A115008 -id:A115008 - cp's OEIS frontend

A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

1, 1, -2, 2, -2, 5, -9, 13, -20, 34, -56, 89, -143, 233, -378, 610, -986, 1597, -2585, 4181, -6764, 10946, -17712, 28657, -46367, 75025, -121394, 196418, -317810, 514229, -832041, 1346269, -2178308, 3524578, -5702888

Offset: 0

Creighton Dement, Feb 23 2006

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

Cf. A000045, A001519, A006498, A056594, A115008, A116698, A116699, A128535.

Cf. A186679 (first differences).

a116697 n = a116697_list !! n
a116697_list = [1,1,-2,2]
               ++ (zipWith (-) a116697_list
                               $ zipWith (+) (tail a116697_list)
                                             (drop 3 a116697_list))
a128535_list = 0 : (map negate $ map a116697 [0,2..])
a001519_list = 1 : map a116697 [1,3..]
a186679_list = zipWith (-) (tail a116697_list) a116697_list
a128533_list = map a186679 [0,2..]
a081714_list = 0 : (map negate $ map a186679 [1,3..])
a075193_list = 1 : -3 : (zipWith (+) a186679_list $ drop 2 a186679_list)
-- Reinhard Zumkeller, Feb 25 2011

A116697:= func< n | (-1)^Floor((n+1)/2)*(1+(-1)^n)/2 -(-1)^n*Fibonacci(n) >;
[A116697(n): n in [0..50]]; // G. C. Greubel, Jun 08 2025

LinearRecurrence[{-1,0,-1,1},{1,1,-2,2},40] (* Harvey P. Dale, Nov 04 2011 *)

def A116697(n): return (-1)^(n//2)*((n+1)%2) - (-1)^n*fibonacci(n)
print([A116697(n) for n in range(51)]) # G. C. Greubel, Jun 08 2025

G.f.: (1 + 2*x - x^2 + x^3)/((1 + x^2)*(1 + x - x^2)).
a(2*n+1) = A000045(2*n+1) = A001519(n).
a(2*n) = - A128535(n+1). - Reinhard Zumkeller, Feb 25 2011
a(n) = A056594(n) - (-1)^n*A000045(n). - Bruno Berselli, Feb 26 2011
E.g.f.: cos(x) + (2/sqrt(5))*exp(-x/2)*sinh(sqrt(5)*x/2). - G. C. Greubel, Jun 08 2025

A116698 Expansion of (1-x+3x^2+x^3) / ((1-x-x^2)(1+2*x^2)).

Original entry on oeis.org

1, 0, 2, 5, 5, 4, 13, 29, 34, 39, 89, 176, 233, 313, 610, 1115, 1597, 2328, 4181, 7277, 10946, 16687, 28657, 48416, 75025, 117297, 196418, 326003, 514229, 815656, 1346269, 2211077, 3524578, 5637351, 9227465

Offset: 0

Creighton Dement, Feb 23 2006

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

Cf. A000045, A001519, A006498, A115008, A116697, A116699.

A116698:= func< n | Fibonacci(n+1) -((n mod 2) -2*0^((n+1) mod 4))*2^Floor(n/2) >;
[A116898(n): n in [0..50]]; // G. C. Greubel, Aug 24 2025

CoefficientList[Series[(1-x+3x^2+x^3)/((1-x-x^2)(1+2x^2)),{x,0,100}],x] (* or *) LinearRecurrence[{1,-1,2,2},{1,0,2,5},100] (* Harvey P. Dale, May 14 2022 *)
Table[Fibonacci[n+1] -I^(n-1)*Mod[n,2]*2^Floor[n/2], {n,0,50}] (* G. C. Greubel, Aug 24 2025 *)

Vec((1-x +3*x^2 +x^3)/((1-x-x^2)*(1+2*x^2)) + O(x^40)) \\ Colin Barker, May 18 2019

def A116898(n): return fibonacci(n+1) - (-1)**((n-1)//2)*(n%2)*2**(n//2)
print([A116898(n) for n in range(51)]) # G. C. Greubel, Aug 24 2025

a(2*n) = A000045(2*n+1) = A001519(n).
a(n) = a(n-1) - a(n-2) + 2*a(n-3) + 2*a(n-4) for n > 3. - Colin Barker, May 18 2019
From G. C. Greubel, Aug 24 2025: (Start)
a(n) = A000045(n+1) - (-1)^floor((n-1)/2) * (n mod 2) * 2^floor(n/2).
E.g.f.: exp(x/2)*(cosh(sqrt(5)*x/2) + (1/sqrt(5))*sinh(sqrt(5)*x/2))  - sin(sqrt(2)*x)/sqrt(2). (End)

A191373 Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 5i+j=n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 4, 1, 2, 2, 4, 2, 3, 3, 5, 1, 3, 2, 5, 2, 3, 4, 6, 1, 3, 2, 6, 2, 3, 4, 6, 2, 4, 3, 7, 3, 5, 5, 8, 1, 4, 3, 8, 2, 3, 5, 8, 2, 4, 3, 8, 4, 6, 6, 9, 1, 5, 3, 9, 2, 3, 6, 9, 2, 4, 3, 9

Offset: 0

Johannes W. Meijer, Jun 05 2011

nonn

The Le1{1,5} and Le2{5,1} triangle sums of Sierpinski’s triangle A047999 equal this sequence; see the formulas for their definitions. The Le1{1,5} and Le2{5,1} triangle sums are similar to the Kn11 and Kn21 sums, the Ca1 and Ca2 sums, and the Gi1 and Gi2 sums, see A180662.

