A020344 -id:A020344 - cp's OEIS frontend

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

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

Offset: 1

Jeff Burch

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

			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.

Ron Knott, Fibonacci Numbers and the Golden Section
Eric Weisstein's World of Mathematics, Fibonacci numbers

Cf. A000045, A020344, A052000, A000350, A050816.

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)

/* 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

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

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

A020345 Smallest Fibonacci number beginning with n.

Original entry on oeis.org

0, 1, 2, 3, 4181, 5, 610, 75025, 8, 987, 10946, 1134903170, 121393, 13, 144, 1597, 165580141, 17711, 1836311903, 196418, 20365011074, 21, 225851433717, 233, 24157817, 2584, 267914296, 27777890035288, 28657, 2971215073, 308061521170129, 317811

Offset: 0

David W. Wilson

The graph of the indices A020344 is much more interesting. - T. D. Noe, Apr 02 2014

a(1382) is the first term with > 1000 digits (1004). - Michael S. Branicky, Jul 08 2022

			a(4) = 4181 is a Fibonacci number starting with 4.

Michael S. Branicky, Table of n, a(n) for n = 0..1389 (terms 1..400 from T. D. Noe)
Ron Knott, Every number starts some Fibonacci Number, The Mathematical Magic of the Fibonacci Numbers.

Cf. A000045, A020344, A023183, A023184.

nn = 31; 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++]]; t = Join[{0}, t] (* T. D. Noe, Apr 02 2014 *)

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]] = f
        f, g, k = g, f+g, k+1
    return [int(ans[str(i)]) for i in range(nn+1)]
print(aupton(31)) # Michael S. Branicky, Jul 08 2022

a(n) = A000045(A020344(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

David W. Wilson

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

Robert Israel, Table of n, a(n) for n = 0..9999

Cf. A000045, A023184, A020344, A020345.

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

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 *)

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

David W. Wilson

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.

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

Cf. A000045, A023185, A020344, A020345.

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);

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

Michel Marcus, Jul 02 2022

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..4000 from Michel Marcus)
Ron Knott, Every number starts some Fibonacci Number, The Mathematical Magic of the Fibonacci Numbers.

Cf. A000032, A355439,

Cf. A020344, A020345.

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;

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

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

Michel Marcus, Jul 02 2022

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

Michael S. Branicky, Table of n, a(n) for n = 1..1381
Ron Knott, Every number starts some Fibonacci Number, The Mathematical Magic of the Fibonacci Numbers.

Cf. A000032, A355438.

Cf. A020344, A020345.

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);

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

A124100 Sum_(x^iy^jz^k) with i + j + k = m and (x, y, z) = the primitive Pythagorean triple (8, 15, 17).

Original entry on oeis.org

1, 40, 1089, 25160, 531521, 10625640, 204744769, 3844391560, 70827391041, 1286290883240, 23101397290049, 411249127989960, 7269184506192961, 127745926316548840, 2234231991096868929, 38920247688751940360

Offset: 0

Giorgio Balzarotti and Paolo P. Lava, Nov 26 2006

nonn

			a(2) = 1089 because x^2 + y^2 + z^2 + x*y + x*z + y*z = 8^2 + 15^2 + 17^2 + 8*15 + 8*17 + 15*17 = 1089 and x^2 + y^2 = z^2.

G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 196.

Harvey P. Dale, Table of n, a(n) for n = 0..811
Index entries for linear recurrences with constant coefficients, signature (40, -511, 2040).

Cf. A019682, A020000, A020340-A020342, A020344-A020346, A021664, A021684, A021844, A025942, A077515.

seq(sum(8^(m-n)*sum(15^p*17^(n-p),p=0..n),n=0..m),m=0..N);

LinearRecurrence[{40,-511,2040},{1,40,1089},30] (* Harvey P. Dale, May 25 2025 *)

a(m) = (x^(m+2)*(z-y) + y^(m+2)*(x-z) + z^(m+2)*(y-x))/((x-y)*(y-z)*(z-x)).
From Chai Wah Wu, Sep 24 2016: (Start)
a(n) = 40*a(n-1) - 511*a(n-2) + 2040*a(n-3) for n > 2.
G.f.: 1/((1 - 8*x)*(1 - 15*x)*(1 - 17*x)). (End)
a(n) = 2^(3*n+6)/63 - 15^(n+2)/14 + 17^(n+2)/18. - Vaclav Kotesovec, Sep 25 2016

A374026 a(n) = the smallest k such that Fibonacci(k) begins and ends with n, where Fibonacci(k) > n, or -1 if there are none.

Original entry on oeis.org

22, 114, 124, 72, 10, 39, 116, 207, 169, 5715, 2428, 5634, 2366, 189, 2620, 4668, 3137, 2673, 5812, 12090, 721, 11583, 20086, 3798, 1975, 999, 10636, 846, 2071, 9105, 1162, 2076, 8287, 11091, 2770, 2928, 12943, 8493, 172, 2220, 5359, 4737, 28126, 11838, 10460, 7479, 10996

Offset: 1

Gonzalo Martínez, Jun 26 2024

Since the smallest number beginning and ending with n is the same n, the condition that Fibonacci(k) > n is imposed. Partial overlap of the start and end is allowed, but not full overlap.

			As Fibonacci(22) = 17711 is the smallest Fibonacci number greater than 1 that begins and ends with 1, a(1) = 22.
As Fibonacci(10) = 55 is the smallest Fibonacci number greater than 5 that begins and ends with 5, a(5) = 10.

Cf. A000045, A272623, A023183, A020344.

isok(s,t) = my(ss=strsplit(s, t)); (#ss >= 3) && (ss[1]=="") && (ss[#ss]=="");
a(n) = my(k=7); while(!isok(Str(fibonacci(k)), Str(n)), k++); k; \\ Michel Marcus, Jun 26 2024

# uses A023183() in A023183
from itertools import count
def a(n):
    if A023183(n) == -1:
        return -1
    f, g, s = 1, 2, str(n)
    pow10 = 10**len(s)
    for k in count(3):
        if g%pow10 == n:
            sfib = str(g)
            if g > n and sfib.startswith(s):
                return k
        f, g = g, f+g
print([a(n) for n in range(1, 51)]) # Michael S. Branicky, Jul 03 2024

a(n) >= max{A023183(n), A020344(n)} except that a(n) = -1 when A023183(n) = -1. - Michael S. Branicky, Jun 27 2024

More terms from Michel Marcus, Jun 26 2024

cp's OEIS Frontend

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

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A020345 Smallest Fibonacci number beginning with n.

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

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A023184 Least Fibonacci number ending with n.

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

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A355439 Smallest Lucas number beginning with n.

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A124100 Sum_(x^i*y^j*z^k) with i + j + k = m and (x, y, z) = the primitive Pythagorean triple (8, 15, 17).

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A374026 a(n) = the smallest k such that Fibonacci(k) begins and ends with n, where Fibonacci(k) > n, or -1 if there are none.

A124100 Sum_(x^iy^jz^k) with i + j + k = m and (x, y, z) = the primitive Pythagorean triple (8, 15, 17).