A050289 -id:A050289 - cp's OEIS frontend

A209280 First difference of A050289 = numbers whose digits are a permutation of (1,...,9).

9, 81, 18, 81, 9, 702, 9, 171, 27, 72, 18, 693, 18, 72, 27, 171, 9, 702, 9, 81, 18, 81, 9, 5913, 9, 81, 18, 81, 9, 1602, 9, 261, 36, 63, 27, 594, 18, 162, 36, 162, 18, 603, 9, 171, 27, 72, 18, 5814, 9, 171, 27, 72, 18, 603, 9, 261, 36, 63, 27, 1584, 27, 63, 36, 261, 9

Offset: 1

M. F. Hasler, Jan 12 2013

This sequence is the natural extension of A107346 (and others, see below) from 5!-1 to 9!-1 terms, which is the natural (since maximal) length, given that OEIS sequence data are stored as decimal numbers. On the other hand, it is quite different from A219664 in many aspects, not only for the reason that the other sequence is infinite and therefore differs from this one in all terms beyond n = 9!-1.
The sequence is finite, with 9!-1 terms, and symmetric: a(n)=a(9!-n).
All terms are multiples of 9, cf. formula.
The subsequence of the first n!-1 terms (n=2,...,9) yields the first differences of the sequence of numbers whose digits are a permutation of (1,...,n):
The first 8!-1 terms yield the first differences of A178478: numbers whose digits are a permutation of 12345678.
The first 7!-1 terms yield the first differences of A178477: numbers whose digits are a permutation of 1234567.
The first 6!-1 terms yield the first differences of A178476: numbers whose digits are a permutation of 123456.
The first 5!-1 terms yield A107346, the first differences of A178475: numbers whose digits are a permutation of 12345.

			The same initial terms are obtained for the permutations of any set of the form {1,...,m}, e.g., {1,2,3} or {1,...,9}: In the first case we have P = (123,132,213,231,312,321) and P(4)-P(3) = 231 - 213 = 18 = a(3), and in the latter case P(4)-P(3) = 123456897 - 123456879 = 18, again. - _M. F. Hasler_, Jan 12 2013

Cf. A030299, A219664.

Take[Differences[Sort[FromDigits/@Permutations[Range[9]]]],70] (* Harvey P. Dale, Mar 31 2018 *)

A209280_list(N=5)={my(v=vector(N,i,10^(N-i))~); v=vecsort(vector(N!,k,numtoperm(N,k)*v)); vecextract(v,"^1")-vecextract(v,"^-1")} \\ return the N!-1 first terms as a vector

A209280(n)={if(a209280=='a209280 || #a209280A209280_list(A090529(n+1)));a209280[n]}

a(n) = A219664(n) = 9*A217626(n) (for n < 9!). - M. F. Hasler, Jan 12 2013

a(n) = a(m!-n) for any m < 10 such that n < m!.

A204140 Numbers m such that both m and 7m are zeroless pandigital numbers (A050289).

Original entry on oeis.org

123456789, 123678459, 123875649, 124563789, 124789563, 127593468, 127638945, 127648953, 132487695, 132596478, 134526789, 135246879, 135926748, 135947826, 136245978, 136597482, 136758924, 136875249, 137645982, 137689245, 137826459, 138752649, 138795462, 138976452

Offset: 1

Zak Seidov, Jan 11 2012

There are precisely 24 such numbers.

The only number m such that m and 8m are zeroless pandigital numbers is A050289(1)=123456789.

Cf. A050289, A204141.

a(n) = A204141(n)/7.

A204141 Numbers m such that both m and m/7 are zeroless pandigital numbers (A050289).

Original entry on oeis.org

864197523, 865749213, 867129543, 871946523, 873526941, 893154276, 893472615, 893542671, 927413865, 928175346, 941687523, 946728153, 951487236, 951634782, 953721846, 956182374, 957312468, 958126743, 963521874, 963824715, 964785213, 971268543, 971568234, 972835164

Offset: 1

Zak Seidov, Jan 11 2012

There are precisely 24 such numbers.

The only number m such that m and m/8 are zeroless pandigital numbers is A050289(362879)=987654312.

Cf. A050289, A204140.

Module[{lst=FromDigits/@Permutations[Range[9]],l1,l2},l1=Select[lst,#<141093475&];l2= Select[ lst,#>864*10^6&];Select[l2,MemberQ[l1,#/7]&]] (* Harvey P. Dale, Sep 25 2023 *)

a(n)=7*A204140(n).

A050278 Pandigital numbers: numbers containing the digits 0-9. Version 1: each digit appears exactly once.

