cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A190949 Odd Fibonacci numbers with odd index.

Original entry on oeis.org

1, 5, 13, 89, 233, 1597, 4181, 28657, 75025, 514229, 1346269, 9227465, 24157817, 165580141, 433494437, 2971215073, 7778742049, 53316291173, 139583862445, 956722026041, 2504730781961, 17167680177565, 44945570212853, 308061521170129, 806515533049393
Offset: 1

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Author

T. D. Noe, May 24 2011

Keywords

Comments

All prime Fibonacci numbers (A005478) except 2 and 3 are in this sequence. All terms equal 1 (mod 4). The indices of these Fibonacci numbers are 6k-1 or 6k+1.
This sequence can be thought of as two interlocking sequences, each of the form a(n) = 18a(n - 1) - a(n - 2).
Proof that all terms equal 1 (mod 4): From the Lucas 1876 identity Fib(2n+1) = Fib(n)^2 + Fib(n+1)^2 (see Weisstein, formula 60, or page 79 of Koshy), we see that odd-indexed Fibonacci numbers are the sum of two squares. Because a square is 0 or 1 (mod 4), the sum of two squares is 0, 1, or 2 (mod 4). All these terms are odd numbers. Hence, the only possibility is that they are 1 (mod 4). This can also be proved from the recursion formula.

References

  • Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001.

Crossrefs

Intersection of A001519 and A014437.
Cf. A000045 (Fibonacci numbers).

Programs

  • Mathematica
    LinearRecurrence[{0,18,0,-1}, {1,5,13,89}, 50]
  • PARI
    a(n)=fibonacci(n\2*6+if(n%2,1,-1)) \\ Charles R Greathouse IV, Jun 08 2011
    
  • PARI
    Vec(x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)) + O(x^30)) \\ Colin Barker, Mar 27 2016

Formula

a(n) = 18*a(n-2) - a(n-4), with a(1)=1, a(2)=5, a(3)=13, and a(4)=89.
G.f.: x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)). - Colin Barker, Jun 19 2012
a(n) = (-(2-sqrt(5))^n + (-2-sqrt(5))^n*(-2+sqrt(5)) + 2*(-2+sqrt(5))^n + sqrt(5)*(-2+sqrt(5))^n + (2+sqrt(5))^n)/(2*sqrt(5)) for n>0. - Colin Barker, Mar 27 2016
a(n) = A000045(A007310(n)). - Amiram Eldar, Jul 25 2024

A011369 a(n+1) = a(n) - F(n) if > 0, otherwise a(n) + F(n), where F() are Fibonacci numbers; a(0) = 0.

Original entry on oeis.org

0, 0, 1, 2, 4, 1, 6, 14, 1, 22, 56, 1, 90, 234, 1, 378, 988, 1, 1598, 4182, 1, 6766, 17712, 1, 28658, 75026, 1, 121394, 317812, 1, 514230, 1346270, 1, 2178310, 5702888, 1, 9227466, 24157818, 1, 39088170, 102334156, 1, 165580142, 433494438, 1, 701408734, 1836311904
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    Module[{n = 0, f}, NestList[If[(f = Fibonacci[n++]) < #, # - f, # + f] &, 0, 49]] (* Paolo Xausa, Nov 08 2024 *)
    Flatten[Join[{0, 0}, Table[{1, Fibonacci[{k, k+2}] + 1}, {k, 2, 49, 3}]]] (* Paolo Xausa, Nov 08 2024 *)
    LinearRecurrence[{1, 0, 4, -4, 0, 1, -1}, {0, 0, 1, 2, 4, 1, 6, 14, 1}, 50] (* Paolo Xausa, Nov 08 2024 *)
  • PARI
    a(n) = if (n==0, 0, my(d=a(n-1)-fibonacci(n-1)); if (d>0, d, d+2*fibonacci(n-1))) \\ Michel Marcus, Dec 29 2018
    
  • PARI
    a(n) = if (n<=1, 0, my(m=(n % 3)); if (m==0, fibonacci(n-1)+1, if (m==1, fibonacci(n)+1, 1))); \\ Michel Marcus, Dec 29 2018

Formula

a(n) = 0, if n <= 1; F(n-1)+1, if n == 0 (mod 3); F(n)+1, if n == 1 (mod 3); 1, if n == 2 (mod 3). - David W. Wilson; corrected by Michel Marcus, Dec 29 2018
For n>=1, a(n) = F(0)<+>F(1)<+>...<+>F(n-1), where operation <+> is defined in comment in A245618. - Vladimir Shevelev, Nov 05 2014
Empirical g.f.: -x^2*(2*x^6 - x^4 + 7*x^3 - 2*x^2 - x - 1) / ((x-1)*(x^2 + x - 1)*(x^4 - x^3 + 2*x^2 + x + 1)). - Colin Barker, Nov 06 2014

Extensions

Name edited by Michel Marcus, Dec 29 2018

A014728 Squares of odd Fibonacci numbers.

