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 1-9 of 9 results.

A067927 Duplicate of A038546.

Original entry on oeis.org

0, 1, 5, 43, 48, 53, 3301, 48515
Offset: 0

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Keywords

A020344 Fibonacci(a(n)) is the least Fibonacci number beginning with n.

Original entry on oeis.org

0, 1, 3, 4, 19, 5, 15, 25, 6, 16, 21, 45, 26, 7, 12, 17, 41, 22, 46, 27, 51, 8, 56, 13, 37, 18, 42, 66, 23, 47, 71, 28, 52, 119, 9, 33, 57, 14, 148, 38, 62, 19, 86, 43, 67, 134, 24, 225, 48, 72, 139, 29, 230, 53, 254, 10, 278, 34, 302, 58, 259, 15, 283, 39, 240, 63, 197, 20, 154, 288
Offset: 0

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Keywords

Comments

Fixed points of this sequence are in A038546. - Alois P. Heinz, Jul 08 2022

Crossrefs

Programs

  • Mathematica
    nn = 100; t = tn = Table[0, {nn}]; found = 0; n = 0; While[found < nn, n++; f = Fibonacci[n]; d = IntegerDigits[f]; i = 1; While[i <= Length[d], k = FromDigits[Take[d, i]]; If[k > nn, Break[]]; If[t[[k]] == 0, t[[k]] = f; tn[[k]] = n; found++]; i++]]; tn = Join[{0}, tn] (* T. D. Noe, Apr 02 2014 *)
  • Python
    def aupton(nn):
        ans, f, g, k = dict(), 0, 1, 0
        while len(ans) < nn+1:
            sf = str(f)
            for i in range(1, len(sf)+1):
                if int(sf[:i]) > nn:
                    break
                if sf[:i] not in ans:
                    ans[sf[:i]] = k
            f, g, k = g, f+g, k+1
        return [int(ans[str(i)]) for i in range(nn+1)]
    print(aupton(70)) # Michael S. Branicky, Jul 08 2022

Formula

A000045(a(n)) = A020345(n).

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

A383320 Lexicographically earliest sequence of distinct terms such that replacing each term k with Fibonacci(k) does not change the succession of digits.

Original entry on oeis.org

0, 1, 5, 43, 3, 4, 9, 44, 37, 2, 33, 470, 140, 8, 7, 332, 41, 57, 81, 71, 35, 24, 578, 74, 93, 86, 58, 6, 61, 14, 242, 47, 46, 936, 9310, 13, 87, 148, 48, 19, 30, 12, 55, 77, 36, 270, 246, 51, 68, 97, 194, 4350, 50, 27, 72, 31, 359, 90, 22, 40, 278, 505, 23
Offset: 1

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Author

Dominic McCarty, Apr 23 2025

Keywords

Examples

			Let b(n) = Fibonacci(a(n))
(a(n)): 0, 1, 5, 43, 3, 4, 9, 44, 37, 2, ...
(b(n)): 0, 1, 5, 433494437,           2, ...
		

Crossrefs

Cf. A383321 (Fibonacci(a(n))), A038546, A383318, A383322, A302656.

Programs

  • Python
    from sympy import fibonacci
    from itertools import count
    a, sa, sb = [0,1,5,43], "01543", "015433494437"
    for _ in range(30):
        a.append(next(n for k in count(1) if not (n := int(sb[len(sa):len(sa)+k])) in a and not (len(sb) > len(sa) + k and sb[len(sa) + k] == "0")))
        sa += str(a[-1]); sb += str(fibonacci(a[-1]))
    print(a)

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 *)

A052000 Fibonacci(k) starting with digits of its index number k.

Original entry on oeis.org

0, 1, 5, 433494437, 4807526976, 53316291173
Offset: 1

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Author

Patrick De Geest, Oct 15 1999

Keywords

Comments

Next term '3301...7801' has 690 digits.

Crossrefs

Formula

a(n) = A000045(A038546(n))

Extensions

Offset changed to 1 by Jon E. Schoenfield, Oct 17 2019

A383321 a(n) = Fibonacci(A383320(n)).

Original entry on oeis.org

0, 1, 5, 433494437, 2, 3, 34, 701408733, 24157817, 1, 3524578, 74938658661142424746936931013871484819301255773627024651689719443505027723135990224027850523592585, 81055900096023504197206408605, 21, 13
Offset: 1

Views

Author

Dominic McCarty, Apr 23 2025

Keywords

Crossrefs

Programs

  • Python
    from sympy import fibonacci
    from itertools import count
    a, b, sa, sb = [0,1,5,43], [0,1,5,433494437], "01543", "015433494437"
    for _ in range(10):
        a.append(next(n for k in count(1) if not (n := int(sb[len(sa):len(sa)+k])) in a and not (len(sb) > len(sa) + k and sb[len(sa) + k] == "0")))
        b.append(fibonacci(a[-1]))
        sa += str(a[-1]); sb += str(b[-1])
    print(b)

A121858 Smallest odd number having prime(n) divisors, where prime(n) is the n-th prime=A000040(n).

Original entry on oeis.org

3, 9, 81, 729, 59049, 531441, 43046721, 387420489, 31381059609, 22876792454961, 205891132094649, 150094635296999121, 12157665459056928801, 109418989131512359209, 8862938119652501095929, 6461081889226673298932241
Offset: 1

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Author

Lekraj Beedassy, Aug 30 2006

Keywords

Comments

a(n) is also the smallest number k with the property that the symmetric representation of sigma(k) has prime(n) subparts. - Omar E. Pol, Oct 08 2022

Crossrefs

Programs

  • Mathematica
    3^(Prime[Range[20]]-1) (* Harvey P. Dale, Mar 19 2013 *)

Formula

a(n) = 3^(prime(n)-1) = 3^A006093(n).
a(n) = A038547(A000040(n)). - Omar E. Pol, Oct 08 2022

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 1-9 of 9 results.