A245306 -id:A245306 - cp's OEIS frontend

A301906 T(n,k)=Number of nXk 0..1 arrays with every element equal to 0, 2 or 3 horizontally or antidiagonally adjacent elements, with upper left element zero.

1, 1, 2, 1, 1, 4, 1, 2, 1, 8, 1, 2, 3, 1, 16, 1, 3, 2, 7, 1, 32, 1, 6, 2, 5, 16, 1, 64, 1, 10, 7, 8, 10, 43, 1, 128, 1, 21, 12, 40, 12, 26, 117, 1, 256, 1, 42, 27, 96, 92, 64, 65, 330, 1, 512, 1, 86, 62, 316, 320, 532, 196, 170, 935, 1, 1024, 1, 179, 160, 1078, 1588, 1934, 1999, 864, 442

Offset: 1

R. H. Hardin, Mar 28 2018

Table starts
...1.1...1...1....1.....1......1.......1........1..........1...........1
...2.1...2...2....3.....6.....10......21.......42.........86.........179
...4.1...3...2....2.....7.....12......27.......62........160.........387
...8.1...7...5....8....40.....96.....316.....1078.......3831.......13331
..16.1..16..10...12....92....320....1588.....7234......34477......171770
..32.1..43..26...64...532...1934...14860....86638.....568382.....4029337
..64.1.117..65..196..1999...8781..104732...770376....7398173....75183520
.128.1.330.170..864.10150..49709..952467..8263384..113474387..1620307756
.256.1.935.442.3236.46226.253844.7931939.81388360.1622304450.32687652025

			Some solutions for n=5 k=4
..0..1..1..0. .0..1..1..0. .0..1..1..1. .0..1..1..0. .0..1..0..1
..1..1..1..1. .1..1..1..0. .1..1..1..0. .1..1..1..0. .0..1..0..1
..0..1..1..0. .0..1..1..0. .0..1..1..0. .0..1..1..0. .0..1..0..1
..1..1..1..1. .0..1..1..1. .0..1..1..1. .0..1..1..1. .0..1..0..1
..0..1..1..0. .0..1..1..0. .0..1..1..0. .1..1..1..0. .0..1..0..1

R. H. Hardin, Table of n, a(n) for n = 1..287

Column 1 is A000079(n-1).
Column 4 is A245306(n-1).
Row 2 is A240513(n-3).

Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = a(n-1)
k=3: a(n) = 3*a(n-1) +a(n-2) -4*a(n-3)
k=4: a(n) = 3*a(n-1) -3*a(n-3) +a(n-4)
k=5: a(n) = 4*a(n-1) +5*a(n-2) -20*a(n-3) -4*a(n-4) +16*a(n-5) for n>6
k=6: [order 20] for n>21
k=7: [order 30] for n>33
Empirical for row n:
n=1: a(n) = a(n-1)
n=2: a(n) = 2*a(n-1) +a(n-2) -a(n-3) -2*a(n-4) +a(n-5)
n=3: [order 30] for n>31

A338762 Greatest prime Fibonacci divisor of F(n)^2 + 1 where F(n) is the n-th Fibonacci number, or 0 if no such prime factor exists.

Original entry on oeis.org

2, 2, 5, 5, 13, 13, 5, 13, 89, 89, 233, 233, 89, 233, 1597, 1597, 5, 1597, 1597, 13, 28657, 28657, 13, 28657, 28657, 5, 514229, 514229, 2, 514229, 514229, 89, 13, 89, 89, 13, 233, 233, 0, 233, 433494437, 433494437, 5, 433494437, 2971215073, 2971215073, 13, 2971215073

Offset: 1

Michel Lagneau, Nov 07 2020

nonn

a(n) = 0 for n = 39, 60, 69, 72, ... .

a(5385) has 1126 decimal digits. - Chai Wah Wu, Nov 19 2020

			a(6) = 13 because F(6)^2 + 1 = 8^2 + 1 = 65 = 5*13 and 13 is the greatest prime Fibonacci divisor.

