cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A217535 Least number having in its decimal representation each digit n times.

Original entry on oeis.org

1023456789, 10012233445566778899, 100011222333444555666777888999, 1000011122223333444455556666777788889999, 10000011112222233333444445555566666777778888899999, 100000011111222222333333444444555555666666777777888888999999
Offset: 1

Views

Author

M. F. Hasler, Oct 05 2012

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> parse(cat(1,0$n,1$(n-1),seq(i$n, i=2..9))):
    seq(a(n), n=1..10);  # Alois P. Heinz, Jun 25 2017
  • PARI
    A217535(n)=sum(d=1,9,10^(n-(d==1))\9*d*10^(n*(9-d)))+10^(10*n-1)

Formula

a(n) ~ 10^(10n-1). See PARI code for an exact formula.

A218770 Smallest number which when multiplied by n gives a pandigital number.

Original entry on oeis.org

1023456789, 511728399, 341152263, 255864474, 204693579, 170576133, 146208114, 127932237, 113717421, 123456789, 93125079, 85288158, 78727446, 73104057, 68231193, 63966123, 60203394, 56858711, 53866251, 61728399, 48736038, 46562544, 44498943, 42644079, 40938759
Offset: 1

Views

Author

Arkadiusz Wesolowski, Nov 05 2012

Keywords

Comments

a(A171102(n)) = 1.

Crossrefs

Cf. A171102.

Programs

  • Mathematica
    lst1 = Select[FromDigits@# & /@ Permutations[Range[0, 9], {10}], # > 10^9 &]; lst2 = {}; Do[m = Select[lst1, Divisible[#, n] &, 1]; If[FreeQ[m, _Integer], Break[], AppendTo[lst2, m/n]], {n, 25}]; Flatten[lst2]

A225295 Pandigital numbers with exactly 4 prime factors (with multiplicity).

Original entry on oeis.org

1023456987, 1023458679, 1023458967, 1023465798, 1023465897, 1023465978, 1023465987, 1023467589, 1023467859, 1023468579, 1023468597, 1023468795, 1023469758, 1023478569, 1023479586, 1023479865, 1023485967, 1023486579, 1023486957, 1023487659, 1023487965, 1023489657, 1023495678
Offset: 1

Views

Author

Jonathan Vos Post, May 04 2013

Keywords

Examples

			a(1) = 1023456987 = 3^2 * 7 * 16245349.
a(2) = 1023458679 = 3^3 * 37905877.
a(3) = 1023458967 = 3^2 * 113 * 1006351.
a(4) = 1023465897 = 3^2 * 6379 * 17827.
a(5) = 1023465987 = 3^2 * 53 * 2145631.
a(511032) = 10123456987 = 7^2 * 9833 * 21011.
		

Crossrefs

Cf. A050288 (pandigital primes), A175845 (pandigital numbers).
Cf. A171102 (3-almost primes).

Programs

Formula

A014613 INTERSECTION A171102.
a(n) ~ 6n log n / (log log n)^3. - Charles R Greathouse IV, May 04 2013

Extensions

a(6)-a(23) from Charles R Greathouse IV, May 04 2013

A232712 Least positive k (not a power of 10) such that k and k^n have the same set of digits.

Original entry on oeis.org

2, 4762, 107624, 35641, 39568, 1380796, 12635940, 40837596, 102349857, 102567384, 106342987, 129046873, 107623945, 231940678, 239607415, 368709154, 1023456789, 164758903, 176384592, 1023456789, 1023456789, 1023456789, 1023456789, 1023456789, 1023456789
Offset: 1

Views

Author

Arkadiusz Wesolowski, Nov 28 2013

Keywords

Comments

a(17) and a(20)-a(40) = A050278(1) = 1023456789, the smallest pandigital number. [Lars Blomberg, Dec 10 2013]

Crossrefs

Programs

  • PARI
    for(n=1, 6, k=1; until(Set(Vec(Str(k)))==Set(Vec(Str(k^n)))&&!(sumdigits(k)==1), k++); print1(k, ", "));

Extensions

a(14)-a(25) from Lars Blomberg, Dec 10 2013

A257001 Numbers such that the largest missing digit is a divisor.

Original entry on oeis.org

18, 27, 36, 45, 54, 63, 72, 81, 96, 98, 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, 192, 207, 216, 225, 234, 243, 252, 261, 270, 288, 296, 306, 315, 324, 333, 342, 351, 360, 378, 387, 392, 405, 414, 423, 432, 441, 450, 468, 477, 486, 496, 504, 513, 522
Offset: 1

Views

Author

Keywords

Comments

a(n) mod A000030(A230959(a(n))) = 0.
Pandigital numbers are not terms: A171102, A050278.
The original definition used the phrase "largest absent digit".

