, Mar 04 2012 *)
CoefficientList[Series[(1+x-x^3)/(-1+x+x^2)^2,{x,0,40}],x] (* or *) LinearRecurrence[{2,1,-2,-1},{1,3,7,14},40] (* Harvey P. Dale, Jul 13 2021 *)

G.f.: (1+x-x^3)/(1-x-x^2)^2. [Corrected by Georg Fischer, Aug 16 2021]
a(n) = Fibonacci(n) + (n+1)*Fibonacci(n+1). - Paul Barry, Apr 20 2004
a(n) = a(n-1) + a(n-2) + Lucas(n). - Yuchun Ji, Apr 23 2023

A022103 Fibonacci sequence beginning 1, 13.

Original entry on oeis.org

1, 13, 14, 27, 41, 68, 109, 177, 286, 463, 749, 1212, 1961, 3173, 5134, 8307, 13441, 21748, 35189, 56937, 92126, 149063, 241189, 390252, 631441, 1021693, 1653134, 2674827, 4327961, 7002788, 11330749, 18333537, 29664286, 47997823, 77662109, 125659932, 203322041, 328981973

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(13;n-1-k,k) for n>=1, a(-1)=12. These are the SW-NE diagonals in P(13;n,k), the (13,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
In general, for b Fibonacci sequence beginning with 1, h, we have:
b(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2*h) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2*h)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Pisano period lengths: 1, 3, 8, 6, 4, 24, 16, 12, 24, 12, 10, 24, 28, 48, 8, 24, 36, 24, 18, 12, ... (is this A106291?). - R. J. Mathar, Aug 10 2012

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

a(n) = A109754(12, n+1) = A101220(12, 0, n+1).

Cf. A000032, A000045.

a0:=1; a1:=13; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013

LinearRecurrence[{1, 1}, {1, 13}, 40] (* or *) Table[LucasL[n + 5] - 5 LucasL[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *)

a(n) = a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=13, and a(-1):=12.
G.f.: (1 + 12*x)/(1 - x - x^2).
a(n) = ((1 + sqrt(5))^n-(1 - sqrt(5))^n)/(2^n*sqrt(5))+ 6*((1 + sqrt(5))^(n-1)-(1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)) for n>0. - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 12*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 14*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
a(n) = Lucas(n+5) - 5*Lucas(n). - Bruno Berselli, Dec 30 2016

A094588 a(n) = n*F(n-1) + F(n), where F = A000045.

Original entry on oeis.org

0, 1, 3, 5, 11, 20, 38, 69, 125, 223, 395, 694, 1212, 2105, 3639, 6265, 10747, 18376, 31330, 53277, 90385, 153011, 258523, 436010, 734136, 1234225, 2072043, 3474029, 5817515, 9730748, 16258910, 27139509, 45258917, 75408775, 125538539

Offset: 0

Paul Barry, May 13 2004

This is the transform of the Fibonacci numbers under the inverse of the signed permutations matrix (see A094587).

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

Cf. A000045, A007502, A045925, A088209, A354044.

a094588 n = a094588_list !! n
a094588_list = 0 : zipWith (+) (tail a000045_list)
                               (zipWith (*) [1..] a000045_list)
-- Reinhard Zumkeller, Mar 04 2012

# The function 'fibrec' is defined in A354044.
function A094588(n)
    n == 0 && return BigInt(0)
    a, b = fibrec(n - 1)
    a*n + b
end
println([A094588(n) for n in 0:34]) # Peter Luschny, May 16 2022

[n*Fibonacci(n-1)+Fibonacci(n): n in [0..60]]; // Vincenzo Librandi, Apr 23 2011

CoefficientList[Series[x (1+x-2x^2)/(1-x-x^2)^2,{x,0,40}],x]  (* Harvey P. Dale, Apr 16 2011 *)

Vec((1+x-2*x^2)/(1-x-x^2)^2+O(x^99)) \\ Charles R Greathouse IV, Mar 04 2012

G.f. : x*(1 + x - 2*x^2)/(1 - x - x^2)^2.
a(n) = A101220(n, 0, n) - Ross La Haye, Jan 28 2005
a(n) = A109754(n, n). - Ross La Haye, Aug 20 2005
a(n) = (sin(c*n)*i - n*sin(c*(n - 1)))/(i^n*sqrt(5/4)) where c = arccos(i/2). - Peter Luschny, May 16 2022

A094688 Convolution of Fibonacci(n) and 3^n.

Original entry on oeis.org

0, 1, 4, 14, 45, 140, 428, 1297, 3912, 11770, 35365, 106184, 318696, 956321, 2869340, 8608630, 25826877, 77482228, 232449268, 697351985, 2092062720, 6276199106, 18828615029, 56485873744, 169457667600, 508373077825, 1525119354868

Offset: 0

Paul Barry, May 19 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).

Cf. A000032, A000045, A101220.

I:=[0,1,4]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2) -3*Self(n-3): n in [1..41]]; // Vincenzo Librandi, Jun 24 2012

LinearRecurrence[{4,-2,-3},{0,1,4},40] (* Vincenzo Librandi, Jun 24 2012 *)
Table[(3^(n+1) -LucasL[n+2])/5, {n,0,40}] (* Vladimir Reshetnikov, Sep 27 2016 *)

a(n)=(3^(n+1)-fibonacci(n+1)-fibonacci(n+3))/5 \\ Charles R Greathouse IV, Jun 28 2011

[(3^(n+1) -lucas_number2(n+2,1,-1))/5 for n in range(41)] # G. C. Greubel, Feb 09 2023

G.f.: x/((1-3*x)*(1-x-x^2)).
a(n) = (1/5)*(3^(n+1) - Lucas(n+2)).
a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
a(n) = A101220(3, 3, n). - Ross La Haye, Jan 28 2005
a(n) = a(n-1) + a(n-2) + 3^(n-1) for n > 1, with a(0) = 0, a(1) = 1. - Ross La Haye, Aug 20 2005
a(n) = 3*a(n-1) + Fibonacci(n), where a(0) = 0. - Taras Goy, Mar 24 2019

cp's OEIS Frontend

A071568 Smallest k>n such that n^3+1 divides k*n^2+1.

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A007502 Les Marvin sequence: a(n) = F(n) + (n-1)*F(n-1), F() = Fibonacci numbers.

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A022099 Fibonacci sequence beginning 1, 9.

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A022100 Fibonacci sequence beginning 1, 10.

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A027973 a(n) = T(n,n) + T(n,n+1) + ... + T(n,2n), T given by A027960.

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A027974 a(n) = Sum_{k=1..n+1} A027960(n+1, n+1+k).

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