Chai Wah Wu, Table of n, a(n) for n = 1..5384 (n = 1..1000 from Alois P. Heinz)

Cf. A000045, A005478, A245306.

a:= proc(n) local F, m, t; F, m, t:=
      [1, 2], 0, (<<0|1>, <1|1>>^n)[2, 1]^2+1;
      while F[2]<=t do if isprime(F[2]) and irem(t, F[2])=0
        then m:=F[2] fi; F:= [F[2], F[1]+F[2]]
      od; m
    end:
seq(a(n), n=1..50);  # Alois P. Heinz, Nov 07 2020

a[n_] := Module[{F, m, t}, F = {1, 2}; m = 0; t = MatrixPower[{{0, 1}, {1, 1}}, n][[2, 1]]^2 + 1; While[F[[2]] <= t, If[PrimeQ[F[[2]]] && Mod[t, F[[2]]] == 0, m = F[[2]]]; F = {F[[2]], F[[1]] + F[[2]]}]; m];
Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Feb 09 2025, after Alois P. Heinz *)

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
a(n) = my(f=factor(fibonacci(n)^2+1)[,1]~, v=select(x->isfib(x), f)); if (#v, vecmax(v), 0); \\ Michel Marcus, Nov 07 2020

A245236 Numbers n such that the Fibonacci number F(n) satisfies F(n)^2+1 = f1*f2 where f1, f2 are prime Fibonacci numbers.

Original entry on oeis.org

4, 5, 6, 9, 12, 15, 45, 432, 570

Offset: 1

Michel Lagneau, Jul 14 2014

Or index i of any Fibonacci number F(i) such that F(i-1) and F(i+1) are primes if i is even or F(i-2) and F(i+2) are primes if i is odd where F(i) is the i-th Fibonacci number.
In the general case, F(i+1)*F(i-1) = F(i)^2 + 1 if i even or F(i+2)*F(i-2) = F(i)^2 + 1 if i odd (Cassini’s identity).
The corresponding Fibonacci numbers are 3, 5, 8, 34, 144, 610, 1134903170,...
If a(10) exists, it is greater than 30000. - Robert Israel, Jul 14 2014

			4 is a term because F(4)^2+1 = F(3)*F(5)=> 3^2+1 = 2*5;
5 is a term because F(5)^2+1 = F(3)*F(7)=> 5^2+1 = 2*13;
6 is a term because F(6)^2+1 = F(5)*F(7)=> 8^2+1 = 5*13;
9 is a term because F(9)^2+1 = F(7)*F(11)=> 34^2+1 = 13*89;
12 is a term because F(12)^2+1 = F(11)*F(13)=> 144^2+1 = 89*233;
15 is a term because F(13)*F(17)=> 610^2+1 = 233* 1597.

Cf. A000045, A005478, A245306.

with(combinat,fibonacci):with(numtheory):nn:=1000:for n from 1 to nn do:if (type(fibonacci(n+1),prime) and type(fibonacci(n-1),prime) and irem(n,2)=0) or (type(fibonacci(n+2),prime) and type(fibonacci(n-2),prime) and irem(n,2)=1) then print(n):else fi:od:
# Alternative:
filter:= proc(n) uses combinat;
    if n::even then isprime(n-1) and isprime(n+1) and isprime(fibonacci(n-1)) and isprime(fibonacci(n+1))
  else isprime(n-2) and isprime(n+2) and isprime(fibonacci(n-2)) and isprime(fibonacci(n+2))
fi end proc:
select(filter, [$1..10^4]); # Robert Israel, Jul 14 2014

A338794 Indices k of Fibonacci numbers F(k) such that F(k)^2 + 1 has no Fibonacci prime factor.

Original entry on oeis.org

39, 60, 69, 72, 99, 102, 105, 108, 111, 150, 165, 180, 192, 195, 198, 225, 228, 231, 240, 270, 279, 282, 309, 312, 315, 348, 351, 381, 399, 420, 441, 459, 462, 465, 489, 501, 522, 588, 591, 600, 615, 618, 642, 645, 660, 675, 702, 741, 759, 771, 810, 822, 825, 828

Offset: 1

Michel Lagneau, Nov 09 2020

nonn

Numbers k such that A338762(k) = 0.

			39 is in the sequence because F(39)^2 + 1 = 63245986^2 + 1 = 73*149*2221*2789*59369 with no Fibonacci prime factors.
38 is not in the sequence because F(38)^2 + 1 = 39088169^2 + 1 =  2*73*149*233*2221*135721. The numbers and 2, 233 are Fibonacci prime factors.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000045, A005478, A168063, A245306, A338762.

a:= proc(n) local F, m, t; F, m, t:=
      [1, 2], 0, (<<0|1>, <1|1>>^n)[2, 1]^2+1;
      while F[2]<=t do if isprime(F[2]) and irem(t, F[2])=0
        then m:=F[2] fi; F:= [F[2], F[1]+F[2]]
      od; m
    end:
for n from 1 to 100 do :
if a(n)=0 then printf(`%d, `,n):else fi:
od: # program from Alois P. Heinz, adapted for the sequence. See A338762.

