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.

Showing 1-3 of 3 results.

A348591 a(n) = L(n)*L(n+1) mod F(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.

Original entry on oeis.org

0, 1, 0, 3, 5, 3, 18, 3, 52, 3, 141, 3, 374, 3, 984, 3, 2581, 3, 6762, 3, 17708, 3, 46365, 3, 121390, 3, 317808, 3, 832037, 3, 2178306, 3, 5702884, 3, 14930349, 3, 39088166, 3, 102334152, 3, 267914293, 3, 701408730, 3, 1836311900, 3, 4807526973, 3, 12586269022, 3, 32951280096, 3, 86267571269, 3
Offset: 0

Views

Author

J. M. Bergot and Robert Israel, Jan 25 2022

Keywords

Examples

			a(5) = L(5)*L(6) mod F(7) = 11*18 mod 13 = 3.

Links

Crossrefs

Cf. A000032, A000045, A215602, A333599, A347861, A348592.

Programs

  • Maple
    F:= combinat:-fibonacci:
L:= n -> F(n-1)+F(n+1):
map(n -> L(n)*L(n+1) mod F(n+2), [$0..30]);
  • Mathematica
    a[n_] := Mod[LucasL[n] * LucasL[n + 1], Fibonacci[n + 2]]; Array[a, 50, 0] (* Amiram Eldar, Jan 26 2022 *)
  • Python
    from gmpy2 import fib, lucas2
def A348591(n): return (lambda x,y:int(x[0]*x[1] % y))(lucas2(n+1),fib(n+2)) # Chai Wah Wu, Jan 26 2022

Formula

a(n) = 3 if n >= 3 is odd.
a(n) = A000045(n+2)-3 if n >= 2 is even.
a(n) + a(n+1) - 3*a(n+2) - 3*a(n+3) + a(n+4) + a(n+5) = 0 for n >= 2.
G.f.: -x*(2*x^5-5*x^3-x-1)/((x+1)*(x^2+x-1)*(x^2-x-1)). - Alois P. Heinz, Jan 26 2022

A348592 a(n) = F(n)*F(n+1) mod L(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.

Original entry on oeis.org

0, 1, 2, 6, 15, 11, 10, 45, 99, 79, 65, 312, 675, 545, 442, 2142, 4623, 3739, 3026, 14685, 31683, 25631, 20737, 100656, 217155, 175681, 142130, 689910, 1488399, 1204139, 974170, 4728717, 10201635, 8253295, 6677057, 32411112, 69923043, 56568929, 45765226, 222149070, 479259663, 387729211, 313679522
Offset: 0

Views

Author

J. M. Bergot and Robert Israel, Jan 25 2022

Keywords

Examples

			a(5) = F(5)*F(6) mod L(7) = 5*8 mod 29 = 11.

Links

Crossrefs

Cf. A000032, A000045, A070352, A333599, A347861, A348591.

Programs

  • Maple
    F:= combinat:-fibonacci:
L:= n -> F(n-1)+F(n+1):
seq(F(n)*F(n+1) mod L(n+2), n=0..20);
  • Mathematica
    a[n_] := Mod[Fibonacci[n] * Fibonacci[n + 1], LucasL[n + 2]]; Array[a, 50, 0] (* Amiram Eldar, Jan 26 2022 *)

Formula

For n >= 1, a(n) = (A070352(n+2)*A000032(n+2) + 3*(-1)^n)/5.
a(n) + 2*a(n + 1) + 3*a(n + 2) + 5*a(n + 3) + a(n + 4) - a(n + 5) - a(n + 7) = 0 for n >= 1.
G.f.: -3 + 3/(5*(1+x)) + (3+x)/(2*(1-x-x^2)) + (9-4*x+6*x^2-x^3)/(10*(1+3*x^2+x^4)).

A357553 a(n) = A000045(n)*A000045(n+1) mod A000032(n).

Original entry on oeis.org

0, 0, 2, 2, 1, 7, 14, 12, 9, 46, 98, 80, 64, 313, 674, 546, 441, 2143, 4622, 3740, 3025, 14686, 31682, 25632, 20736, 100657, 217154, 175682, 142129, 689911, 1488398, 1204140, 974169, 4728718, 10201634, 8253296, 6677056, 32411113, 69923042, 56568930, 45765225, 222149071, 479259662, 387729212
Offset: 0

Views

Author

J. M. Bergot and Robert Israel, Oct 02 2022

Keywords

Comments

a(n) is the product of the n-th and (n+1)th Fibonacci numbers mod the n-th Lucas number.

Examples

			a(3) = A000045(3)*A000045(4) mod A000032(3) = 2*3 mod 4 = 2.

Links

Crossrefs

Cf. A000032, A000045, A333599, A347861.

Programs

  • Maple
    luc:= n -> combinat:-fibonacci(n+1) + combinat:-fibonacci(n-1):
f:= proc(n) local m;
  m:= n mod 4;
  if m = 0 then (luc(n)-2)/5
  elif m = 1 then (3*luc(n)+2)/5
  elif m = 2 then (4*luc(n)-2)/5
  else (2*luc(n)+2)/5
  fi
end proc:
f(1):= 0:
map(f, [$0..50]);
  • Mathematica
    a[n_] := Mod[Fibonacci[n] * Fibonacci[n + 1], LucasL[n]]; Array[a, 50, 0] (* Amiram Eldar, Oct 03 2022 *)

Formula

G.f. (2 + 2*x + 3*x^2 + 7*x^3 + 3*x^4 + 2*x^5 + x^6)*x^2/(1 + x^2 - x^3 - 5*x^4 - 3*x^5 - 2*x^6 - x^7).
For n == 0 (mod 4), a(n) = (A000032(n) - 2)/5.
For n == 1 (mod 4) and n > 1, a(n) = (3*A000032(n) + 2)/5.
For n == 2 (mod 4), a(n) = (4*A000032(n) - 2)/5.
For n == 3 (mod 4), a(n) = (2*A000032(n) + 2)/5.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.