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

Offset

1,1

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

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

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_

T. Sillke, Powers of 2 and 5 Puzzle

Example

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

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 *)

Crossrefs

Cf. A088935, A010467.

Sequence in context: A350874 A053980 A215299 * A093749 A197474 A332695

Adjacent sequences: A088992 A088993 A088994 * A088996 A088997 A088998

Keyword

base,nonn

Author

Lekraj Beedassy, Dec 01 2003

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

Status

approved