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 1-6 of 6 results.

A055391 Palindromic Lucas numbers (in the order of appearance).

Original entry on oeis.org

2, 1, 3, 4, 7, 11, 167761
Offset: 1

Views

Author

M. Harminc (harminc(AT)duro.science.upjs.sk), Jul 06 2000

Keywords

Comments

Perhaps 167761 is the last term?
No other terms through L(10000) - James Sellers, Jul 07 2000
No further terms through L(200000). - Lior Manor, Oct 18 2007

Examples

			11 is the sixth Lucas number and it is also palindromic in base 10.
		

Crossrefs

Programs

  • Mathematica
    Select[LucasL[Range[0,200]],PalindromeQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 26 2018 *)

A372729 a(n) is the smallest k such that the first n digits of Fibonacci(k) are the reverse of its last n digits.

Original entry on oeis.org

0, 10, 317, 1235, 28898, 120742, 1411753, 201095722, 306312948, 12306316582, 32679761048, 806327047899, 9600042921304, 172192972068022
Offset: 1

Views

Author

Jon E. Schoenfield, May 11 2024

Keywords

Comments

(Inspired by D. S. McNeil's comment at A045504 that typically a Fibonacci number can be ruled out as a possible palindrome by checking only a few digits at the start and end of the number.)

Examples

			a(1) = 0 (as Fibonacci(0) = 0 is the smallest Fibonacci number).
a(2) = 10 (Fibonacci(10) = 55 is the only 2-digit Fibonacci number that is palindromic, and almost certainly the only multidigit palindromic Fibonacci number; see A045504).
a(3) = 317 because Fibonacci(317) (a 66-digit number) is the smallest Fibonacci number whose first 3 digits (793) are the reverse of its last 3 digits (397).
The table below lists the first 8 terms and the corresponding Fibonacci numbers (abbreviated, for n > 2):
.
  n       a(n)    Fibonacci(a(n))
  -  ---------  -------------------
  1          0           0
  2         10          55
  3        317       793...397
  4       1235      5626...6265
  5      28898     94480...08449
  6     120742    172255...552271
  7    1411753   3789665...5669873
  8  201095722  11367389...98376311
		

Crossrefs

Programs

  • C
    /* See links. */

Extensions

a(9)-a(14) from Kevin Ryde, Aug 22 2024

A294688 Palindromes in Narayana's cows sequence (A000930).

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 88, 595
Offset: 1

Views

Author

Chai Wah Wu, Nov 06 2017

Keywords

Comments

595 is probably the last term.

Crossrefs

Programs

  • Maple
    a := proc(n) if n < 4 then n else
    simplify(hypergeom([-n/3, (1-n)/3, (-1-n)/3], [-n/2, (-1-n)/2], -27/4));
    convert(%, base, 10); ListTools:-Reverse(%);
    if %% = % then add(%[k]*10^(k-1), k=1..nops(%)) else NULL fi fi end:
    seq(a(n), n=1..200); # Peter Luschny, Nov 07 2017

A372754 a(n) is the least base in which the Fibonacci number A000045(n) is a palindrome.

Original entry on oeis.org

2, 2, 3, 2, 2, 3, 3, 2, 4, 4, 8, 11, 3, 11, 9, 11, 15, 29, 25, 29, 29, 29, 40, 17, 121, 76, 76, 69, 147, 39, 148, 199, 199, 199, 311, 361, 10876, 428, 521, 521, 1026, 1364, 1025, 1364, 1364, 1364, 2100, 2018, 4973, 3571, 3571, 3571, 5802, 6461, 11343, 9349, 9349, 9349, 31952, 24476, 15885, 24476
Offset: 1

Views

Author

Robert Israel, May 12 2024

Keywords

Comments

With F = A000045 the Fibonacci numbers and L = A000032 the Lucas numbers, for j odd we have F(3*j+k) = F(j+k)*L(j)^2 + F(k)*L(j) + F(j+k), thus this is a palindrome mod L(j) if F(k) >= 0 and 0 <= F(j+k) < L(j). Therefore a(6*n), a(6*n+2), a(6*n+3) and a(6*n+4) all <= L(2*n+1).

Examples

			A000045(6) = 8.  In base 2 this is 1000, not a palindrome, but in base 3 it is 22, a palindrome.  Thus a(6) = 3.
		

Crossrefs

Programs

  • Maple
    f:= proc(x) local b,L;
      for b from 2 do
        L:= convert(x,base,b);
        if L = ListTools:-Reverse(L) then return b fi
      od
    end proc:
    map(f, [seq(combinat:-fibonacci(n), n=1..70)]);
  • Mathematica
    A372754[n_] := Block[{b = 1}, While[!PalindromeQ[IntegerDigits[#, ++b]]] & [Fibonacci[n]]; b]; Array[A372754, 70] (* Paolo Xausa, May 18 2024 *)
  • Python
    from itertools import count
    from sympy import fibonacci
    from sympy.ntheory.factor_ import digits
    def A372754(n): return next(b for b in count(2) if (s := digits(fibonacci(n),b)[1:])[:(t:=len(s)+1>>1)]==s[:-t-1:-1]) # Chai Wah Wu, May 13 2024

Formula

a(n) = A016026(A000045(n)).

A233524 Numbers n such that the binary expansion of Fibonacci(n) is a palindrome (leading zero digits are not permitted).

Original entry on oeis.org

0, 1, 2, 4, 5, 8
Offset: 1

Views

Author

L. Edson Jeffery, Dec 11 2013

Keywords

Comments

The next term of the sequence, if it exists, is greater than 10^5.
No more terms < 10^7. - Lars Blomberg, Jun 19 2014

Crossrefs

Cf. A000045 (Fibonacci), A004685 (binary Fibonacci), A007088 (binary nonnegative integers).
Cf. A045504.

Programs

  • Mathematica
    t = {}; Do[b = IntegerDigits[Fibonacci[n], 2]; If[b == Reverse[b], AppendTo[t, n]], {n, 0, 1000}]; t (* T. D. Noe, Dec 14 2013 *)
    Select[Range[0,10],PalindromeQ[IntegerDigits[Fibonacci[#],2]]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 04 2019 *)

A294720 Palindromes in tribonacci sequence with initial values 0,1,0 (A001590).

Original entry on oeis.org

0, 1, 2, 3, 6, 11, 7256527
Offset: 1

Views

Author

Chai Wah Wu, Nov 07 2017

Keywords

Comments

7256527 is probably the last term.

Crossrefs

Programs

  • Mathematica
    a = c = 0; b = 1; lst = {0}; While[a < 10^7, d = a + b + c; If[ Reverse[idn = IntegerDigits[d]] == idn, AppendTo[lst, d]]; a = b; b = c; c = d]; lst (* Robert G. Wilson v, Nov 09 2017 *)
Showing 1-6 of 6 results.