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 1-5 of 5 results.

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

Views

Author

David W. Wilson

Keywords

Comments

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

Links

Crossrefs

Cf. A000045, A020345, A023183, A023184, A038546.

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).

A023183 a(n) = least k such that Fibonacci(k) ends with n, or -1 if there are none.

Original entry on oeis.org

0, 1, 3, 4, 9, 5, 21, 14, 6, 11, 15, 22, 216, 7, 111, 130, 168, 37, 27, 112, 60, 8, 117, 64, 198, 25, 99, 136, 204, 29, 105, 88, 174, 13, 9, 70, 222, 43, 93, 172, 30, 41, 63, 124, 12, 55, 21, 154, 186, 49, 75, 148, 36, 67, 129, 10, 162, 23, 87, 118, 180, 61, 57, 166, 72, 20
Offset: 0

Views

Author

David W. Wilson

Keywords

Comments

It appears that if n is greater than 99 and congruent to 4 or 6 (mod 8) then there is no Fibonacci number ending in that n. - Jason Earls, Jun 19 2004
This is because there is no Fibonacci number == 4 or 6 (mod 8). - Robert Israel, Sep 11 2020

Links

Crossrefs

Cf. A000045, A023184, A020344, A020345.

Programs

  • Maple
    V:= Array(0..999,-1):
V[0]:= 0: u:= 1: v:= 0:
for n from 1 to 1500 do
  t:= v;
  v:= u+v mod 1000;
  u:= t;
  if V[v] = -1 then V[v]:= n fi;
  if V[v mod 100] = -1 then V[v mod 100] := n fi;
  if V[v mod 10] = -1 then V[v mod 10]:= n fi;
od:
seq(V[i],i=0..999); # Robert Israel, Sep 11 2020
  • Mathematica
    d[n_]:=IntegerDigits[n]; Table[j=0; While[Length[d[Fibonacci[j]]]<(le=Length[y=d[n]]), j++]; i=j; While[Take[d[Fibonacci[i]],-le]!=y,i++]; i,{n,0,65}] (* Jayanta Basu, May 18 2013 *)
  • Python
    from itertools import count
def A023183(n):
    if n < 2: return n
    if n > 99 and n%8 in {4, 6}: return -1
    k, f, g, s = 3, 1, 2, str(n)
    pow10, seen = 10**len(s), set()
    while (f, g) not in seen:
        seen.add((f, g))
        if g%pow10 == n:
            return k
        f, g, k = g, (f+g)%pow10, k+1
    return -1
print([A023183(n) for n in range(66)]) # Michael S. Branicky, Jun 27 2024

A023184 Least Fibonacci number ending with n.

Original entry on oeis.org

0, 1, 2, 3, 34, 5, 10946, 377, 8, 89, 610, 17711, 619220451666590135228675387863297874269396512, 13, 70492524767089125814114, 659034621587630041982498215, 57602132235424755886206198685365216, 24157817, 196418, 114059301025943970552219, 1548008755920
Offset: 0

Views

Author

David W. Wilson

Keywords

Comments

The Fibonacci index of the 12th to 25th terms respectively are 216, 7, 111, 130, 168, 37, 27, 112, 90, 8, 183, 286, 252 and 25.

Examples

			a(11) = 17711 is the smallest Fibonacci number ending in 11.

Crossrefs

Cf. A000045, A023185, A020344, A020345.

Programs

  • Maple
    with(combinat):for n from 1 to 40 do e := 1; g := ceil(log(n+1)/log(10)-0.00001): while((fibonacci(e) mod 10^g)n) do e := e+1:end do: q[n] := fibonacci(e):end do:seq(q[i],i=1..40);
  • Mathematica
    d[n_]:=IntegerDigits[n]; Table[j=0; While[Length[d[Fibonacci[j]]]<(le=Length[y=d[n]]), j++]; i=j; While[Take[d[x=Fibonacci[i]],-le]!=y,i++]; x,{n,0,20}] (* Jayanta Basu, May 18 2013 *)

A355438 Lucas(a(n)) is least Lucas number beginning with n.

Original entry on oeis.org

1, 0, 2, 3, 13, 23, 4, 14, 19, 24, 5, 10, 15, 39, 20, 25, 49, 6, 11, 35, 59, 16, 64, 21, 45, 69, 26, 50, 7, 31, 55, 12, 36, 60, 17, 151, 41, 65, 22, 156, 46, 70, 27, 94, 51, 252, 8, 32, 166, 56, 190, 13, 281, 37, 305, 61, 18, 85, 42, 109, 310, 66, 267, 23, 224, 47, 181, 71, 138, 339
Offset: 1

Views

Author

Michel Marcus, Jul 02 2022

Keywords

Links

Crossrefs

Cf. A000032, A355439,
Cf. A020344, A020345.

Programs

  • PARI
    L(n) = real((2 + quadgen(5)) * quadgen(5)^n); \\ A000032
isok(k, dn) = my(dk=digits(L(k))); if (#dk >= #dn, Vec(dk, #dn) == dn);
a(n) = my(k=0, dn=digits(n)); while (!isok(k, dn), k++); k;
  • Python
    def aupton(nn):
    ans, f, g, k = dict(), 2, 1, 0
    while len(ans) < nn:
        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(1, nn+1)]
print(aupton(70)) # Michael S. Branicky, Jul 08 2022

Formula

Trivially a(n) >= log_phi(n-1) for n > 1. Probably upper bounds are obtainable using the theory of linear forms in logarithms. - Charles R Greathouse IV, Jul 08 2022

A355439 Smallest Lucas number beginning with n.

Original entry on oeis.org

1, 2, 3, 4, 521, 64079, 7, 843, 9349, 103682, 11, 123, 1364, 141422324, 15127, 167761, 17393796001, 18, 199, 20633239, 2139295485799, 2207, 23725150497407, 24476, 2537720636, 263115950957276, 271443, 28143753123, 29, 3010349, 312119004989, 322, 33385282, 3461452808002, 3571
Offset: 1

Views

Author

Michel Marcus, Jul 02 2022

Keywords

Comments

a(1382) is the first term with > 1000 digits (1156). - Michael S. Branicky, Jul 08 2022

Links

Crossrefs

Cf. A000032, A355438.
Cf. A020344, A020345.

Programs

  • PARI
    L(n) = real((2 + quadgen(5)) * quadgen(5)^n); \\ A000032
isok(k, dn) = my(dk=digits(L(k))); if (#dk >= #dn, Vec(dk, #dn) == dn);
a(n) = my(k=0, dn=digits(n)); while (!isok(k, dn), k++); L(k);
  • Python
    def aupton(nn):
    ans, f, g, k = dict(), 2, 1, 0
    while len(ans) < nn:
        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]] = f
        f, g, k = g, f+g, k+1
    return [int(ans[str(i)]) for i in range(1, nn+1)]
print(aupton(35)) # Michael S. Branicky, Jul 08 2022
Showing 1-5 of 5 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.