Crossrefs

Programs

  • Haskell
    import Data.List ((\\)); import Data.Char (digitToInt)
    a257001 n = a257001_list !! (n-1)
    a257001_list = filter f [1..] where
       f x = h > 0 && mod x h == 0 where h = a000030 $ a230959 x
  • Mathematica
    f[x_]:=Union[Sort[IntegerDigits[x]]];
    d={1,2,3,4,5,6,7,8,9};
    Select[Range[525],And[f[#]!=d,Length[f[#]]<10,IntegerQ[#/Max[Complement[d,f[#]]]]]&] (* Ivan N. Ianakiev, Apr 14 2015 *)

A260900 Table read by rows: n-th row lists all positive integers k such that exactly half the integers in 1, 2, ..., k are pandigital in base n.

Original entry on oeis.org

2, 4, 174, 20056, 9026066, 9612284, 9612296, 9612298, 9612308, 9612310, 9612312, 9612946, 9612954, 9612962, 9612966, 9612968, 9613074, 9613394, 9667944, 10138460, 10144636, 10144638, 10144640, 10144650, 10144712, 10144756, 10144758, 10144760, 10144770
Offset: 1

Views

Author

Jon E. Schoenfield, Aug 04 2015, expanded Aug 07 2015

Keywords

Comments

Numbers that are pandigital in base 2 (i.e., numbers whose digits include at least one each of 0 and 1) are 2=10_2, 4=100_2, 5=101_2, 6=110_2, 8=1000_2, etc. (i.e., all positive integers not of the form 2^j-1); exactly 2/2=1 of the first 2 positive integers and exactly 4/2=2 of the first 4 positive integers are base-2 pandigital, so 2 and 4 are in the sequence. For all k > 4, there are more base-2 pandigital numbers in 1..k than base-2 nonpandigital numbers, so there are no more terms in the n=2 row.
In base 3, exactly half of the integers in 1..174 are pandigital, so 174 is in the sequence. Fewer than half of the integers in 1..k are pandigital for all k < 174, and more than half of the integers in 1..k are pandigital for all k > 174, so 174 is the only term in the n=3 row.
The 27-digit number 245836727707164139860503406, which is a(134), is the only term in the n=10 row: in base 10, exactly half of the integers in 1..a(134) are pandigital, fewer than half of the integers in 1..k are pandigital for all k < a(134), and more than half of the integers in 1..k are pandigital for all k > a(134).
For each of rows 2 through 10, the number of terms and a list of those terms (abridged for rows 5 and 6) are as follows:
Row # terms list of terms
==== ======= ===========================================
2 2 2, 4
3 1 174
4 1 20056
5 46 9026066, 9612284, ..., 10384656;
6 80 12436651810, 12438872740, ..., 13770404636;
7 1 45381851638748
8 1 282633399694638258
9 1 9255986333928835642154
10 1 245836727707164139860503406

Crossrefs

Cf. A171102.

A276590 Pandigital numbers n such that sum of all permutations of digits of n is also a pandigital number. Sequence lists the least ones of corresponding permutational classes.

Original entry on oeis.org

10234567889, 100223456789, 100234566789, 100234567889, 101234556789, 101234567789, 102234456789, 102234566789, 102334556789, 102334567899, 102344456789, 102344567889, 102345567789, 102345666789, 102345677789, 102345678899, 1000223456789, 1000234456789, 1000234566789
Offset: 1

Views

Author

Robert Israel and Altug Alkan, Sep 06 2016

Keywords

Comments

The least pandigital number in A276510 is 10234567889.
Each member of the sequence has digits in increasing order except that the first digit is 1.
The sequence has 1 member with 11 digits, 15 with 12 digits, 90 with 13 digits, 261 with 14 digits and 1190 with 15 digits.

Examples

			100223456789 is a term because A045876(100223456789) = 52113599999947886400 is pandigital.
		

Crossrefs

Programs

  • Maple
    pandig:= n -> evalb(nops(convert(convert(n,base,10),set))=10):
    sump:= proc(x) local L, D, n, M, s, j;
      L:= convert(x, base, 10);
      D:= [seq(numboccur(j, L), j=0..9)];
      n:= nops(L);
      M:= n!/mul(d!, d=D);
      s:= add(j*D[j+1], j=0..9);
      (10^n-1)*M/9/n*s
    end proc:
    n0:= 1023456789:
    rep:= proc(n) local L,n0,i;
      L:= sort(convert(n,base, 10));
      n0:= numboccur(0, L);
      L:= subsop(1=1,n0+1=0,L);
      add(L[-i-1]*10^(i),i=0..nops(L)-1); end proc:
    sort(convert(map(rep,
    select(pandig @ sump, {seq(seq(n0*10^d+x,x=0..10^d-1),d=0..3)})),list));

A277028 Numbers that are never pandigital for any base b > 1.