Original entry on oeis.org

1023456789, 1023456798, 1023456879, 1023456897, 1023456978, 1023456987, 1023457689, 1023457698, 1023457869, 1023457896, 1023457968, 1023457986, 1023458679, 1023458697, 1023458769, 1023458796, 1023458967, 1023458976, 1023459678, 1023459687, 1023459768

Offset: 1

Eric W. Weisstein, Dec 11 1999

This is a finite sequence with 9*9! = 3265920 terms: a(9*9!) = 9876543210.
A171102 is the infinite version, where each digit must appear at least once.
More precisely, this is exactly the subset of the first 9*9! terms of A171102. - M. F. Hasler, Jan 05 2020
Subsequence of A134336 and of A178403; A178401(a(n)) = 1. - Reinhard Zumkeller, May 27 2010
Smallest prime factors: A178775(n) = A020639(a(n)). - Reinhard Zumkeller, Jun 11 2010
A178788(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2010
All these numbers are composite because the sum of the digits, 45, is divisible by 9. - T. D. Noe, Nov 09 2011
This is the 10th row of the array T(k,n) = n-th number in which the number of distinct base-10 digits is k. A031969 is the 4th row. A220063 is the 5th row. A220076 is the 6th row. A218019 is the 7th row. A219743 is the 8th row. - Jonathan Vos Post, Dec 05 2012
From Hieronymus Fischer, Feb 13 2013: (Start)
The sum of all terms is 9!*49444444440 = 17942399998387200.
General formula for the sum of all terms of the finite sequence of the corresponding base-p pandigital numbers with p places: sum = ((p^2 - p - 1)*(p^p - 1) + p - 1)*(p-2)!/2.
General formula for the sum of all terms (interpreted as decimal permutational numbers with exactly d+1 different digits from the range 0..d < 10): sum = (d+1)!*((10d - 1)*10^d - d + 1)/18, d > 1.
(End)

Robert G. Wilson v, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Pandigital Number
Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 1.

Cf. A171102, A050288, A050289.
Cf. A199630, A199631, A114260, A199632, A199633.
Cf. A031969, A050278, A220063, A220076, A218019, A219743.

Select[ FromDigits@# & /@ Permutations[ Range[0, 9]], # > 10^9 &, 20] (* Robert G. Wilson v, May 30 2010, Jan 17 2012 *)

A050278(n)={ my(b=vector(9,k,1+(n+9!-1)%(k+1)!\k!), t=b[9]-1, d=vector(9,i,i+(i>t)-1)); for(i=1,8, t=10*t+d[b[9-i]]; d=vecextract(d,Str("^"b[9-i]))); t*10+d[1]} \\ M. F. Hasler, Jan 15 2012

is_A050278(n)={ 9<#vecsort(Vecsmall(Str(n)),,8) & n<1e10 } /* assuming that n is a nonnegative integer */ /* M. F. Hasler, Jan 10 2012 */

a(n)=my(d=numtoperm(10,n+9!-1));sum(i=1,#d,(d[i]-1)*10^(#d-i)) \\ David A. Corneth, Jun 01 2014

from itertools import permutations
A050278_list = [int(''.join(d)) for d in permutations('0123456789',10) if d[0] != '0'] # Chai Wah Wu, May 25 2015

A050278 = 9*A171571. - M. F. Hasler, Jan 12 2012

A050278(n) = A171102(n) for n <= 9*9!.

Edited by N. J. A. Sloane, Sep 25 2010 to clarify that this is a finite sequence

A171102 Pandigital numbers: numbers containing the digits 0-9. Version 2: each digit appears at least once.

Original entry on oeis.org

Offset: 1

N. J. A. Sloane, Sep 25 2010

This is the infinite version. See A050278 for the finite version.
The first 9*9!=3265920 terms of this sequence are permutations of the digits 0-9 with a(9*9!)=9876543210 (see Version 1, A050278). - Jeremy Gardiner, May 29 2010
Subsequence of A134336 and of A178403; A178401(a(n))>0. - Reinhard Zumkeller, May 27 2010
Smallest prime factors: A178775(n) = A020639(a(n)). - Reinhard Zumkeller, Jun 11 2010
A178788(a(n)) = 1, for n <= 9*9!, else A178788(a(n)) = 0. - Reinhard Zumkeller, Jun 30 2010 [corrected by Hieronymus Fischer, Feb 02 2013]
A230959(a(n)) = 0. - Reinhard Zumkeller, Nov 02 2013
The first term of the sequence absent in A050278 is a(3265921) = 10123456789. Also, the  first prime is a(3306373) = 10123457689 = A050288(1). - Zak Seidov, Sep 23 2015
Almost all numbers are in this sequence, in the sense that it has asymptotic density equal to 1. Indeed, the fraction of n-digit numbers which don't have a given digit d is roughly 0.9^n (not exactly because the first digit is chosen among {1..9}) which tends to zero as n -> oo. - M. F. Hasler, Jan 05 2020

Robert G. Wilson v, Table of n, a(n) for n = 1..1000 . [From _Robert G. Wilson v_, May 30 2010]
Eric Weisstein's World of Mathematics, Pandigital Number.
Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 1.

