A005013 -id:A005013 - cp's OEIS frontend

A210209 GCD of all sums of n consecutive Fibonacci numbers.

0, 1, 1, 2, 1, 1, 4, 1, 3, 2, 11, 1, 8, 1, 29, 2, 21, 1, 76, 1, 55, 2, 199, 1, 144, 1, 521, 2, 377, 1, 1364, 1, 987, 2, 3571, 1, 2584, 1, 9349, 2, 6765, 1, 24476, 1, 17711, 2, 64079, 1, 46368, 1, 167761, 2, 121393, 1, 439204, 1, 317811, 2, 1149851, 1, 832040

Offset: 0

Alonso del Arte, Mar 18 2012

Early on in the Posamentier & Lehmann (2007) book, the fact that the sum of any ten consecutive Fibonacci numbers is a multiple of 11 is presented as an interesting property of the Fibonacci numbers. Much later in the book a proof of this fact is given, using arithmetic modulo 11. An alternative proof could demonstrate that 11*F(n + 6) = Sum_{i=n..n+9} F(i).

			a(3) = 2 because all sums of three consecutive Fibonacci numbers are divisible by 2 (F(n) + F(n-1) + F(n-2) = 2F(n)), but since the GCD of 3 + 5 + 8 = 16 and 5 + 8 + 13 = 26 is 2, no number larger than 2 divides all sums of three consecutive Fibonacci numbers.
a(4) = 1 because the GCD of 1 + 1 + 2 + 3 = 7 and 1 + 2 + 3 + 5 = 11 is 1, so the sums of four consecutive Fibonacci numbers have no factors in common.

Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, New York (2007) p. 33.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Dan Guyer and aBa Mbirika, GCD of sums of k consecutive Fibonacci, Lucas, and generalized Fibonacci numbers, Journal of Integer Sequences, 24 No.9, Article 21.9.8 (2021), 25pp; arXiv preprint, arXiv:2104.12262 [math.NT], 2021.
I. D. Ruggles, Elementary problem B-1, Fibonacci Quarterly, Vol. 1, No. 1, February 1963, p. 73; Solution by Marjorie R. Bicknell, published in Vol. 1, No. 3, October 1963, pp. 76-77.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Cf. A000045, A000071, sum of the first n Fibonacci numbers, A001175 (Pisano periods). Cf. also A229339.
Bisections give: A005013 (even part), A131534 (odd part).
Sums of m consecutive Fibonacci numbers: A055389 (m = 3, ignoring the initial 1); A000032 (m = 4, these are the Lucas numbers); A013655 (m = 5); A022087 (m = 6); A022096 (m = 7); A022379 (m = 8).

a:= n-> (Matrix(7, (i, j)-> `if`(i=j-1, 1, `if`(i=7, [1, 0, -3, -1, 1, 3, 0][j], 0)))^iquo(n, 2, 'r'). `if`(r=0, <<0, 1, 1, 4, 3, 11, 8>>, <<1, 2, 1, 1, 2, 1, 1>>))[1, 1]: seq(a(n), n=0..80);  # Alois P. Heinz, Mar 18 2012

Table[GCD[Fibonacci[n + 1] - 1, Fibonacci[n]], {n, 1, 50}] (* Horst H. Manninger, Dec 19 2021 *)

a(n)=([0,1,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0,0,0,0,0,0; 0,0,0,0,1,0,0,0,0,0,0,0,0,0; 0,0,0,0,0,1,0,0,0,0,0,0,0,0; 0,0,0,0,0,0,1,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,1,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,1,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,1,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,1; 1,0,0,0,-3,0,-1,0,1,0,3,0,0,0]^n*[0;1;1;2;1;1;4;1;3;2;11;1;8;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2017

G.f.: -x*(x^12-x^11+2*x^10-x^9-2*x^8-x^7-6*x^6+x^5-2*x^4+x^3+2*x^2+x+1) / (x^14-3*x^10-x^8+x^6+3*x^4-1) = -1/(x^4+x^2-1) + (x^2+1)/(x^4-x^2-1) + (x+2)/(6*(x^2+x+1)) + (x-2)/(6*(x^2-x+1)) - 2/(3*(x+1)) - 2/(3*(x-1)). - Alois P. Heinz, Mar 18 2012
a(n) = gcd(Fibonacci(n+1)-1, Fibonacci(n)). - Horst H. Manninger, Dec 19 2021
From Aba Mbirika, Jan 21 2022: (Start)
a(n) = gcd(F(n+1)-1, F(n+2)-1).
a(n) = Lcm_{A001175(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)

More terms from Alois P. Heinz, Mar 18 2012

A075536 a(n) = ((1+(-1)^n)T(n+1) + (1-(-1)^n)S(n))/2, where T(n) = tribonacci numbers A000073, S(n) = generalized tribonacci numbers A001644.