A338762[n_] := Module[{F, m, t}, F = {1, 2}; m = 0; t = MatrixPower[{{0, 1}, {1, 1}}, n][[2, 1]]^2 + 1; While[F[[2]] <= t, If[PrimeQ[F[[2]]] && Mod[t, F[[2]]] == 0, m = F[[2]]]; F = {F[[2]], F[[1]] + F[[2]]}]; m];
Reap[For[k = 1, k <= 1000, k++, If[A338762[k] == 0, Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Mar 16 2025, after Alois P. Heinz *)

isok(n) = {my(i=0, f=0, x=fibonacci(n)^2+1, m=0); while(f < x, i++; f = fibonacci(i); if (ispseudoprime(f) && (x%f) == 0, return (0));); return(1);} \\ Michel Marcus, Nov 13 2020

A281618 Fibonacci numbers F such that all the prime factors of F^2 + 1 are also Fibonacci numbers.

Original entry on oeis.org

1, 2, 3, 5, 8, 34, 144, 610, 1134903170

Offset: 1

Michel Lagneau, Jan 25 2017

The corresponding indices of F are 1 or 2, 3, 4, 5, 6, 9, 12, 15, 45, ... and A245236 is in this sequence.

			a(9)^2+1 = Fibonacci(45)^2+1 = 1134903170^2+1 = 1288005205276048901 = 433494437 * 2971215073 = Fibonacci(43)*Fibonacci(47).

Cf. A000045, A245306, A245236.

with(numtheory):with(combinat,fibonacci):nn:=100:
for n from 1 to nn do:
  f:=fibonacci(n)^2+1:x:=factorset(f):n0:=nops(x):it:=0:
    for m from 1 to n0 do:
    c:=x[m]:
    x1:=sqrt(5*c^2-4):x2:=sqrt(5*c^2+4):
    if x1=floor(x1) or x2=floor(x2)
     then
     it:=it+1:
     else
    fi:
od:
if it=n0 then print(fibonacci(n)):else fi:od:

With[{s = Rest@ Fibonacci@ Range@ 120}, Select[s, Times @@ Boole@ Map[MemberQ[s, #] &, FactorInteger[#^2 + 1][[All, 1]]] > 0 &]] (* Michael De Vlieger, Jan 27 2017 *)

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
isokf(n) = {my(f = factor(fibonacci(n)^2+1)); for (k=1, #f~, if (!isfib(f[k,1]), return(0));); return(1);}
for (n=2, 50, if (isokf(n), print1(fibonacci(n), ", "))) \\ Michel Marcus, Jan 28 2017

A335568 a(n) is the number m such that F(m) is the greatest prime Fibonacci divisor of F(n)^2 + 1 where F(n) is the n-th Fibonacci number, or 0 if no such prime factor exists.

Original entry on oeis.org

3, 3, 5, 5, 7, 7, 5, 7, 11, 11, 13, 13, 11, 13, 17, 17, 5, 17, 17, 7, 23, 23, 7, 23, 23, 5, 29, 29, 3, 29, 29, 11, 7, 11, 11, 7, 13, 13, 0, 13, 43, 43, 5, 43, 47, 47, 7, 47, 47, 17, 7, 17, 17, 11, 3, 11, 11, 3, 3, 0, 7, 7, 13, 13, 7, 13, 23, 23, 0, 23, 23, 0, 5

Offset: 1

Chai Wah Wu, Nov 20 2020

nonn

Fibonacci index of the terms in A338762.

All terms are prime or 0. - Alois P. Heinz, Nov 21 2020

			a(10) = 11 because F(10)^2 + 1 = 55^2 + 1 = 3026 = 2*17*89 and 89 = F(11) is the greatest prime Fibonacci divisor of 3026.

Chai Wah Wu, Table of n, a(n) for n = 1..20000

Cf. A000040, A000045, A005478, A245306, A338762, A338794 (indices of the 0's).

a:= proc(n) local i, F, m, t; F, m, t:=
      [1, 2], 0, (<<0|1>, <1|1>>^n)[2, 1]^2+1;
      for i from 3 while F[2]<=t do if isprime(F[2]) and
        irem(t, F[2])=0 then m:=i fi; F:= [F[2], F[1]+F[2]]
      od; m
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 21 2020

a[n_] := Module[{i, F = {1, 2}, m = 0, t}, t = MatrixPower[{{0, 1}, {1, 1}}, n][[2, 1]]^2 + 1; For[i = 3, F[[2]] <= t, i++, If[PrimeQ[F[[2]]] && Mod[t, F[[2]]] == 0, m = i]; F = {F[[2]], F[[1]] + F[[2]]}]; m];
Array[a, 100] (* Jean-François Alcover, Dec 01 2020, after Alois P. Heinz *)

A000045(a(n)) = A338762(n).

