A088995 Least k > 0 such that the first n digits of 2^k and 5^k are identical.

5, 98, 1068, 1068, 127185, 2728361, 15917834, 73482154, 961700165, 961700165, 83322853582, 1404948108914, 7603192018819, 167022179253602, 3550275020220728, 5729166542536373, 106675272785875442, 331484151442699072, 2330288635428230258177, 2330288635428230258177

Offset: 1

Author

Lekraj Beedassy, Dec 01 2003

Keywords

• base
• nonn

Comments

The number of matching first digits of 2^n and 5^n increases with n and forms the sequence 3,1,6,2,2,7,7,6,6,... which approaches sqrt(10).

Numbers are half of the denominator of some convergent to log_10(2). - J. Mulder (jasper.mulder(AT)planet.nl), Feb 03 2010 [WARNING: This holds only for n < 6, it is wrong from a(6) = 2728361 on: The denominators are ..., 325147, 6107016, ... which would yield the larger solution 3053508 for n = 6. - M. F. Hasler, Mar 20 2025]

Xianwen Wang guesses that if the length of the continued fraction of m/k is h (where m is the difference between the numbers of digits of 2^k and 5^k), the first h-1 items of the continued fractions of m/k and log_10(2.5) agree. But this guess is not true for the similar sequence A359698. - Zhao Hui Du, Jun 06 2023

The terms grow like 10^n, oscillating by a few orders of magnitude around this value. I conjecture that log(a(n)/10^n) = O(log log n). - M. F. Hasler, Mar 22 2025

Examples

			a(2) = 98: 2^98 = 316912650057057350374175801344 and 5^98 = 315544362088404722164691426113114491869282574043609201908111572265625.
		

Links

• Zhao Hui Du, Table of n, a(n) for n = 1..1000
• Zhao Hui Du, Chinese page to introduce algorithm for a(n) to prove the guess of _Xianwen Wang_
• Robert Munafo, Sequence A88995: The 32 and 3125 Property
• T. Sillke, Powers of 2 and 5 Puzzle

Crossrefs

Cf. A088935, A010467, A359698, A073733.

Programs

Mathematica

L2 = N[ Log[ 10, 2 ], 50 ]; L5 = N[ Log[ 10, 5 ], 50 ]; k = 1; Do[ While[ Take[ RealDigits[ 10^FractionalPart[ L2*k ] ][[ 1 ] ], n ] != Take[ RealDigits[ 10^FractionalPart[ L5*k ] ][[ 1 ] ], n ], k++ ]; Print[ k ], {n, 1, 10} ]
L2 = N[ Log[ 10, 2 ], 50 ]; L5 = N[ Log[ 10, 5 ], 50 ]; k = 1; Do[ While[ Take[ RealDigits[ 10^FractionalPart[ L2*k ]][[ 1 ]], n ] != Take[ RealDigits[ 10^FractionalPart[ L5*k ]][[ 1 ]], n ], k++ ]; Print[ k ], {n, 1, 7} ]
f[n_, k_] := {Floor[ 10^(k - 1 + N[FractionalPart[n Log[5]/Log[10]], 20])], Floor[10^(k - 1 + N[FractionalPart[n Log[2]/Log[10]], 20])]} Flatten@Block[{$MaxExtraPrecision = \[Infinity]}, Block[{l = Denominator /@ Convergents[Log10[2], 1000]}, Array[k \[Function] l[[Flatten@Position[f[ #/2, k] & /@ l, {x_, x_}, {1}, 1]]]/2, 20]]] (* J. Mulder (jasper.mulder(AT)planet.nl), Feb 03 2010 *)
(* alternate program *)
n = 100; $MaxExtraPrecision = n; ans =
 ContinuedFraction[Log10[5/2], n]; data =
 Denominator /@
  Flatten[Table[
    FromContinuedFraction[Join[ans[[1 ;; p - 1]], {#}]] & /@
     Range[1, ans[[p]]], {p, 2, n}]]; sol =
 Select[Table[{k, a = N[FractionalPart[{k Log10[2], k Log10[5]}], n];
    10^a, b = RealDigits[10^a][[All, 1]];
    LengthWhile[Range[Length[b[[1]]]], b[[1, #]] == b[[2, #]] &],
    10^a . {-1, 1}, RealDigits[10^a . {-1, 1}][[-1]]}, {k, data}],
  Abs[#[[-2]]] < 1 &];
acc = Association[{}]; s = sol[[All, {1, 3}]]; For[i = 1,
 i < Length[s], i++,
 If[Lookup[acc, s[[i, 2]], 0] == 0,
  acc[s[[i, 2]]] = s[[i, 1]]]]; final =
 Rest[Sort[Normal[acc]]] /. Rule -> List;
bcc = Association[{}]; For[i = Max[Keys[acc]], i >= Min[Keys[acc]], i--,
  j = i; While[Lookup[acc, j, 0] == 0 && j < Max[Keys[acc]], j++];
 bcc[i] = acc[j]; j = i; While[bcc[j] >= bcc[j + 1], j++];
 bcc[i] = Min[bcc[i], bcc[j]]]; bb =
 Rest[Sort[Normal[Reverse[bcc]]]] /. Rule -> List (* Xianwen Wang, Jun 02 2023 *)

PARI

apply( {A088995(n) = localprec(max(n/.4,38)); my(L1=log(10), L2=log(2)/L1, L5=1-L2, c=contfrac(L5-L2), T=sqrtint(10^(2*n-1)), a=log(T)/L1%1, b=log(T+1)/L1%1, d(x)=a3, while(t(c-cv[i]), c-=cv[i]))+return(c))}, [1..15]) \\ - M. F. Hasler, Mar 22 2025

Python

# NOTE: Although sympy's frac() may give incorrect results in some cases, care has been taken to ensure there should be no issue here. - M. F. Hasler, Mar 26 2025
from sympy import S, sqrt, log, frac
def A088995(n):
    T = 10**n//sqrt(10); prec = (n+9)/.4; cf = 'continued_fraction'
    L2, L5, a, b = (frac(log(x,10).n(prec)) for x in (2, 5, T, T+1))
    L = []; test = lambda k: a < frac(k*L2) < b > frac(k*L5) > a
    for K in S(f"{cf}_convergents({cf}_iterator(log(5/2,10)))"):
        if test(k := K.q):    # = K.denominator but K.denominator() in old versions
            for step in L[::-1]:
              while test(k - step): k -= step
            return k
        if k > 99: L.append(k) # M. F. Hasler, Mar 22 2025, edited Mar 26 2025
print(first30 := [A088995(n) for n in range(1,30)]) # M. F. Hasler, Mar 18 2025, edited Mar 24 2025, Mar 28 2025

Extensions

Edited by Robert G. Wilson v, Dec 02 2003

More terms from J. Mulder (jasper.mulder(AT)planet.nl), Feb 03 2010

a(6) and a(7) corrected by Keith F. Lynch, May 25 2023

a(11), a(13)-a(15), a(17) corrected by Zhao Hui Du, Jun 07 2023