A349646 Nonnegative integers which produce a record maximum MD5 hash.

0, 3, 44, 65, 83, 373, 575, 1126, 12673, 25670, 30268, 30525, 40691, 48240964, 63327632, 298506737, 369490840, 1113434519, 1647703600, 4958115803, 64657664035, 86155378906, 184280298746, 400812644253, 411723964986, 714853066875, 1627993432495, 2607864795784

Offset

1,2

Comments

a(1) = 0; a(n) is the smallest k such that MD5(k) > MD5(a(n-1)), where integer parameters to MD5 are encoded as base-10 ASCII strings.

If a(1) were defined as 1 instead of 0, the sequence would begin 1, 2, 3, 44, ... and then continue in the same way.

If we assume that MD5 behaves like a random function from N to {0, ..., 2^128-1}, the expected length of this sequence is the harmonic number H(2^128) ~= 89.3.

a(31) > 10^15.

Links

Ben Whitmore, Table of n, a(n) for n = 1..30

Wikipedia, MD5

Example

a(5) = 83 because MD5("83") = fe9fc289c3ff0af142b6d3bead98a923_16 = 338453431832254946862081270079334951203, which is larger than all previous values MD5("0"), ..., MD5("82").

Mathematica

recordsBy[l_, P_] :=
Module[{max = -Infinity, x, i, recs = {}},
For[i = 1, i <= Length[l], i++,
  x = P[l[[i]]];
  If[x > max,
   max = x;
   AppendTo[recs, l[[i]]];
  ]
];
recs
];
recordsBy[Range[1000], Hash[ToString[#], "MD5"] &]

Prog

(Python)
from hashlib import md5
def afind(limit):
    record = ""
    for k in range(limit+1):
        hash = md5(str(k).encode('utf-8')).hexdigest()
        if hash > record:
            print(k, end=", ")
            record = hash
afind(10**5) # Michael S. Branicky, Nov 24 2021

Crossrefs

Record minima: A349647.

Sequence in context: A326970 A076361 A130408 * A296279 A133073 A055539

Adjacent sequences: A349643 A349644 A349645 * A349647 A349648 A349649

Keyword

nonn,base,fini,hard

Author

Ben Whitmore, Nov 23 2021

Status

approved