Original entry on oeis.org

0, 1, 1, 7, 4, 21, 13, 71, 44, 241, 149, 815, 504, 2757, 1705, 9327, 5768, 31553, 19513, 106743, 66012, 361109, 223317, 1221623, 755476, 4132721, 2555757, 13980895, 8646064, 47297029, 29249425, 160004703, 98950096, 541292033, 334745777

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 23 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, 1, 0, 1).

Cf. A000073, A001644, A005013, A005247.

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+x+4*x^2+x^3-x^4)/(1-3*x^2-x^4-x^6) )); // G. C. Greubel, Apr 21 2019

A075536 := proc(n)
    if type(n,'even') then
        A000073(n+1) ;
    else
        A001644(n) ;
    end if;
end proc:
seq(A075536(n),n=0..80) ; # R. J. Mathar, Aug 05 2021

CoefficientList[Series[(x+x^2+4x^3+x^4-x^5)/(1-3x^2-x^4-x^6), {x, 0, 40}], x]
LinearRecurrence[{0,3,0,1,0,1},{0,1,1,7,4,21},40] (* Harvey P. Dale, Jul 10 2012 *)

my(x='x+O('x^40)); concat([0], Vec(x*(1+x+4*x^2+x^3-x^4)/(1-3*x^2-x^4-x^6))) \\ G. C. Greubel, Apr 21 2019

(x*(1+x+4*x^2+x^3-x^4)/(1-3*x^2-x^4-x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(2n) = A073717(n) = A000073(2n+1).
a(2n+1) = A001644(2n+1).
a(n) = 3*a(n-2) + a(n-4) + a(n-6), a(0)=0, a(1)=1, a(2)=1, a(3)=7, a(4)=4, a(5)=21.
O.g.f.: x*(1 + x + 4*x^2 + x^3 - x^4)/(1 - 3*x^2 - x^4 - x^6).

Index in definition corrected. - R. J. Mathar, Aug 05 2021

A330051 a(n) = 1 + F(2n+1) - (F(n+4) - (-1)^nF(n-2))/2 where F=A000045.

Original entry on oeis.org

0, 0, 2, 7, 25, 72, 208, 564, 1530, 4059, 10769, 28336, 74560, 195576, 513010, 1344063, 3521385, 9221688, 24149456, 63230860, 165558250, 433454835, 1134845857, 2971111392, 7778592000, 20364739632, 53315898338, 139583151799, 365434267705, 956720165544

Offset: 0

Michael Somos, Dec 01 2019

			G.f. = 2*x^2 + 7*x^3 + 25*x^4 + 72*x^5 + 208*x^6 + 564*x^7 + 1530*x^8 + ...

Vaclav Kotesovec, Why is this product equal to zero, when the correct result is 2+GoldenRatio, Mathematica StackExchange, Sep 22 2019.
Index entries for linear recurrences with constant coefficients, signature (4,-1,-11,11,1,-4,1).

Cf. A000045, A005013, A329421, A330050.

a[n_] := 1 + Fibonacci[2 n + 1] - (Fibonacci[n + 4] - (-1)^n Fibonacci[n - 2])/2

{a(n) = 1 + fibonacci(2*n + 1) - (fibonacci(n + 4) - (-1)^n*fibonacci(n - 2))/2};

a(n) = 1 + F(2*n+1) - F(n+2) - (F(-n+2) + F(n+1))/2.
G.f.: (2*x^2 - x^3 - x^4 + x^5) / (1 - 4*x + x^2 + 11*x^3 - 11*x^4 - x^5 + 4*x^6 - x^7).
b(n) + a(n) * sqrt(5) = F(2*n+2) * Product_{k=2..n} 1 / (1 - q^k/(1 - q^(2*k))) where q = (sqrt(5)-1)/2 and b=A330050.
a(n) = A005013(floor(n/2)) * A329421(n).

Definition corrected by N. J. A. Sloane, May 29 2022 following a suggestion from Kevin Ryde.

Additional corrections by Eric Rowland, May 31 2022

A075676 Sequences A001644 and A000073 interleaved.

