A033890 -id:A033890 - cp's OEIS frontend

A360467 a(n) = Fibonacci(4n+2) + 3Fibonacci(2*n+1)^2.

4, 20, 130, 884, 6052, 41474, 284260, 1948340, 13354114, 91530452, 627359044, 4299982850, 29472520900, 202007663444, 1384581123202, 9490060198964, 65045840269540, 445830821687810, 3055769911545124, 20944558559128052, 143556140002351234, 983948421457330580

Offset: 0

Alexander M. Domashenko, Feb 08 2023

nonn

Values of x + 3*y in solutions of x^2 = 5*y^2 - 4*y in positive integers. In the solutions, the values of x and y are given by Fibonacci(4*n + 2) and Fibonacci(2*n + 1)^2 respectively.

The above Diophantine equation arises out of the following problem regarding the subdivision of a square into four triangles of integer area. For n >= 1, the sequence gives the areas of the squares in the solutions (see illustration in Links). Two lines are drawn from a corner of a square to points on the opposing sides. A third line is added between the two points so that the square is divided into four triangles. The area of each triangle is required to be an integer and those of the right triangles to form an arithmetic progression with difference 1. The smallest right triangle by area is the one formed by the third line. In the solutions, the area of the inner triangle is given by Fibonacci(4*n + 2) and the total area of the three right triangles is 3*Fibonacci(2*n + 1)^2. The area of the square is then equal to a(n).

			a(2) = F(4*2+2) + 3*F(2*2 +1)^2 = F(10) + 3*F(5)^2 = 55 + 3*5^2 = 130.
a(4) = F(4*4+2) + 3*F(2*4 +1)^2 = F(18) + 3*F(9)^2 = 2584+ 3*34^2 = 6052.
G.f. = 4 + 20*x + 130*x^2 + 884*x^3 + 6052*x^4 + ... - _Michael Somos_, Mar 02 2023

Nicolay Avilov, Problem 2450. The sixth parallelogram (in Russian),
Nicolay Avilov, Problem 2447. Four triangles in a parallelogram (in Russian),
Nicolay Avilov, Problem 2442. Herringbone in a parallelogram (in Russian).
Alexander M. Domashenko, Illustration of square divided into four triangles.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A033890, A064170, A081068, A295683.

a := proc(n) option remember; if n < 3 then return [4, 20, 130][n + 1] fi;
a(n-3) - 8 * (a(n-2) - a(n-1)) end: seq(a(n), n = 0..22); # Peter Luschny, Feb 17 2023

LinearRecurrence[{8, -8, 1}, {4, 20, 130}, 22] (* Amiram Eldar, Feb 17 2023 *)
a[ n_] := 2 * Fibonacci[2*n+1] * Fibonacci[2*n+3]; (* Michael Somos, Mar 02 2023 *)

Vec(2*(2 - 6*x + x^2)/((1 - x)*(1 - 7*x + x^2)) + O(x^25)) \\ Andrew Howroyd, Feb 16 2023

print([2*(lucas_number2(n+1, 7, 1) + 3) // 5 for n in range(23)]) # Peter Luschny, Feb 17 2023

a(n) = A033890(n) + 3*A081068(n)^2.
a(n) = Fibonacci(2*n+1)*(Fibonacci(2*n) + Fibonacci(2*n+2) + 3*Fibonacci(2*n+1)).
a(n) = 2*A064170(n+3).
G.f.: 2*(2 - 6*x + x^2)/((1 - x)*(1 - 7*x + x^2)). - Andrew Howroyd, Feb 16 2023
a(n) = a(n-3) - 8 * (a(n-2) - a(n-1)) for n >= 3. - Peter Luschny, Feb 17 2023
a(n) = a(-2-n) = 2*F(2*n+1) * F(2*n+3) = A295683(4*(n+1)) for all n in Z. - Michael Somos, Mar 02 2023

A081004 a(n) = Fibonacci(4n+2) + 1, or Fibonacci(2n+2)*Lucas(2n).

Original entry on oeis.org

2, 9, 56, 378, 2585, 17712, 121394, 832041, 5702888, 39088170, 267914297, 1836311904, 12586269026, 86267571273, 591286729880, 4052739537882, 27777890035289, 190392490709136, 1304969544928658, 8944394323791465, 61305790721611592, 420196140727489674

Offset: 0

R. K. Guy, Mar 01 2003

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Nathaniel Johnston, Table of n, a(n) for n = 0..600
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A056854 (first differences).

List([0..30], n-> Fibonacci(4*n+2)+1); # G. C. Greubel, Jul 15 2019

[Fibonacci(4*n+2)+1: n in [0..30]]; // Vincenzo Librandi, Apr 15 2011

with(combinat): for n from 0 to 30 do printf(`%d,`,fibonacci(4*n+2)+1) od: # James Sellers, Mar 03 2003

Table[Fibonacci[4n+2] +1, {n,0,30}] (* Wesley Ivan Hurt, Nov 20 2014 *)

vector(30, n, n--; fibonacci(4*n+2)+1) \\ G. C. Greubel, Jul 15 2019

[fibonacci(4*n+2)+1 for n in (0..30)] # G. C. Greubel, Jul 15 2019

a(n) = A033890(n)+1.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (2-7*x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Jun 24 2012

More terms from James Sellers, Mar 03 2003

A159950 Dividends where Fibonacci products/sums yield integral quotients.

Original entry on oeis.org

240, 122522400, 137932073613734400, 342696507457909818131702784000, 1879127177606120717127879344567470740879360000, 22740756589119797763590969093409514524935686067027158720512000000

Offset: 1

Enoch Haga, Apr 27 2009

In looking at the Fibonacci sequence I happened to notice that after each pair of terms >1 the product of terms divided by the sum of terms produced an integral quotient every other time. Example 240/20=12, integral.