A262342 Area of Lewis Carroll's paradoxical F(2n+1) X F(2n+3) rectangle.

Original entry on oeis.org

10, 65, 442, 3026, 20737, 142130, 974170, 6677057, 45765226, 313679522, 2149991425, 14736260450, 101003831722, 692290561601, 4745030099482, 32522920134770, 222915410843905, 1527884955772562, 10472279279564026, 71778070001175617, 491974210728665290, 3372041405099481410

Offset: 1

Jonathan Sondow, Oct 16 2015

Warren Weaver (1938): "In a familiar geometrical paradox a square of area 8 X 8 = 64 square units is cut into four parts which may be refitted to form a rectangle of apparent area 5 X 13 = 65 square units.... Lewis Carroll generalized this paradox...."
Carroll cuts a F(2n+2) X F(2n+2) square into four parts, where F(n) is the n-th Fibonacci number. Two parts are right triangles with legs F(2n) and F(2n+2); two are right trapezoids three of whose sides are F(2n), F(2n+1), and F(2n+1). (Thus n > 0.) The paradox (or dissection fallacy) depends on Cassini's identity F(2n+1) * F(2n+3) = F(2n+2)^2 + 1.
For an extension of the paradox to a F(2n+1) X F(2n+1) square using Cassini's identity F(2n) * F(2n+2) = F(2n+1)^2 - 1, see Dudeney (1970), Gardner (1956), Horadam (1962), Knott (2014), Kumar (1964), and Sillke (2004). Sillke also has many additional references and links.

			F(3) * F(5) = 2 * 5 = 10 = 3^2 + 1 = F(4)^2 + 1, so a(1) = 10.
G.f. = 10*x + 65*x^2 + 442*x^3 + 3026*x^4 + 20737*x^5 + 142130*x^6 + 974170*x^7 + ...

W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th edition, Dover, 1987, p. 85.
Henry E. Dudeney, 536 Puzzles and Curious Problems, Scribner, reprinted 1970, Problems 352-353 and their answers.
Martin Gardner, Mathematics, Magic and Mystery, Dover, 1956, Chap. 8.
Edward Wakeling, Rediscovered Lewis Carroll Puzzles, Dover, 1995, p. 12.
David Wells, The Penguin Book of Curious and Interesting Puzzles, Penguin, 1997, Puzzle 143.

Colin Barker, Table of n, a(n) for n = 1..1000
Margherita Barile, Dissection Fallacy, MathWorld.
A. F. Horadam, Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 35, No. 1 (1962), pp. 1-11.
Ron Knott, Fibonacci jigsaw puzzles, 2014.
Santosh Kumar, On Fibonacci Sequences and a Geometrical Paradox, Math. Mag., Vol. 37, No. 4 (1964), pp. 221-223.
Oskar Schlömilch, Ein geometrisches Paradoxon, Zeitschrift für Mathematik und Physik, Vol. 13 (1868), p. 162.
Torsten Sillke, Jigsaw paradox, 2004.
David Singmaster, Vanishing area puzzles, Recreational Math. Mag., Vol. 1 (2014), pp. 10-21.
Warren Weaver, Lewis Carroll and a Geometrical Paradox, American Math. Monthly, Vol. 45, No. 4 (1938), pp. 234-236.
Wikipedia, Cassini and Catalan identities.
Wikipedia, Fibonacci number.
Wikipedia, Missing square puzzle; also see External Links.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A001519, A059929, A064170, A070550, A166516, A176055, A190018, A236165, A245306.