Original entry on oeis.org

1, 1, 9, 25, 169, 441, 3025, 7921, 54289, 142129, 974169, 2550409, 17480761, 45765225, 313679521, 821223649, 5628750625, 14736260449, 101003831721, 264431464441, 1812440220361, 4745030099481, 32522920134769
Offset: 0

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Author

Keywords

Crossrefs

Cf. A014437.

Programs

Formula

G.f.: (-x^5-x^4+8x^3-8x^2+x+1)/[(1+x^2)(1+4x-x^2)(1-4x-x^2)].
a(n) = A014437(n)^2. - Sean A. Irvine, Nov 18 2018

Extensions

More terms from James Sellers

A181156 Odd Fibonacci numbers F which have a proper Fibonacci divisor G such that F/G is a Lucas number or a product of Lucas numbers.

Original entry on oeis.org

3, 21, 55, 377, 987, 6765, 17711, 121393, 317811, 2178309, 5702887, 39088169, 102334155, 701408733, 1836311903, 12586269025, 32951280099, 225851433717, 591286729879, 4052739537881, 10610209857723, 72723460248141, 190392490709135, 1304969544928657, 3416454622906707
Offset: 1

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Author

Vladimir Shevelev, Oct 07 2010

Keywords

Comments

A conjectural statement that, for odd prime p, the ratio F_{p^2}/F_{p} is never a Lucas number or a product of some Lucas numbers, yields that
a) an odd Fibonacci number F is in the sequence iff for its maximal proper Fibonacci divisor G, we have: ind G does not equal sqrt(ind F) and F/G does not have a proper Fibonacci divisor > 3;
b) an odd Fibonacci number F is in the sequence iff its index has one of the forms: 6k+2 or 6k+4 (see A047235).

Examples

			F = 3 has the proper Fibonacci divisor G=1, and F/G = 3 is a Lucas number.
F = 317811 has the proper Fibonacci divisors 3, 13, and 377, and F/377 = 843 is a Lucas number.
		

Crossrefs

A193666 F(2) odd Fib. numbers, F(3) even Fib. numbers, F(4) odd Fib. numbers..., F(n) = A000045(n), a(n) < a(n+1).

Original entry on oeis.org

1, 2, 8, 13, 21, 55, 144, 610, 2584, 10946, 46368, 75025, 121393, 317811, 514229, 1346269, 2178309, 5702887, 9227465, 14930352, 63245986, 267914296, 1134903170, 4807526976, 20365011074, 86267571272, 365435296162, 1548008755920, 6557470319842, 27777890035288
Offset: 1

Views

Author

Ctibor O. Zizka, Aug 08 2011

Keywords

Examples

			Taking (F(2)=) 1 odd Fib. number gives a(1)=1.
Then taking F(3)=2 even Fib. numbers starting with 2 gives a(2)=2, a(3)=8.
Then taking F(4)=3 odd Fib. numbers starting with 13 gives a(4)=13, a(5)=21, a(6)=55.
Then taking F(5)=5 even Fib. numbers starting with 144 gives a(7)=144, a(8)=610, a(9)=2584, a(10)=10946, a(11)=46368, etc...
		

Crossrefs

A384219 Areas of triangles whose three vertices are consecutive ordered pairs of consecutive odd Fibonacci numbers such that an ordered pair’s y-value is the next ordered pair’s x-value.

Original entry on oeis.org

2, 6, 24, 104, 442, 1870, 7920, 33552, 142130, 602070, 2550408, 10803704, 45765226, 193864606, 821223648, 3478759200, 14736260450, 62423800998, 264431464440, 1120149658760, 4745030099482, 20100270056686, 85146110326224, 360684711361584, 1527884955772562
Offset: 1

Views

Author

Angela L. Brobson, May 22 2025

Keywords

Comments

As described above, the vertices of each triangle are made up exclusively of odd integers in both coordinates. The area of a parallelogram spanned by (x_1, y_1), (x_2, y_2), and (x_3, y_3) is given by abs(x_1*(y_2 - y_3) + x_2*(y_3 - y_1) + x_3*(y_1 - y_2)). This value is even since each y_i - y_j is a difference of odd integers and is thereby even. As the area of a triangle formed from three given vertices is half the area of the parallelogram spanned by those vertices, it follows that the area of each triangle formed according to the description is an integer.

Examples

			The first term is created by finding the area of the triangle formed by the ordered pairs (1,1), (1,3), and (3,5), which is 2.
The second term is created by finding the area of the triangle formed by the ordered pairs (1,3), (3,5), and (5,13), which is 6.
The third term is created by finding the area of the triangle formed by the ordered pairs (3,5), (5,13), and (13,21), which is 24.
		

Crossrefs

Programs

  • Maple
    a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|4|0|4>>^n.<<0, 2, 6, 24>>)[1,1]:
    seq(a(n), n=1..25);  # Alois P. Heinz, May 22 2025

Formula

G.f.: 2*x*(x-1)/((x^2+1)*(x^2+4*x-1)). - Alois P. Heinz, May 22 2025
E.g.f.: (exp(2*x)*(cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x)) - cos(x) + 3*sin(x))/5. - Stefano Spezia, May 25 2025
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