Cf. A050278, A050288, A050289.
Cf. A051264, A051018, A051019, A051020.
Subsequence of A253172.

Take[ Select[ FromDigits@# & /@ Permutations[ Range[0, 9], {10}], # > 10^9 &], 20] (* Robert G. Wilson v, May 30 2010 *)

is_A171102(n)=9<#vecsort(Vecsmall(Str(n)),,8) /* assuming that n is a nonnegative integer. In PARI/GP V.2.4 - 2.9 this is faster than other possibilities involving Set(),Vec(),eval() or digits() */ \\ M. F. Hasler, Jan 10 2012, Sep 19 2017

A171102=A050278 /*** valid for n <= 9*9! ***/ \\ M. F. Hasler, Jan 10 2012

a(n) = 1011111111 + A178478(n) for n = 1,...,8!. - M. F. Hasler, Jan 10 2012

A171102(n) = A050278(n) for n <= 9*9!.

A144688 "Magic" numbers: all numbers from 0 to 9 are magic; a number >= 10 is magic if it is divisible by the number of its digits and the number obtained by deleting the final digit is also magic.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 102, 105, 108, 120, 123, 126, 129, 141, 144, 147, 162, 165, 168, 180

Offset: 1

N. J. A. Sloane, based on email from Roberto Bosch Cabrera, Feb 02 2009

Roberto Bosch Cabrera finds that there are exactly 20457 terms. (Total corrected by Zak Seidov, Feb 08 2009.)
The 20457th and largest term is the 25-digit number 3608528850368400786036725. - Zak Seidov, Feb 08 2009
a(n) is also the number such that every k-digit substring ( k <= n ) taken from the left, is divisible by k. - Gaurav Kumar, Aug 28 2009
A probabilistic estimate for the number of terms with k digits for the corresponding sequence in base b is b^k/k!, giving an estimate of e^b total terms. For this sequence, the estimate is approximately 22026, compared to the actual value of 20457. - Franklin T. Adams-Watters, Jul 18 2012
Numbers such that their first digit is divisible by 1, their first two digits are divisible by 2, and so on. - Charles R Greathouse IV, May 21 2013
These numbers are also called polydivisible numbers, because so many of their digits are divisible. - Martin Renner, Mar 05 2016
The unique zeroless pandigital (A050289) term, also called penholodigital, is a(7286) = 381654729 (see Penguin reference); so, the unique pandigital term (A050278) is a(9778) = 3816547290. - Bernard Schott, Feb 07 2022

			102 has three digits, 102 is divisible by 3, and 10 is also magic, so 102 is a member.

Robert Bosch, Tale of a Problem Solver, Arista Publishing, Miami FL, 2016.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Revised Edition), Penguin Books, 1997, entry 381654729, page 185.

Zak Seidov, The full table of n, a(n) for n=1..20457
James Grime and Brady Haran, Why 381,654,729 is awesome, Numberphile video (2013).
Shyam Sunder Gupta, On Some Special Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 22, 527-565.
Wikipedia, Polydivisible number

A subsequence of A098952.
Cf. A082399, A051883, A143671, A214437.
Cf. A050278, A050289.

P1:={seq(i,i=1..9)}:
for i from 2 to 25 do
  P||i:={}:
  for n from 1 to nops(P||(i-1)) do
    for j from 0 to 9 do
      if P||(i-1)[n]*10+j mod i = 0 then P||i:={op(P||i),P||(i-1)[n]*10+j}: fi:
    od:
  od:
od:
`union`({0},seq(P||i,i=1..25)); # Martin Renner, Mar 05 2016

divQ[n_]:=Divisible[n,IntegerLength[n]];
lessQ[n_]:=FromDigits[Most[IntegerDigits[n]]];
pdQ[n_]:=If[Or[n<10,And[divQ[n],divQ[lessQ[n]]]],True];
Select[Range[0,180],pdQ[#]&] (* Ivan N. Ianakiev, Aug 23 2016 *)

def agen(): # generator of terms
    yield 0
    magic, biggermagic, digits = list(range(1, 10)), [], 2
    while len(magic) > 0:
        yield from magic
        for i in magic:
            for d in range(10):
                t = 10*i + d
                if t%digits == 0:
                    biggermagic.append(t)
        magic, biggermagic, digits = biggermagic, [], digits+1
print([an for an in agen()][:70]) # Michael S. Branicky, Feb 07 2022

A215940 Difference between the n-th and the first (identity) permutation of (0,...,m-1), interpreted as a decimal number, divided by 9 (for any m for which 10! >= m! >= n).