[Fibonacci(2*n+1)*Fibonacci(2*n+3) : n in [1..30]]; // Wesley Ivan Hurt, Oct 16 2015

with(combinat): A262342:=n->fibonacci(2*n+1)*fibonacci(2*n+3): seq(A262342(n), n=1..30); # Wesley Ivan Hurt, Oct 16 2015

Table[Fibonacci[2 n + 1] Fibonacci[2 n + 3], {n, 22}]
LinearRecurrence[{8,-8,1},{10,65,442},30] (* Harvey P. Dale, Aug 06 2024 *)

Vec(-x*(2*x^2-15*x+10)/((x-1)*(x^2-7*x+1)) + O(x^30)) \\ Colin Barker, Oct 17 2015

a(n) = fibonacci(2*n+1) * fibonacci(2*n+3) \\ Altug Alkan, Oct 17 2015

a(n) = Fibonacci(2n+1) * Fibonacci(2n+3) = Fibonacci(2n+2)^2 + 1 for n > 0.
From Colin Barker, Oct 17 2015: (Start)
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: -x*(2*x^2-15*x+10) / ((x-1)*(x^2-7*x+1)).
(End)
a(3*k-2) mod 2 = 0; a(3*k-1) mod 2 = 1; a(3*k) mod 2 = 0, k > 0. - Altug Alkan, Oct 17 2015
a(n) = A059929(2*n+1) = A070550(4*n+1) = A166516(2*n+2) = A190018(8*n) = A236165(4*n+4) = A245306(2*n+2). - Bruno Berselli, Oct 17 2015
a(n) = A064170(n+3). - Alois P. Heinz, Oct 17 2015
E.g.f.: (1/5)*((1/phi*r)*exp(b*x) + (phi^4/r)*exp(a*x) + 3*exp(x) - 10), where r = 2*phi+1, 2*a=7+3*sqrt(5), 2*b=7-3*sqrt(5). - G. C. Greubel, Oct 17 2015
a(n) = (A337928(n+1) - A337929(n+1)) / 2. - Flávio V. Fernandes, Feb 06 2021
Sum_{n>=1} 1/a(n) = sqrt(5)/2 - 1 = A176055 - 2. - Amiram Eldar, Mar 04 2021

A339082 a(n) is the number m such that F(prime(m)) is the greatest prime Fibonacci divisor of F(n)^2 + 1 where F(n) is the n-th Fibonacci number, or 0 if no such prime factor exists.

Original entry on oeis.org

2, 2, 3, 3, 4, 4, 3, 4, 5, 5, 6, 6, 5, 6, 7, 7, 3, 7, 7, 4, 9, 9, 4, 9, 9, 3, 10, 10, 2, 10, 10, 5, 4, 5, 5, 4, 6, 6, 0, 6, 14, 14, 3, 14, 15, 15, 4, 15, 15, 7, 4, 7, 7, 5, 2, 5, 5, 2, 2, 0, 4, 4, 6, 6, 4, 6, 9, 9, 0, 9, 9, 0, 3, 3, 5, 5, 3, 5, 5, 2, 23, 23, 7

Offset: 1

Chai Wah Wu, Nov 24 2020

nonn

If a(n) > 0, then prime(a(n)) = A335568(n).

			a(15) = 7 because F(15)^2 + 1 = 610^2 + 1 = 372101 = 233*1597, 1597 = F(17) is the greatest prime Fibonacci divisor of 372101 and 17 is the 7th prime.

Chai Wah Wu, Table of n, a(n) for n = 1..30000

Cf. A000040, A000045, A005478, A245306, A335568, A338762, A338794 (indices of the 0's).

a:= proc(n) local i, F, m, t; F, m, t:=
      [1, 2], 0, (<<0|1>, <1|1>>^n)[2, 1]^2+1;
      for i from 3 while F[2]<=t do if isprime(F[2]) and
        irem(t, F[2])=0 then m:=i fi; F:= [F[2], F[1]+F[2]]
      od; numtheory[pi](m)
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 25 2020

a[n_] := Module[{i, F = {1, 2}, m = 0, t}, t = MatrixPower[{{0, 1}, {1, 1}}, n][[2, 1]]^2 + 1; For[i = 3, F[[2]] <= t, i++, If[PrimeQ[F[[2]]] && Mod[t, F[[2]]] == 0, m = i]; F = {F[[2]], F[[1]] + F[[2]]}]; PrimePi[m]];
Array[a, 100] (* Jean-François Alcover, Dec 01 2020, after Alois P. Heinz *)

If A335568(n) = 0, then a(n) = 0, otherwise a(n) = A000720(A335568(n)).

A340542 Number of Fibonacci divisors of Fibonacci(n)^2 + 1.

