A354044 -id:A354044 - cp's OEIS frontend

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

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

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

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

A178521 The cost of all leaves in the Fibonacci tree of order n.

Original entry on oeis.org

0, 0, 3, 7, 17, 35, 70, 134, 251, 461, 835, 1495, 2652, 4668, 8163, 14195, 24565, 42331, 72674, 124354, 212155, 360985, 612743, 1037807, 1754232, 2959800, 4985475, 8384479, 14080601, 23614931, 39556030, 66181310, 110608187, 184670693, 308030923, 513334855

Offset: 0

Emeric Deutsch, Jun 14 2010

A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node. In a Fibonacci tree the cost of a left (right) edge is defined to be 1 (2). The cost of a leaf of a Fibonacci tree is defined to be the sum of the costs of the edges that form the path from the root to this leaf.

D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Colin Barker, Table of n, a(n) for n = 0..1000
Y. Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, 20, No. 2, 1982, 168-178.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A136376, A354044.

# The function 'fibrec' is defined in A354044.
function A178521(n)
    n < 2 && return BigInt(0)
    a, b = fibrec(n - 1)
    a*n + (n - 1)*b
end
println([A178521(n) for n in 0:35]) # Peter Luschny, May 16 2022

with(combinat); seq(n*fibonacci(n+1)-fibonacci(n), n = 0 .. 35);

Table[n Fibonacci[n + 1] - Fibonacci[n], {n, 0, 40}]  (* Harvey P. Dale, Apr 21 2011 *)
Table[(n - 1) Fibonacci[n] + n Fibonacci[n - 1], {n, 0, 40}] (* Bruno Berselli, Dec 06 2013 *)

concat(vector(2), Vec(x^2*(x+3)/(x^2+x-1)^2 + O(x^50))) \\ Colin Barker, Jul 26 2017

a(n) = n*F(n+1) - F(n), where F(k) = A000045(k).
G.f.: x^2*(x+3)/(x^2+x-1)^2. - Colin Barker, Nov 11 2012
a(n) = Sum_{k=1..n-1} F(k) * L(n-k+1) where F(n) = A000045(n), L(n) = A000032(n). - Gary Detlefs, Dec 29 2012
a(n) = (n-1)*F(n) + n*F(n-1). - Bruno Berselli, Dec 06 2013
a(0) = 0, a(n) = A023607(n-1) + A099920(n-1). - Collin Berman, Dec 12 2016
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4). - Wesley Ivan Hurt, Dec 14 2016
a(n) = (n+1)*F(n+1) - F(n+2). - Bruno Berselli, Jul 26 2017
a(n) = (2^(-1-n)*(2*sqrt(5)*((1-sqrt(5))^n - (1+sqrt(5))^n) + (-(1-sqrt(5))^n*(-5+sqrt(5)) + (1+sqrt(5))^n*(5+sqrt(5)))*n))/5. - Colin Barker, Jul 26 2017
a(n) = (-n*sin(c*(-n - 1)) - sin(c*n)*i)/((-i)^(-n)*sqrt(5/4)) where c = arccos(i/2). - Peter Luschny, May 16 2022

A136391 a(n) = nF(n) - (n-1)F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

Original entry on oeis.org

1, 1, 4, 6, 13, 23, 43, 77, 138, 244, 429, 749, 1301, 2249, 3872, 6642, 11357, 19363, 32927, 55861, 94566, 159776, 269469, 453721, 762793, 1280593, 2147068, 3595422, 6013933, 10048559, 16773139, 27971549, 46605186, 77587084, 129063117, 214531397, 356346557

Offset: 1

Gary W. Adamson, Dec 28 2007

By definition, the arithmetic mean of a(1) ... a(n) is equal to A000045(n).

Proof of the three-term recurrence formula: a(n+1) - a(n) - a(n-1) = ((n+1)*F(n+1) - n*F(n)) - (n*F(n) - (n-1)*F(n-1)) - ((n-1)*F(n-1) - (n-2)*F(n-2)) = (n+1)*F(n+1) - 2*n*F(n) + (n-2)*F(n-2) = (n+1)*(2*F(n) - F(n-2)) - 2*n*F(n) + (n-2)*F(n-2) = 2*F(n) - 3*F(n-2) = F(n-1) + F(n-3) = L(n-2). - Giuseppe Coppoletta, Sep 01 2014

			a(6) = 23 = 6*F(6) - 5*F(5) = 6*8 - 5*5 = 48 - 25.

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

Cf. A000045, A246715, A354044.