Original entry on oeis.org

0, 1, 3, 7, 31, 255, 32767
Offset: 1

Views

Author

Rémy Sigrist, Sep 25 2016

Keywords

Comments

Here, a number is considered pandigital in base b if any digit from 0 to b-1 appears at least once in its base b representation (ignoring leading zeros).
This is a subsequence of A000225, conjectured to be finite (no other term below 2^1000000).
No other terms below 2^10^10. - Charles R Greathouse IV, Sep 26 2016

Examples

			32767 is not pandigital in any base b between 2 and 6:
  b  32767 in base b  Missing digits
  -  ---------------  --------------
  2  111111111111111  0
  3  1122221121       0
  4  13333333         0, 2
  5  2022032          1, 4
  6  411411           0, 2, 3, 5
Moreover, 32767 is too small to be pandigital in any base b > 6, hence 32767 is in the sequence.
		

Crossrefs

Programs

  • PARI
    See Sigrist link.

A292833 Lexicographically earliest sequence of distinct positive numbers such that the sum of any two consecutive terms is a pandigital number in base 5.

Original entry on oeis.org

1, 693, 5, 689, 9, 685, 13, 681, 17, 677, 21, 673, 25, 669, 29, 665, 33, 661, 37, 657, 41, 653, 45, 649, 49, 645, 53, 641, 57, 637, 61, 633, 65, 629, 69, 625, 73, 621, 77, 617, 81, 613, 85, 609, 89, 605, 93, 601, 97, 597, 101, 593, 105, 589, 109, 585, 113, 581
Offset: 1

Views

Author

Rémy Sigrist, Sep 24 2017

Keywords

Comments

Similarly to A171102, we say that a number is pandigital in base 5 iff all digits in the set {0, 1, 2, 3, 4} appear at least once in the base 5 representation of n (leading zeros being ignored); hence we have infinitely many pandigital numbers in base 5, and this sequence is infinite.
The choice of base 5 is motivated by the fact that it allows the apprehension of the graphical features of the variants of this sequence in other bases, using only a few thousand terms (see also scatterplots in Links section).
This sequence is likely a permutation of the positive numbers.

Examples

			The first terms of the sequence, alongside the sum of consecutive terms in base 5, are:
n   a(n)    a(n) + a(n+1) in base 5
--  ----    -----------------------
1   1       10234
2   693     10243
3   5       10234
4   689     10243
5   9       10234
6   685     10243
7   13      10234
8   681     10243
9   17      10234
10  677     10243
11  21      10234
12  673     10243
		

Crossrefs

Cf. A171102.

Programs

  • PARI
    See Links section.

A302096 a(n) is the smallest pandigital number divisible by n, or 0 if no such pandigital number exists.

Original entry on oeis.org

1023456789, 1023456798, 1023456789, 1023457896, 1023467895, 1023456798, 1023456798, 1023457896, 1023456789, 1234567890, 1024375869, 1023457896, 1023456798, 1023456798, 1023467895, 1023457968, 1023457698, 1023456798, 1023458769, 1234567980, 1023456798, 1024375968
Offset: 1

Views

Author

Rodolfo Kurchan, May 06 2018

Keywords

Comments

Note: in this sequence, "pandigital" numbers are defined as in A050278 (i.e., with each of the ten digits 0..9 appearing exactly once).
The first 99 terms coincide with those of A061604. - Giovanni Resta, May 15 2018
From Jon E. Schoenfield, May 19 2018: (Start)
Record high values exceeding 2*10^9 begin a(10001) = 2650134987, a(20002) = 2750134986, a(27775) = 3012948675, a(40004) = 3760215984, a(44440) = 4123987560, a(50005) = 6820431975, ...
a(n)=0 for every n divisible by 100. Other than multiples of 100, the smallest values of n for which a(n)=0 are 37037 and 55550. The last nonzero term is a(9876543210) = 9876543210. (End)
There are 44021407 nonzero terms. - Michael S. Branicky, Mar 05 2025

Examples

			a(11) = 1024375869 = 11 * 93125079 because it is the smallest pandigital number that is divisible by 11;
a(100) = 0 because there is no pandigital number that is divisible by 100.
		

Crossrefs

Programs

  • Mathematica
    s = Select[FromDigits /@ Permutations[Range[0, 9]], # > 10^9 &]; Table[ SelectFirst[ s, Mod[#, n] == 0 &, 0], {n, 22}] (* Giovanni Resta, May 15 2018 *)
  • Python
    # see link for another program
    from itertools import permutations
    def a(n): return next((t for p in permutations("0123456789") if p[0] != "0" and (t:=int("".join(p)))%n == 0), 0)
    print([a(n) for n in range(1, 23)]) # Michael S. Branicky, Mar 05 2025
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