Original entry on oeis.org

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

Offset: 0

Michel Lagneau, Jan 12 2021

nonn

A Fibonacci divisor of a number k is a Fibonacci number that divides k.
It is interesting to compare this sequence with A339669.
We observe that a(2n) = A339669(2n) if n = 5*k + 2 or n = 5*k + 3, with k >= 0, because Lucas(2n)^2 = 5*Fibonacci(2n)^2 + 4 (see A005248: all nonnegative integer solutions of the Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)= A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004).
So, Lucas(2n)^2 + 1 = 5*(Fibonacci(2n)^2 + 1). Lucas(2n)^2 + 1 and Fibonacci(2n)^2 + 1 have the same Fibonacci divisors for n = 5*k + 2 or n = 5*k + 3. For the other values of n = 5*k, 5*k + 1 or 5*k + 4, 5 is a Fibonacci divisor of Lucas(2n)^2 + 1 but not of Fibonacci(2n)^2 + 1. So for these last three values of n, a(2n) = A339669(2n) - 1 (except for m = 1 and 2, 5*F(m) is never a Fibonacci number).

			a(13) = 5 because the 5 Fibonacci divisors of Fibonacci(13)^2 + 1 = 233^2 + 1 are 1, 2, 5, 89 and 610.
a(16) = 5 because the 5 Fibonacci divisors of Fibonacci(16)^2 + 1 = 987^2 + 1 are 1, 2, 5, 610, and 1597.
Remark: the 5 Fibonacci divisors of Lucas(16)^2 + 1 = 2207^2 + 1 are 1, 2, 5, 610, and 1597, the index 16 = 2*8 with 8 of the form 5*k + 3.

Michel Marcus, Table of n, a(n) for n = 0..200

Cf. A000032, A000045, A005248, A010056, A339461, A339669.

Cf. A005086, A245306.

with(combinat,fibonacci):nn:=100:F:={}:
for k from 0 to nn do:
  F:=F union {fibonacci(k)}:
od:
   for m from 0 to 90 do:
    f:=fibonacci(m)^2+1:d:=numtheory[divisors](f):
    lst:= F intersect d: n1:=nops(lst):printf(`%d, `,n1):
   od:

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ A010056
a(n) = sumdiv(fibonacci(n)^2+1, d, isfib(d)); \\ Michel Marcus, Jan 12 2021

a(n) = A005086(A245306(n)). - Michel Marcus, Aug 10 2022

A360107 Numbers k such that sigma_2(Fibonacci(k)^2 + 1) == 0 (mod Fibonacci(k)).

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 11, 13, 15, 19, 21, 25, 27, 31, 41, 45, 49, 81, 85, 129, 133, 135, 139, 357, 361, 429, 431, 433, 435, 447, 451, 507, 511, 567, 569, 571, 573

Offset: 1

Michel Lagneau, Jan 26 2023

			7 is in the sequence because the divisors of Fibonacci(7)^2 + 1 = 13^2 + 1 = 170 are {1, 2, 5, 10, 17, 34, 85, 170}, and 1^2 + 2^2 + 5^2 + 10^2 + 17^2 + 34^2 + 85^2 + 170^2 = 37700 = 13*2900 == 0 (mod 13).

Cf. A000045, A001157, A067719, A245236, A245306, A338762, A360105.

Select[Range[140],Divisible[DivisorSigma[2,Fibonacci[#]^2+1],Fibonacci[#]]&]

isok(k) = my(f=fibonacci(k)); sigma(f^2 + 1, 2) % f == 0; \\ Michel Marcus, Jan 26 2023

a(24)-a(37) from Daniel Suteu, Jan 27 2023

cp's OEIS Frontend

A301906 T(n,k)=Number of nXk 0..1 arrays with every element equal to 0, 2 or 3 horizontally or antidiagonally adjacent elements, with upper left element zero.

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A338762 Greatest prime Fibonacci divisor of F(n)^2 + 1 where F(n) is the n-th Fibonacci number, or 0 if no such prime factor exists.

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A245236 Numbers n such that the Fibonacci number F(n) satisfies F(n)^2+1 = f1*f2 where f1, f2 are prime Fibonacci numbers.

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A338794 Indices k of Fibonacci numbers F(k) such that F(k)^2 + 1 has no Fibonacci prime factor.

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A281618 Fibonacci numbers F such that all the prime factors of F^2 + 1 are also Fibonacci numbers.

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A335568 a(n) is the number m such that F(m) is the greatest prime Fibonacci divisor of F(n)^2 + 1 where F(n) is the n-th Fibonacci number, or 0 if no such prime factor exists.

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A262342 Area of Lewis Carroll's paradoxical F(2n+1) X F(2n+3) rectangle.

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