Some A191373(2^n-p) sequences, 0<=p<=32, lead to known sequences, see the crossrefs.

Sam Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.

Cf. A001316 (1,1), A002487 (2,1), A120562 (3,1), A112970 (4,1), A191373 (5,1)

Cf. A000012 (p=0), A006498 (p=1, p=2, p=4, p=8, p=16, p=32), A070550 (p=3, p=6, p=12, p=24), A000071 (p=15, p=30), A115008 (p=23).

A191373:=proc(n) option remember; if n <0 then A191373(n):=0 fi: if (n=0 or n=1) then 1 elif n mod 2 = 0 then A191373(n/2) else A191373((n-1)/2) + A191373(((n-1)/2)-2); fi; end: seq(A191373(n),n=0..75);

a(2*n) = a(n) and a(2*n+1) = a(n) + a(n-2) with a(0) = 1, a(1) = 1 and a(n)=0 for n<=-1.
a(n) = Le1{1,5}(n) = add(T(n-4*k,k),k=0..floor(n/5))
a(n) = Le1{1,5}(n) = sum(binomial(i + j, i) mod 2 | (i + 5*j) = n)
a(n) = Le2{5,1}(n) = add(T(n-4*k,n-5*k),k=0..floor(n/5))
a(n) = Le2{5,1}(n) = sum(binomial(i + j, i) mod 2 | (5*i + j) = n)
G.f.: Product_{n>=0} (1+x^(2^n)+x^(5*2^n)).
G.f. A(x) satisfies: A(x) = (1 + x + x^5) * A(x^2). - Ilya Gutkovskiy, Jul 09 2019

A236144 a(n) = F(floor( (n+3)/2 )) * L(floor( (n+2)/2 )) where F=Fibonacci and L=Lucas numbers.

Original entry on oeis.org

2, 2, 1, 2, 6, 9, 12, 20, 35, 56, 88, 143, 234, 378, 609, 986, 1598, 2585, 4180, 6764, 10947, 17712, 28656, 46367, 75026, 121394, 196417, 317810, 514230, 832041, 1346268, 2178308, 3524579, 5702888, 9227464, 14930351, 24157818, 39088170, 63245985, 102334154

Offset: 0

Michael Somos, Jan 19 2014

			G.f. = 2 + 2*x + x^2 + 2*x^3 + 6*x^4 + 9*x^5 + 12*x^6 + 20*x^7 + 35*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..2500
Index entries for linear recurrences with constant coefficients, signature (1,0,1,1).

Cf. A000032, A000045, A115008, A115339, A128534, A128535.

m:=60; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(2-x^2-x^3)/(1-x-x^3-x^4)); // G. C. Greubel, Aug 07 2018

a[ n_] := Fibonacci[ Quotient[ n + 3, 2]] LucasL[ Quotient[ n, 2]];
CoefficientList[Series[(2-x^2-x^3)/(1-x-x^3-x^4), {x, 0, 60}], x] (* G. C. Greubel, Aug 07 2018 *)

{a(n) = fibonacci( (n+3)\2 ) * (fibonacci( n\2+1 ) + fibonacci( n\2-1 ))};

x='x+O('x^60); Vec((2-x^2-x^3)/(1-x-x^3-x^4)) \\ G. C. Greubel, Aug 07 2018

G.f.: (2 - x^2 - x^3) / (1 - x - x^3 - x^4) = (1 - x) * (2 + 2*x + x^2) / ((1 + x^2) * (1 - x - x^2)).
a(n) = a(n-1) + a(n-3) + a(n-4) for all n in Z.
0 = a(n)*a(n+2) + a(n+1)*(+a(n+2) -a(n+3)) for all n in Z.
a(n) = A115008(n+2) - A115008(n+1).
a(n) = A115339(n) * A115339(n-1).
a(2*n - 1) = F(n+1) * L(n-1) = A128535(n+1). a(2*n) = F(n+1) * L(n) = A128534(n+1).
a(n) = A000045(n+1)+A057077(n). - R. J. Mathar, Sep 24 2021

cp's OEIS Frontend

A116697 a(n) = -a(n-1) - a(n-3) + a(n-4).

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A116698 Expansion of (1-x+3*x^2+x^3) / ((1-x-x^2)*(1+2*x^2)).

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A191373 Sum of binomial coefficients C(i+j,i) modulo 2 over all pairs (i,j) of positive integers satisfying 5i+j=n.

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A236144 a(n) = F(floor( (n+3)/2 )) * L(floor( (n+2)/2 )) where F=Fibonacci and L=Lucas numbers.

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Magma

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A116698 Expansion of (1-x+3x^2+x^3) / ((1-x-x^2)(1+2*x^2)).