Original entry on oeis.org

0, 1, 10, 12, 21, 22, 100, 101, 120, 123, 131, 133, 210, 212, 220, 223, 242, 243, 321, 322, 331, 333, 342, 343, 1000, 1001, 1010, 1012, 1021, 1022, 1200, 1201, 1230, 1234, 1241, 1244, 1310, 1312, 1330, 1334, 1352, 1354, 1421, 1422, 1441, 1444, 1452, 1454, 2100

Offset: 1

R. J. Cano, Sep 21 2012

Original definition: "Quotients of the polynomial remainder theorem for Diophantine equations among permutations."
The set built from the first terms of this sequence {0, 1, 10, 12, 21, 22, ....} contains the general solutions for a class of Diophantine linear equations among permutations, if one takes into account the unlimited number of distinct bases where these may be read.
From M. F. Hasler, Jan 12 2013, edited by R. J. Cano, May 08 2017: (Start)
Let P be the sequence of permutations of (0,...,m-1) interpreted as decimal numbers, P(n) = Sum_{i=1..m} 10^(m-i)*s(i) where s=(s(1),...,s(m)) is the n-th permutation (in lexicographical order), n <= m!. Then the difference P(n)-P(1) is independent of the choice of m, and divisible by 9. (Since 10 == 1 (mod 9), the numbers P(n) are all congruent (mod 9) to the sum s(1)+...+s(m).) This yields well-defined terms a(n)=(P(n)-P(1))/9.
For n>10!, P(n) will no longer be the concatenation of the "digits" (some of which will exceed 9). The pattern present in the decimal representation of the first terms will also be lost since there will be digits as large as d+1.
Note that the same a(n) is obtained independently of the chosen base b, provided that (i) 10 and 9 in the above are replaced with b and b-1, (ii) the result is (and can be) written in base b. (This implies the restriction to terms which can be written using the digits 0-9 to which the OEIS is limited.) See EXAMPLES for an illustration. (End)
We have P(n)-P(1)=a(n)*g(n), with g(n) = 9 = 10-1. Considering this and P(n) as polynomials in x=10, one can see an analogy with the polynomial remainder theorem. [Given as "formula" by R. J. Cano, rephrased by M. F. Hasler, Jan 12 2013]
Contribution by R. J. Cano, Feb 09 2013, (Start)
The maximum of the first m! terms of this sequence is given in base R by the explicit formula (please see A211869): max(m,R)=Sum_{k=1..m} k*(m-k)*R^(m-k-1);
If the first m! terms of this sequence were computed reading the permutations in base A033638(m), dividing their differences by A033638(m)-1, the resulting quotients would be written in the same way (with the same digits) in every base > A033638(m).
(End)
From R. J. Cano, Apr 29 2016, (Start)
Although in the sequence name it reads: "permutation of (0,...,m-1)", the most general statement that could replace it is: "permutation of any m-tuple of integers all of them in arithmetic progression", obtaining a multiple of this sequence, lambda*a(n), where lambda is the common difference for the progression. It works in such way because only the differences is what matters here.
Given x>1 and k>=0, if a polynomial G(x) of degree k is divided by x-1 then the remainder will be the sum of all the coefficients in G. Let us consider the case in which those coefficients are the differences among the letters ("digits") of two permutations for the same set of letters (0..x-1): The sum of all those differences must vanish. This explains how the difference between two of such permutations expressed in base x is 0 mod x-1, particularly why differences for pairs of permutations are divisible by 9.
Another way of introducing this sequence takes advantage of the fact that for n>1, n! is even. Consider for n>1 to obtain only the first n! terms. This can be done by subtracting the last permutation from the first, the penultimate permutation from the second, and so on by following the pattern (P(k)-P(n!-k+1))/9 with 1<=k<=n!. Such procedure generates an antisymmetric sequence f(k) from which a(k)=(f(k)+f(n!))/2. This partially explains why A217626 is symmetric. Also a base-independent treatment is possible using linear algebra: Column vectors and the strictly lower triangular matrix instead of the division by (r-1) where r is the base (and r=10 here for this sequence). This approach leads one to conclude that terms in this sequence are the differences between pairs of vectors made from the first n-1 partial sums of letters ("digits") taken from permutations for n consecutive letters, when components in these vectors are viewed as coefficients for a power series in r=10.
(End)

