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 1-4 of 4 results.

A038546 Numbers n such that n-th Fibonacci number has initial digits n.

Original entry on oeis.org

0, 1, 5, 43, 48, 53, 3301, 48515, 348422, 406665, 1200207, 6698641, 190821326, 2292141445, 257125021372, 5843866639660, 45173327533483, 46312809996150, 59358981837795, 129408997210988, 1450344802530203, 5710154240910003
Offset: 1

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Keywords

Comments

The Mathematica coding used by Robert G. Wilson v implements Binet's Fibonacci number formula as suggested by David W. Wilson and incorporates Benoit Cloitre's use of logarithms to achieve a further increase in speed.
Fixed points of A020344. - Alois P. Heinz, Jul 08 2022

Examples

			a(3)=43 since 43rd Fibonacci number starts with 43 -> {43}3494437.
Fibonacci(53) is 53316291173, which begins with 53, so 53 is a term in the sequence.
		

Crossrefs

Programs

  • Mathematica
    a = N[ Log[10, Sqrt[5]/5], 24]; b = N [Log[10, GoldenRatio], 24]; Do[ If[ IntegerPart[10^FractionalPart[a + n*b]*10^Floor[ Log[10, n]]] == n, Print[n]], {n, 225000000}] (* Robert G. Wilson v, May 09 2005 *)
    (* confirmed with: *) fQ[n_] := (FromDigits[ Take[ IntegerDigits[ Fibonacci[n]], Floor[ Log[10, n] + 1]]] == n)
  • PARI
    /* To obtain terms > 5: */ a=(1+sqrt(5))/2; b=1/sqrt(5); for(n=1,3500, if(n==floor(b*(a^n)/10^( floor(log(b *(a^n))/log(10))-floor(log(n)/log(10)))),print1(n,","))) \\ Benoit Cloitre, Feb 27 2002

Formula

n>5 is in the sequence if a=(1+sqrt(5))/2 b=1/sqrt(5) and n==floor(b*(a^n)/10^(floor((log(b) +n*log(a))/log(10))-floor(log(n)/log(10))) ). - Benoit Cloitre, Feb 27 2002

Extensions

Term a(6) from Patrick De Geest, Oct 15 1999
a(7) from Benoit Cloitre, Feb 27 2002
a(8)-a(11) from Robert G. Wilson v, May 09 2005
a(12) from Robert G. Wilson v, May 11 2005
More terms from Robert Gerbicz, Aug 22 2006

A000350 Numbers m such that Fibonacci(m) ends with m.

Original entry on oeis.org

0, 1, 5, 25, 29, 41, 49, 61, 65, 85, 89, 101, 125, 145, 149, 245, 265, 365, 385, 485, 505, 601, 605, 625, 649, 701, 725, 745, 749, 845, 865, 965, 985, 1105, 1205, 1249, 1345, 1445, 1585, 1685, 1825, 1925, 2065, 2165, 2305, 2405, 2501, 2545, 2645, 2785, 2885
Offset: 1

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Comments

Conjecture: Other than 1 and 5, there is no m such that Fibonacci(m) in binary ends with m in binary. The conjecture holds up to m=50000. - Ralf Stephan, Aug 21 2006
The conjecture for binary numbers holds for m < 2^25. - T. D. Noe, May 14 2007
Conjecture is true. It is easy to prove (by induction on k) that if Fibonacci(m) ends with m in binary, then m == 0, 1, or 5 (mod 3*2^k) for any positive integer k, i.e., m must simply be equal to 0, 1, or 5. - Max Alekseyev, Jul 03 2009

Examples

			Fibonacci(25) = 75025 ends with 25.
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    import Data.List (isSuffixOf, elemIndices)
    import Data.Function (on)
    a000350 n = a000350_list !! (n-1)
    a000350_list = elemIndices True $
                   zipWith (isSuffixOf `on` show) [0..] a000045_list
    -- Reinhard Zumkeller, Apr 10 2012
    
  • Mathematica
    a=0;b=1;c=1;lst={}; Do[a=b;b=c;c=a+b;m=Floor[N[Log[10,n]]]+1; If[Mod[c,10^m]==n,AppendTo[lst,n]],{n,3,5000}]; Join[{0,1},lst] (* edited and changed by Harvey P. Dale, Sep 10 2011 *)
    fnQ[n_]:=Mod[Fibonacci[n],10^IntegerLength[n]]==n; Select[Range[ 0,2900],fnQ] (* Harvey P. Dale, Nov 03 2012 *)
  • PARI
    for(n=0,1e4,if(((Mod([1,1;1,0],10^#Str(n)))^n)[1,2]==n,print1(n", "))) \\ Charles R Greathouse IV, Apr 10 2012
    
  • Python
    from sympy import fibonacci
    [i for i in range(1000) if str(fibonacci(i))[-len(str(i)):]==str(i)] # Nicholas Stefan Georgescu, Feb 27 2023

A050816 Fibonacci(k) ending with digits of its index number k.

Original entry on oeis.org

0, 1, 5, 75025, 514229, 165580141, 7778742049, 2504730781961, 17167680177565, 259695496911122585, 1779979416004714189, 573147844013817084101, 59425114757512643212875125, 898923707008479989274290850145
Offset: 0

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Author

Patrick De Geest, Oct 15 1999

Keywords

Crossrefs

Programs

  • Mathematica
    fQ[n_]:=Module[{idni=IntegerDigits[First[n]],indf=IntegerDigits[Last[n]]}, idni== Take[indf,-Length[idni]]]; Transpose[Select[Table[ {n, Fibonacci[n]}, {n,0,150}],fQ]][[2]] (* Harvey P. Dale, Apr 21 2011 *)

A132365 Least number k such that the Lucas number A000032(k) contains n.

Original entry on oeis.org

1, 0, 2, 3, 13, 9, 4, 6, 7, 24, 5, 10, 15, 26, 20, 25, 49, 6, 11, 16, 13, 12, 10, 21, 45, 40, 20, 36, 7, 31, 50, 12, 35, 19, 17, 15, 41, 36, 22, 23, 39, 39, 14, 21, 41, 60, 8, 32, 19, 56, 20, 13, 45, 37, 51, 44, 17, 56, 42, 22, 25, 62, 35, 15, 71, 47, 25, 24, 43, 32, 17, 45, 49, 38
Offset: 1

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Author

Jonathan Vos Post, Nov 08 2007

Keywords

Comments

Values such that a(n)=n (fixed points) are 1, 62. I don't know if there are any other fixed points. The first time a(n)=a(n+1) occurs because L(39)=141422324 which includes both 41 and 42 (and later on in the sequence, because it contains 141 and 142). [Sean A. Irvine, Nov 30 2009]

Crossrefs

Programs

  • Python
    def A132365(n):
        a, b, m, s = 2, 1, 0, str(n)
        while True:
            if s in str(a):
                return m
            m += 1
            a, b = b, a+b # Chai Wah Wu, Jun 06 2017

Formula

a(n) = Min{k such that A000032(k) contains the decimal digit substring which represents the integer n}.

Extensions

Incorrect comment removed by Sean A. Irvine, Nov 30 2009
Corrected and extended by Sean A. Irvine, Nov 30 2009
Showing 1-4 of 4 results.