# The function 'fibrec' is defined in A354044.
function A136391(n)
    a, b = fibrec(n - 1)
    n*b - (n - 1)*a
end
println([A136391(n) for n in 1:35]) # Peter Luschny, May 18 2022

with(combinat): seq(n*fibonacci(n)-(n-1)*fibonacci(n-1),n=1..30); # Emeric Deutsch, Jan 01 2008

Table[n Fibonacci[n] - (n-1) Fibonacci[n-1], {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)

Vec(x*(1-x)*(1+x^2)/(1-x-x^2)^2 + O(x^100)) \\ Altug Alkan, Oct 28 2015

Equals A128064 * A000045.
From R. J. Mathar, Nov 25 2008: (Start)
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4) = A045925(n) - A045925(n-1).
G.f.: x*(1 - x)*(1 + x^2)/(1 - x - x^2)^2.
a(n) = A014286(n-1) - A014286(n-2), n>3. (End)
Recurrence: a(n+1) = a(n) + a(n-1) + L(n-2) for n>1, where L = A000032 (see proof in Comments section). - Giuseppe Coppoletta, Sep 01 2014
E.g.f.: (exp(x*phi)/phi+exp(-x/phi)*phi)*(x+1)/sqrt(5)-1, where phi=(1+sqrt(5))/2. - Vladimir Reshetnikov, Oct 28 2015
a(n) = F(n-1) + n*F(n-2). - Bruno Berselli, Jul 26 2017

More terms from Emeric Deutsch, Jan 01 2008

A264147 a(n) = nF(n+1) - (n+1)F(n), where F = A000045.

Original entry on oeis.org

0, -1, 1, 1, 5, 10, 22, 43, 83, 155, 285, 516, 924, 1639, 2885, 5045, 8773, 15182, 26162, 44915, 76855, 131119, 223101, 378696, 641400, 1084175, 1829257, 3081193, 5181893, 8702290, 14594830, 24446971, 40902299, 68359619, 114132765, 190373580, 317258388, 528265207

Offset: 0

Bruno Berselli, Nov 04 2015

a(n) is prime for n = 4, 7, 8, 26, 28, 52, 86, 87, 93, 97, 158, 196, 303, 2908, 3412, 4111, 4208, 6183, 6337, 9878, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).

Cf. A000045, A001629, A007502, A354044.
Cf. A178521: n*F(n+1) + (n+1)*F(n).
Cf. A094588: n*F(n-1) + F(n).
Cf. A099920: Sum_{i=0..n} F(i)*L(n-i).
Cf. A023607: Sum_{i=0..n} F(i)*L(n+1-i).

# The function 'fibrec' is defined in A354044.
function A264147(n)
    n == 0 && return BigInt(0)
    a, b = fibrec(n)
    n*b - a*(n + 1)
end # Peter Luschny, May 16 2022

[n*Fibonacci(n+1)-(n+1)*Fibonacci(n): n in [0..40]];

A264147 := proc(n)
    n*combinat[fibonacci](n+1)-(n+1)*combinat[fibonacci](n) ;
end proc:
seq(A264147(n),n=0..10) ; # R. J. Mathar, Jun 02 2022

Table[n Fibonacci[n + 1] - (n + 1) Fibonacci[n], {n, 0, 40}]

makelist(n*fib(n+1)-(n+1)*fib(n), n, 0, 40);

for(n=0, 40, print1(n*fibonacci(n+1)-(n+1)*fibonacci(n)", "));

concat(0, Vec(-x*(1 - 3*x) / (1 - x - x^2)^2 + O(x^50))) \\ Colin Barker, Jul 27 2017

[n*fibonacci(n+1)-(n+1)*fibonacci(n) for n in (0..40)]

G.f.:  x*(-1 + 3*x)/(1 - x - x^2)^2.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) - a(n-4).
a(n) = n*F(n-1) - F(n).
a(n) = Sum_{i=0..n} F(i)*L(n-1-i), where L() is a Lucas number (A000032).
a(n) = 3*A001629(n) - A001629(n+1).
a(n) = -(-1)^n*A178521(-n).
a(n+2) - a(n) = A007502(n+1).
Sum_{i>0} 1/a(i) = 1.39516607051636028893879220294180374...
a(n) = (-((1+sqrt(5))/2)^n*(2*sqrt(5) + (-5+sqrt(5))*n) + ((1-sqrt(5))/2)^n*(2*sqrt(5) + (5+sqrt(5))*n)) / 10. - Colin Barker, Jul 27 2017
a(n) = (-i)^n*(n*sin(c*(n+1)) - (n+1)*sin(c*n)*i)/sqrt(5/4) where c = arccos(i/2). - Peter Luschny, May 16 2022

cp's OEIS Frontend

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

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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], ...

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

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A178521 The cost of all leaves in the Fibonacci tree of order n.

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A136391 a(n) = n*F(n) - (n-1)*F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

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A264147 a(n) = n*F(n+1) - (n+1)*F(n), where F = A000045.

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A136391 a(n) = nF(n) - (n-1)F(n-1), where the F(j)'s are the Fibonacci numbers (F(0)=0, F(1)=1).

A264147 a(n) = nF(n+1) - (n+1)F(n), where F = A000045.