			From _M. F. Hasler_, Jan 12 2013, edited by _R. J. Cano_, May 09 2017: (Start)
The permutations of {0,1,2}, read as numbers, are {12, 21, 102, 120, 201, 210}. Subtracting the first one (12) from each of these numbers yields the differences {0, 9, 90, 108, 189, 198}. These are all multiples of 9, see comment and links. Dividing the differences by 9 yields {0, 1, 10, 12, 21, 22}, which are by definition the first six terms of this sequence.
Using all permutations of 0123 would yield 4!=24 terms, where the first 6 would be identical to those above. For n>10! we must consider permutations of (0,...,m-1) with m>10. These are no longer valid digits in base 10, and the numbers P(n) as defined in the comment are no longer equal to the concatenation. However, the first 10! terms obtained as (P(n)-P(1))/9, are still the same as for m=10;
In order to illustrate that the result is independent of the base chosen to make the calculation, let us consider permutations of 012 in base 3. The 3rd resp. 5th term ((102-012)/9=10 resp. (201-012)/9=21) would be ((1-0)*3^2+(0-1)*3+(2-2)*1)/(3-1) = 3 = 10[3], resp. ((2-0)*3^2+(0-1)*3+(1-2)*1)/(3-1) = 7 = 21[3]. The same terms would also be "10" and "21" if the calculation were made in base b=11. In that base, with digit "A" having the value b-1, we have (1023456789A - 0123456789A)/A = 0A000000000[11], (2013456789A - 0123456789A)/A = 02100000000[11], and (0A123456789 - 0123456789A)/A = 0A987654321[11] (the analog of (40123[5]-01234[5])/4[5] = 04321[5]). (End)

R. J. Cano, Table of n, a(n) for n = 1..10000
R. J. Cano, Additional information.
R. J. Cano, Template for a base-independent sequencer in C.
Wikipedia, Polynomial remainder theorem.

Cf. A207324, A217626, A211869.

Cf. A030299, A107346, A220664, A219664, A083449, A019566, A209280.

C
```
// See links.
```

N:= 100: # to get a(1)..a(N)
for M from 3 while M! <= N do od:
p0:= [$1..M]: p:= p0: A[1]:= 0:
for n from 2 to N do
  p:= combinat:-nextperm(p);
  d:= p - p0;
  A[n]:= add(10^(i-1)*d[-i],i=1..M)/9;
od:
seq(A[i],i=1..N); # Robert Israel, Apr 19 2017

maxm = 5; Table[dd = FromDigits /@ Permutations[Range[m]]; (Drop[dd, If[m == 1, 0, (m - 1)!]] - First[dd])/9, {m, 1, maxm}] // Flatten (* Jean-François Alcover, Apr 25 2013 *)

A215940(n)=for(k=2, n+1, k!M. F. Hasler, Jan 12 2013

first_n_factorial_terms(n)={my(u=n!);my(x=numtoperm(n,0),y,z=vector(u),i:small);i=0;for(j=0,u-1,y=numtoperm(n,j)-x;z[i++]=fromdigits(vector(#x-1,k,vecsum(y[1..k]))));z} \\ R. J. Cano, Apr 19 2017

a(n) = Sum_{k=1..n-1} A217626(k).

a(n) = (A050289(n)-A050289(1))/9, for n <= 9!. - M. F. Hasler, Jan 12 2013

Edited by M. F. Hasler, Jan 12 2013

Minor edits by N. J. A. Sloane, Feb 19 2013

cp's OEIS Frontend

A209280 First difference of A050289 = numbers whose digits are a permutation of (1,...,9).

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A204140 Numbers m such that both m and 7m are zeroless pandigital numbers (A050289).

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A204141 Numbers m such that both m and m/7 are zeroless pandigital numbers (A050289).

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A050278 Pandigital numbers: numbers containing the digits 0-9. Version 1: each digit appears exactly once.

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A171102 Pandigital numbers: numbers containing the digits 0-9. Version 2: each digit appears at least once.

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A144688 "Magic" numbers: all numbers from 0 to 9 are magic; a number >= 10 is magic if it is divisible by the number of its digits and the number obtained by deleting the final digit is also magic.

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A215940 Difference between the n-th and the first (identity) permutation of (0,...,m-1), interpreted as a decimal number, divided by 9 (for any m for which 10! >= m! >= n).

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