Eder Vanzei - cp's OEIS frontend

A355458 Numbers k that set a new record m where m is the largest left-truncatable prime up to the final k (stop on reaching the final k).

1, 7, 111, 3367, 7787, 8517, 9071, 54079, 54451, 138657, 262157, 759461, 857817, 4662317, 21754021, 25400729, 41171387, 50304331, 368119693, 799245603, 938577991

Offset: 1

Eder Vanzei, Jul 02 2022

If instead of comparing the values of m, we compare the number of digits concatenated to k, then there are only 3 known terms: 1, 7 and 50304331 with 19, 23 and 24 digits respectively.

			a(1) = 1 because 1 sets a record m = 89726156799336363541 and 89726156799336363541, 9726156799336363541, 726156799336363541, 26156799336363541, 6156799336363541, 156799336363541, 56799336363541, 6799336363541, 799336363541, 99336363541, 9336363541, 336363541, 36363541, 6363541, 363541, 63541, 3541, 541, 41 are all primes (the truncation stops when the final k is reached).
a(2) = 7 because for k = 2..6 no m exceeds 89726156799336363541, but for k = 7, m = 357686312646216567629137.

Cf. A024785.

from sympy import isprime
def findNewCandidates(candidates):
  newCandidates = []
  for candidate in candidates:
    for k in range(1,10):
      p = int(str(k) + str(candidate))
      if (isprime(p)):
        newCandidates.append(p)
  return newCandidates
record = 0
for k in range(1, 10**6):
  if (k % 2 == 0 or k % 5 == 0):
    continue
  toCheck = [k]
  while len(toCheck) > 0:
    lastToCheck = toCheck
    toCheck = findNewCandidates(toCheck)
  result = lastToCheck[-1]
  if (result > record):
    record = result
    print(str(k))

a(15)-a(18) from Michael S. Branicky, Jul 02 2022

a(19)-a(21) from Michael S. Branicky, Jul 04 2022

A351653 a(n) is the concatenation of the length of each run of digits in the decimal representation of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 2, 11, 11

Offset: 0

Eder Vanzei, Feb 16 2022

			a(100) = 12, because the 1st run (1) has length 1 and the 2nd run (00) has length 2.
a(13211) = 1112, because the 1st run (1) has length 1, the 2nd run (3) has length 1, the 3rd run (2) has length 1 and the 4th run (11) has length 2.

Cf. A001462, A318921, A318927 (binary version).

a[n_] := FromDigits[IntegerDigits[Length /@ Split[IntegerDigits[n]]] // Flatten]; Array[a, 100, 0] (* Amiram Eldar, Feb 16 2022 *)

a(n) = {my(result = "", count = 0, dig = digits(n), last = dig[1]); for(i = 1, length(dig), current = dig[i]; if (current == last, count++, result = concat(result, Str(count)); count = 1); last = current); result = concat(result, Str(count)); return(eval(result))};

apply( {A351653(n,d=digits(n+!n))=d=concat([0,[k|k<-[1..#n=d[^1]-d[^-1]], n[k]], #d]); fromdigits(d[^1]-d[^-1])}, [0..77]) \\ M. F. Hasler, Aug 21 2025

from itertools import groupby
def A351653(n): return int(''.join(str(len(list(g))) for k, g in groupby(str(n)))) # Chai Wah Wu, Mar 11 2022

A346674 Nonnegative numbers k such that MD5(k interpreted as a string) contains only decimal digits.

Original entry on oeis.org

1518375, 6637317, 16781059, 20274157, 20680348, 22080638, 24026537, 25394302, 26059015, 28926467, 40459791, 42057668, 42390227, 42943634, 43983547, 47788382, 49974597, 51201829, 59344568, 63236613, 63341298, 70557689, 74946923, 76642382, 77213479, 77641296

Offset: 1

Eder Vanzei, Jul 28 2021

See A234849 for more information about MD5.

This sequence is probably infinite, because MD5 returns a fixed-length output, but we don't know if outputs containing only decimal digits appear infinitely many times. Even if this is the case, it would not be enough, as we would not know if this holds when we consider the input as a number (interpreted as a string).

			a(1) = 1518375, because 1518375 is the smallest k >= 0 such that MD5(k) contains only decimal digits: MD5("1518375") = "93240121540327474319550261818423".

Michael S. Branicky, Table of n, a(n) for n = 1..10000
Wikipedia, MD5

Cf. A234849.

Monitor[Do[If[!StringContainsQ[Hash[ToString@k,"MD5","HexString"],Alphabet[]],Print@k],{k,10^9}],k] (* Giorgos Kalogeropoulos, Jul 28 2021 *)

from hashlib import md5
i=0
while True:
    m = md5()
    m.update(str(i).encode('utf-8'))
    result = m.hexdigest()
    if result.isdecimal():
        print(i)
    i+=1

A337844 a(1) = 1; a(n) = a(n-1) concatenated with the smallest number k>0 such that tau(a(n)) > tau(a(n-1)).