Original entry on oeis.org

3, 1, 3, 2, 11, 7, 39, 24, 131, 81, 443, 274, 1499, 927, 5071, 3136, 17155, 10609, 58035, 35890, 196331, 121415, 664183, 410744, 2246915, 1389537, 7601259, 4700770, 25714875, 15902591, 86992799, 53798080, 294294531, 181997601

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 24 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, 1, 0, 1).

Cf. A000073, A001644, A005013, A005247, A075536.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (3+x- 6*x^2-x^3-x^4)/(1-3*x^2-x^4-x^6) )); // G. C. Greubel, Apr 21 2019

CoefficientList[Series[(3+x-6x^2-x^3-x^4)/(1-3x^2-x^4-x^6), {x, 0, 40}], x]
LinearRecurrence[{0,3,0,1,0,1},{3,1,3,2,11,7},40] (* Harvey P. Dale, May 01 2014 *)

my(x='x+O('x^40)); Vec((3+x-6*x^2-x^3-x^4)/(1-3*x^2-x^4-x^6)) \\ G. C. Greubel, Apr 21 2019

((3+x-6*x^2-x^3-x^4)/(1-3*x^2-x^4-x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(n) = A000073(n) if n odd, a(n) = A001644(n) if n even.
a(n) = ((1-(-1)^n)*T(n) + (1+(-1)^n)*S(n))/2, where T(n) = A000073(n), S(n) =  A001644(n).
a(n) = 3*a(n-2) + a(n-4) + a(n-6), a(0)=3, a(1)=1, a(2)=3, a(3)=2, a(4)=11, a(5)=7.
O.g.f.: (3 + x - 6*x^2 - x^3 - x^4)/(1 - 3*x^2 - x^4 - x^6).
a(n) = T(n) + (1+(-1)^n)*(T(n-1) + (3/2)*T(n-2)).

A138123 Antidiagonal sums of a triangle of coefficients of recurrences of the Fibonacci sequence.

Original entry on oeis.org

1, 1, 3, 0, 3, 0, 7, 1, 11, 0, 17, 0, 29, 1, 47, 0, 75, 0, 123, 1, 199, 0, 321, 0, 521, 1, 843, 0, 1363, 0, 2207, 1, 3571, 0, 5777, 0, 9349, 1, 15127, 0, 24475, 0, 39603, 1, 64079, 0, 103681, 0, 167761, 1, 271443, 0, 439203, 0, 710647, 1, 1149851, 0, 1860497, 0

Offset: 1

Paul Curtz, May 04 2008

nonn

Consider the irregular sparse triangle T(p,p) = A000204(p), T(p,2p)= -A033999(p)=(-1)^(p+1), T(p,m) =0 else; 1<=m<=2p, p>=1. Then a(n)=sum_{m=1..[2(n+1)/3]} T(1+n-m,m).
The T are coefficients in recurrences f(n)=sum_{m=1..2p} T(p,m)*f(n-m).
The recurrence for p=1, f(n)=f(n-1)+f(n-2), is satisfied by the Fibonacci sequence A000045. The recurrence for p=2, f(n)=3f(n-2)-f(n-4), is satisfied by A005013, A005247, A075091, A075270, A108362 and A135992.
Conjecture: The Fibonacci sequence F obeys all the recurrences: A000045(n)=F(n)= L(p)*F(n-p)-(-1)^p*F(n-2p), any p>0, L=A000204.
[Proof: conjecture is equivalent to the existence of a g.f. of F with denominator 1-L(p)x^p+(-1)^p*x^(2p). Since 1-x-x^2 is known to be a denominator of such a g.f. of A000045, the conjecture is that 1-L(p)*x^p+(-1)^p*x^(2p) can be reduced to 1-x-x^2. One finds: {1-L(p)*x^p+(-1)^p*x^(2p)}/(1-x-x^2) = sum{n=0..p-1}F(n+1)x^n-sum{n=0..p-2} (-1)^(n+p)F(n+1)x^(2p-n-2) is a polynomial with integer coefficients, which is proved by multiplication with 1-x-x^2 and via F(n)+F(n+1)=F(n+2) and L(n)=F(n-1)+F(n+1). - R. J. Mathar, Jul 10 2008].
Conjecture: The Lucas sequence L also obeys all the recurrences: L(n)= L(p)*L(n-p)-(-1)^p*L(n-2p), any p>0, L=A000204.