			This table illustrates the alternating nature of the first three integral quotients: 1 1 2 3 -- 6/7=.85+ 5 8 -- 240/20=12 Integral 13 21 -- 65520/54=1213.33+ 34 55 -- 122522400/143=856800 Integral 89 144 -- 1570247078400/376=4176189038.29+ 233 377 -- 137932073613734400/986=139890541190400 Integral etc.

Carlos Rivera, Problem 55. Fibonacci dividing terms, The Prime Puzzles & Problems Connection.

Cf. A159951, A001519, A001906, A003481, A033890, A003266, A052449.

seq(mul(fibonacci(k), k = 1..4*n+2), n = 1..10); # Peter Bala, Nov 04 2021

10 'Fibo 20 'R=SUM:S=PRODUCT 30 'T integral every other pair 40 A=1:S=1:print A;:S=S*1 50 B=1:print B;:S=S*B 60 C=A+B:print C;:R=R+C:S=S*C 70 D=B+C:print D;:R=R+D:R=R+2:print R:S=S*D:print S 80 T=S/R:if T=int(S/R) then print T:stop 90 A=C:B=D:R=R-2:goto 60

a(1)=240 because in the Fibonacci sequence up to 8 : 1 1 2 3 5 8, the product is 240 1*1*2*3*5*8. The sum is 1+1+2+3+5+8=20 (see A003481). The integral quotient is 12. From then on, every other pair produces an integral quotient.

a(n) = Product_{k = 1..4*n+2} Fibonacci(k) = A003266(4*n+2) = A052449(4*n+2) - 1. - Peter Bala, Nov 04 2021

A142880 a(n) = 7*a(n-3) - a(n-6).

Original entry on oeis.org

0, 1, 2, 3, 8, 13, 21, 55, 89, 144, 377, 610, 987, 2584, 4181, 6765, 17711, 28657, 46368, 121393, 196418, 317811, 832040, 1346269, 2178309, 5702887, 9227465, 14930352, 39088169, 63245986, 102334155, 267914296, 433494437, 701408733, 1836311903

Offset: 0

Roger L. Bagula and Gary W. Adamson, Sep 28 2008

Index entries for linear recurrences with constant coefficients, signature (0,0,7,0,0,-1).

Cf. A119016, A002965, A002530, A048788.

a[0] = 0; a[1] = 1;
a[n_] := a[n] = If[Mod[n, 3] == 1, 2*a[n - 1] + a[n - 2], If[Mod[n, 3] == 0, a[n - 1] + a[n - 2], 2*a[n - 1] - a[n - 2]]];
Table[a[n], {n, 0, 50}]
LinearRecurrence[{0,0,7,0,0,-1},{0,1,2,3,8,13},40] (* Harvey P. Dale, Jul 17 2021 *)

G.f.: -x*(1+x)*(x^3 - 2*x^2 - x - 1) / ( 1 - 7*x^3 + x^6 ).
a(3n) = A033888(n).
a(3n+1) = A033890(n).
a(3n+2)= A033891(n).
a(n) = 2*a(n-1) + a(n-2) if n == 1 (mod 3).
a(n) = a(n-1) + a(n-2) if n == 0 (mod 3).
a(n) = 2*a(n-1) - a(n-2) if n == 2 (mod 3).

A351222 Decimal expansion of Sum_{k>=0} (-1)^k/Fibonacci(4*k+2).

Original entry on oeis.org

8, 9, 0, 8, 6, 7, 0, 2, 1, 9, 7, 2, 1, 1, 8, 2, 6, 0, 0, 4, 8, 8, 5, 1, 5, 2, 9, 2, 4, 1, 5, 6, 8, 0, 2, 0, 4, 3, 0, 5, 1, 2, 8, 4, 4, 1, 5, 8, 2, 0, 4, 3, 4, 5, 6, 6, 2, 0, 8, 0, 2, 7, 1, 9, 7, 5, 5, 2, 1, 5, 5, 6, 7, 2, 2, 1, 9, 9, 7, 5, 7, 6, 0, 5, 3, 1, 7, 8, 8, 3, 4, 9, 1, 6, 6, 2, 6, 7, 9, 5, 8, 5, 9, 2, 6

Offset: 0

Amiram Eldar, Feb 05 2022

			0.89086702197211826004885152924156802043051284415820...

Wray G. Brady, Problem B-319, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 13, No. 4 (1975), p. 373; Rerun, Solution to Problem B-319, ibid., Vol. 14, No. 5 (1976), p. 472.
L. Carlitz, Problem B-111, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; Another Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 470-471.

Cf. A000032, A000045, A033890, A246453.

RealDigits[NSum[(-1)^n/Fibonacci[4*n + 2], {n, 0, Infinity}, WorkingPrecision -> 1200], 10, 100][[1]]

sumpos(k=0, (-1)^k/fibonacci(4*k+2)) \\ Michel Marcus, Feb 05 2022

Equals sqrt(5) * Sum_{k>=0} 1/Lucas(4*k+2) (Carlitz, 1967).

cp's OEIS Frontend

A360467 a(n) = Fibonacci(4*n+2) + 3*Fibonacci(2*n+1)^2.

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A081004 a(n) = Fibonacci(4n+2) + 1, or Fibonacci(2n+2)*Lucas(2n).

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A159950 Dividends where Fibonacci products/sums yield integral quotients.

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A142880 a(n) = 7*a(n-3) - a(n-6).

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A351222 Decimal expansion of Sum_{k>=0} (-1)^k/Fibonacci(4*k+2).

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A360467 a(n) = Fibonacci(4n+2) + 3Fibonacci(2*n+1)^2.