Original entry on oeis.org

1, 11, 111, 1112, 11124, 1112410, 111241020, 11124102080, 11124102080130, 11124102080130102, 11124102080130102180, 11124102080130102180480, 11124102080130102180480160, 11124102080130102180480160116

Offset: 1

Eder Vanzei, Sep 25 2020

This sequence is infinite because k can be of the form m*10^n for some m and n > 0.

			a(4) = 1112, because a(3) = 111 and tau(111) = 4, so a(n) must be equal to 111//k ("//" denotes concatenation) and tau(a(4)) > 4, therefore the smallest k that satisfies this condition is 2.

Cf. A000005.

f[1] = {1, 1}; f[n_] := f[n] = Module[{k = 1, t = f[n - 1][[1]], d = IntegerDigits[f[n - 1][[2]]]}, While[(t1 = DivisorSigma[0, (n1 = FromDigits @ Join[d, IntegerDigits[k]])]) <= t, k++]; {t1, n1}]; a[n_] := f[n][[2]]; Array[a, 14] (* Amiram Eldar, Sep 26 2020 *)

a(n) = if(n == 1, l = m = vector(100); return(l[1] = 1), for(k = 1,oo, j = eval(concat(Str(l[n-1]),k)); if((m[n] = numdiv(j)) > m[n-1], return(l[n] = j))))
for(k=1,10,print1(a(k),", "));

a(1) = 1; a(n) = a(n-1) concatenated with the smallest number k>0, such that A000005(a(n)) > A000005(a(n-1)).

A334620 a(n) is the smallest multiple of n formed by the concatenation 1,2,3,...,k for some k.

Original entry on oeis.org

1, 12, 12, 12, 12345, 12, 1234567891011, 123456, 12345678, 12345678910, 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106, 12

Offset: 1

Eder Vanzei, Sep 09 2020

			a(3) = 12, because 12 is the smallest multiple of 3 that appears in A007908.

Robert Israel, Table of n, a(n) for n = 1..144

Cf. A007908, A075002.

f:= proc(n) local x,i;
x:= 0;
for i from 1 do
  x:= x*10^(1+ilog10(i))+i;
  if x mod n = 0 then return x fi
od
end proc:
map(f, [$1..20]); # Robert Israel, Oct 25 2020

smn[n_]:=Module[{k=1,c=1},While[!Divisible[c,n],k++;c= c*10^IntegerLength[ k]+ k];c]; Array[ smn,20] (* Harvey P. Dale, Apr 04 2022 *)

a(n) = j="";for(k=1, oo, j=eval(concat(Str(j), k)); if(j%n==0, return(j)))

a(n) is the smallest multiple of n appearing in A007908.

A336401 a(n) = a(n-1) concatenated with the smallest number k, such that a(n) is divisible by lcm(1..n).

Original entry on oeis.org

1, 10, 102, 1020, 10200, 102000, 102000360, 1020003600, 10200036001680, 102000360016800, 10200036001680035280, 102000360016800352800, 102000360016800352800332640, 1020003600168003528003326400

Offset: 1

Eder Vanzei, Jul 20 2020

Cf. A045874, A078282, A078283, A214437, A336399.

a(n)={if(n==1,return(1));for(n1=0,oo,k=eval(concat(Str(a(n-1)),n1));n2=0;for(n3=1,n,if(k%n3==0,n2+=1;if(n2==n,return(k)))))};

A336399 a(1) = 1, a(n) is the smallest number such that the concatenation a(1)a(2)...a(n) is divisible by lcm(1..n).

Original entry on oeis.org

1, 0, 2, 0, 0, 0, 360, 0, 1680, 0, 35280, 0, 332640, 0, 0, 0, 8648640, 0, 306306000, 0, 0, 0, 232792560, 0, 0, 0, 26771144400, 0, 481880599200, 0, 41923612130400, 0, 0, 0, 0, 0, 5487335009956800, 0, 0, 0, 245774847024907200, 0, 8105227020364874400, 0, 0, 0, 452140231622516236800, 0, 3984485791173424336800, 0

Offset: 1

Eder Vanzei, Jul 20 2020

			a(7) = 360 as the smallest positive integer k such that the concatenation a(1)a(2)..a(6)k is divisible by lcm(1..7) = 420. - _David A. Corneth_, Jul 21 2020

Robert Israel, Table of n, a(n) for n = 1..2297
David A. Corneth, PARI program

Cf. A336401 (corresponding numbers), A003418 (LCM's).

Cf. A045874, A051883, A078282, A078283, A214437.