			The triangle T(p,m) with Lucas numbers on the diagonal starts
  1, 1;
  0, 3, 0,-1;
  0, 0, 4, 0, 0, 1;
  0, 0, 0, 7, 0, 0, 0,-1;
  0, 0, 0, 0,11, 0, 0, 0, 0, 1;
The antidiagonal sums are a(1)=1. a(2)=0+1=1. a(3)=0+3=3. a(4)=0+0+0=0. a(5)=0+0+4-1=3.

Row sums: Sum_{m=1..2p} T(p,m) = A098600(p).
Conjectures from Chai Wah Wu, Apr 15 2024: (Start)
a(n) = a(n-2) - a(n-3) + a(n-4) + a(n-5) + a(n-7) for n > 7.
G.f.: x*(-x^5 - 2*x^2 - x - 1)/((x + 1)*(x^2 - x + 1)*(x^4 + x^2 - 1)). (End)

Edited and extended by R. J. Mathar, Jul 10 2008

A259875 Irregular triangle read by rows: coefficients (highest degree first) of polynomials defined by p_0(x)=0, p_1(x)=p_2(x)=1, p_3(x)=x+1; p_n(x)=x*p_{n-2}(x)-p_{n-4}(x).

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 0, 1, 1, -1, 1, 0, -1, 1, 1, -2, -1, 1, 0, -2, 0, 1, 1, -3, -2, 1, 1, 0, -3, 0, 1, 1, 1, -4, -3, 3, 1, 1, 0, -4, 0, 3, 0, 1, 1, -5, -4, 6, 3, -1, 1, 0, -5, 0, 6, 0, -1, 1, 1, -6, -5, 10, 6, -4, -1, 1, 0, -6, 0, 10, 0, -4, 0, 1, 1, -7, -6, 15, 10, -10, -4, 1

Offset: 0

N. J. A. Sloane, Jul 09 2015

			Triangle begins:
0;
1;
1;
1, 1;
1, 0;
1, 1, -1;
1, 0, -1;
1, 1, -2, -1;
1, 0, -2,  0;
1, 1, -3, -2, 1;
1, 0, -3,  0, 1;
1, 1, -4, -3, 3, 1;
1, 0, -4,  0, 3, 0;
...

Alois P. Heinz, Rows n = 0..200, flattened
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. [Annotated scanned copy]
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.

p_n(3) gives A005013.

p:= proc(n) option remember; expand(`if`(n=0, 0,
     `if`(n<3, 1, `if`(n=3, x+1, x*p(n-2)-p(n-4)))))
    end:
T:= n-> `if`(n=0, 0, (s-> seq(coeff(s, x, degree(s)-i)
        , i=0..degree(s)))(p(n))):
seq(T(n), n=0..20);  # Alois P. Heinz, Jul 10 2015

p[0] = 0&; p[1] = p[2] = 1&; p[3] = #+1&; p[n_][x_] := p[n, x] = x*p[n-2][x] - p[n-4][x];
row[0] = {0}; row[n_] := CoefficientList[p[n][x], x] // Reverse;
Table[row[n], {n, 0, 20}] // Flatten (* Jean-François Alcover, Jun 12 2018 *)

More terms from Alois P. Heinz, Jul 10 2015

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A210209 GCD of all sums of n consecutive Fibonacci numbers.

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A075536 a(n) = ((1+(-1)^n)*T(n+1) + (1-(-1)^n)*S(n))/2, where T(n) = tribonacci numbers A000073, S(n) = generalized tribonacci numbers A001644.

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A330051 a(n) = 1 + F(2*n+1) - (F(n+4) - (-1)^n*F(n-2))/2 where F=A000045.

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A075676 Sequences A001644 and A000073 interleaved.

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PARI

Sage

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A138123 Antidiagonal sums of a triangle of coefficients of recurrences of the Fibonacci sequence.

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Extensions

A259875 Irregular triangle read by rows: coefficients (highest degree first) of polynomials defined by p_0(x)=0, p_1(x)=p_2(x)=1, p_3(x)=x+1; p_n(x)=x*p_{n-2}(x)-p_{n-4}(x).

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A075536 a(n) = ((1+(-1)^n)T(n+1) + (1-(-1)^n)S(n))/2, where T(n) = tribonacci numbers A000073, S(n) = generalized tribonacci numbers A001644.

A330051 a(n) = 1 + F(2n+1) - (F(n+4) - (-1)^nF(n-2))/2 where F=A000045.