N:= 1: R:= 1: C:= 1:
for n from 2 to 60 do
  N:= ilcm(N,n);
  for d from 1 do
    x:= -C*10^d mod N;
    if x = 0 then lx:= 1 else lx:= 1+ilog10(x) fi;
    if lx = d then
       R:= R,x;
       C:= C*10^d+x;
       break
    elif lx < d then
       k:= ceil((10^(d-1)-x)/N);
       x:= x + k*N;
       if x < 10^d then
         R:= R,x;
         C:= C*10^d+x;
         break
    fi fi
od; od:
R; # Robert Israel, Sep 16 2020

a(n) = {if(n==1,return(1));for(n1 = 0, oo, ; k[n]=eval(concat(Str(k[n-1]), n1)); n2=0; for(n3 = 1, n, if(k[n] % n3 == 0, n2+=1; if(n2==n, return(k[n])))))};
k = vector(10000);print1(k[1]=1,", ");for(j=1, 20, print1(a(j+1) - a(j)*10^(length(Str(a(j+1))) - length(Str(a(j)))), ", "))

\\ See Corneth link. David A. Corneth, Jul 21 2020

a(27)-a(50) from David A. Corneth, Jul 20 2020

A333368 Primes of the form km^(km) - 1 with m > 1.

Original entry on oeis.org

3, 31, 191, 5119, 131071, 524287, 3758096383, 4353564671, 1356446145697, 1618481116086271, 2058911320946489, 1046695266054721074427023041, 847823165504324070285888664019, 5359447279004780799548150067050349330431, 2817103802133904744169307240538184064530443801964688726052818649087

Offset: 1

Eder Vanzei, Mar 17 2020

			31 appears in this sequence because 31=2*2^(2*2)-1 and 31 is prime.

Sean A. Irvine, Java program (github)

Cf. A000040, A050918.

lista(nn) = {my(k, n=2, v=List([]), x=4, y); while(xJinyuan Wang, Mar 18 2020

Corrected by Michel Marcus and Jinyuan Wang, Mar 17 2020

A333770 Smallest palindromic number >= 3^n.

Original entry on oeis.org

1, 3, 9, 33, 88, 252, 737, 2222, 6666, 19691, 59095, 177771, 532235, 1594951, 4783874, 14355341, 43055034, 129141921, 387424783, 1162332611, 3486886843, 10460406401, 31381118313, 94143234149, 282429924282, 847288882748

Offset: 0

Eder Vanzei, Apr 04 2020

			a(10) = 59095, because 3^10 = 59049 and 59095 is the smallest palindromic number >= 59049.

Robert Israel, Table of n, a(n) for n = 0..2093

Cf. A000244, A002113, A262038, A333016.

digrev:= proc(n) local L,i;
  L:= convert(n,base,10);
  add(L[-i]*10^(i-1),i=1..nops(L))
end proc:
f:= proc(n) local d,x,y,t;
  d:= ilog10(n)+1;
  if d::even then
    x:= floor(n/10^(d/2));
    t:= x*10^(d/2)+digrev(x);
    if t >= n then return t fi;
    (x+1)*10^(d/2)+digrev(x+1);
  else
    x:= floor(n/10^((d-1)/2));
    t:= x*10^((d-1)/2)+digrev(floor(x/10));
    if t >= n then return t fi;
    y:= x mod 10;
    if y < 9 then return t + 10^((d-1)/2) fi;
    x:= x+1;
    x*10^((d-1)/2)+digrev(floor(x/10));
  fi
end proc:
seq(f(3^i),i=0..30); # Robert Israel, May 04 2020

a(n) = for(k=3^n, oo, if(Vecrev(v=digits(k))==v, return(k)));

a(n) = A262038(A000244(n)). - Michel Marcus, May 04 2020

A333578 Numbers k such that a division occurs in A330139(k).

Original entry on oeis.org

5, 12, 21, 172, 15277, 703981, 2060911, 6663113

Offset: 1

Eder Vanzei, Mar 27 2020

Is this sequence infinite?

			5 is in this sequence because a division occurs in A330139(5).

Cf. A330139.

seq = {}; n1 = n2 = 1; Do[n3 = n1 + n2; If[Divisible[n3, n], n3/=n; AppendTo[seq, n]]; n1 = n2; n2 = n3, {n, 3, 2*10^4}]; seq (* Amiram Eldar, Mar 28 2020 *)

lista(nn) = {my(x = 1, y = 1, z); for (n=3, nn, z = x + y; if ((z % n) == 0, z = z/n; print1(n, ", ")); x = y; y = z;);} \\ Michel Marcus, Mar 27 2020

cp's OEIS Frontend

User: Eder Vanzei

A355458 Numbers k that set a new record m where m is the largest left-truncatable prime up to the final k (stop on reaching the final k).

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A351653 a(n) is the concatenation of the length of each run of digits in the decimal representation of n.

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A346674 Nonnegative numbers k such that MD5(k interpreted as a string) contains only decimal digits.

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A337844 a(1) = 1; a(n) = a(n-1) concatenated with the smallest number k>0 such that tau(a(n)) > tau(a(n-1)).

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A334620 a(n) is the smallest multiple of n formed by the concatenation 1,2,3,...,k for some k.

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A336401 a(n) = a(n-1) concatenated with the smallest number k, such that a(n) is divisible by lcm(1..n).

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A336399 a(1) = 1, a(n) is the smallest number such that the concatenation a(1)a(2)...a(n) is divisible by lcm(1..n).

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A333368 Primes of the form k*m^(k*m) - 1 with m > 1.

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A333368 Primes of the form km^(km) - 1 with m > 1.