A083449 -id:A083449 - cp's OEIS frontend

A277830 Number of digits '0' in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

1, 1, 2, 23, 344, 4665, 58986, 713307, 8367628, 96021949, 1083676272, 12071330614, 133058985146, 1454046641578, 15775034317010, 170096022182442, 1824417011947874, 19478738020713306, 207133059219478738, 2194787382318244170, 23182441724417009624, 244170096256515775267

Offset: 0

M. F. Hasler, Nov 01 2016

The first 10 terms are given by a simple explicit formula and linear recurrence, which does not hold for n > 9. Note that A007908 (concat(1..n)) differs from A014824 (a(n) = a(n-1)*10 + n) for n > 9. - M. F. Hasler, Nov 07 2020

M. F. Hasler, Table of n, a(n) for n = 0..100

Cf. A277831 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=1);N=0;for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==0,digits(k))))) \\ For purpose of illustration.

apply( A277830(n)={A061217(A014824(n)+!n)+1}, [0..22]) \\ Thanks to Kevin Ryde's formula. - M. F. Hasler, Nov 07 2020

a(n) = A083449(n) + 1 for n <= 9.

a(n) = 1 + A061217(A014824(n)), taking A061217(0)=0. - Kevin Ryde, Nov 07 2020

Incorrect data, b-file, links, formulas and programs deleted by M. F. Hasler, following observations by Kevin Ryde, Nov 07 2020

A277849 Number of digits '9' in the set of all numbers from 0 to A014824(n) = sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713306, 8367627, 96021949, 1083676281, 12071330713, 133058986145, 1454046651577, 15775034417009, 170096023182441, 1824417021947873, 19478738120713305, 207133060219478737, 2194787392318244180, 23182441824417009723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n = 2 there is only one digit '9' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '9' in { 19, 29, ..., 89, 90, ..., 99, 109, 119 }, where 99 accounts for two '9's.

Lars Blomberg, Table of n, a(n) for n = 0..997

Cf. A277830 - A277838, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==9,digits(k)))))

A014824(n)=(10^n-1)*(10/81)-n/9;
A102684(n)=my(pow,f,g,h);sum(j=1,#Str(n),pow=10^j;f=floor(n/pow);g=floor(n/pow+1/10);h=(4/5+g)*pow;g*(2*n+2-h)-f*(2*n+2-(1+f)*pow))/2;
A277849(n)=A102684(A014824(n));
vector(50,n,A277849(n-1)) \\ Lars Blomberg, Nov 11 2020

a(n) = A083449(n) = A277830(n) - 1 for 0 < n < 9.
a(n) = A277838(n) for n < 8, and a(8) = A277838(8) - 1.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Replaced incorrect b-file by Lars Blomberg, Nov 11 2020

A277635 Number of 7's appearing in the sequence of consecutive natural numbers from 1 to A007908(n), where A007908 = (1, 12, 123, 1234, ...).

Original entry on oeis.org

0, 1, 22, 343, 4664, 58985, 713307, 8367637, 96022049

Offset: 1

Alexander R. Povolotsky, Oct 24 2016

First 6 terms are the same as in A083449, also see A272525. [See the OEIS wiki page for more details. - M. F. Hasler, Dec 29 2020]
a(n) gives the number of times the digit 7 occurs in all terms of A000027 in the interval [A000027(1), A007908(n)]. - Felix Fröhlich, Oct 28 2016
The sequence was initially defined only up to n = 9 and then extended using A007908 = concat(1..n); see A277837 for the extension using A014824 (a(n) = 10 a(n-1) + n) leading to a smoother growth, in particular at powers of 10. - M. F. Hasler, Nov 01 2016, edited Dec 29 2020

			22 is the third term of the sequence because there are 22 occurrences of the digit '7' contained in numbers within the range of 1 to 123.
96022049 is the 9th term of the sequence because there are 96022049 occurrences of the digit '7' contained in numbers within the range of 1 to 123456789.

Puzzling Stack Exchange, How many sevens?
M. F. Hasler, Digits d in 0 through 123...n, OEIS Wiki, Nov. 2016.

Cf. A083449, A272525, A014824.

Cf. A277830 - A277838 and A277849: analog for digits 0 .. 9, but based on A014824 instead of A083449.

Table[a[n] = Count[Flatten@ Map[IntegerDigits, Range@ FromDigits@ Range@ n], k_ /; k == 8]; Print@ a@ n; an = a[n]; an, {n, 0, 9}] (* Michael De Vlieger, Oct 30 2016 *)

print1(c=0);N=1;for(n=2,8,print1(","c+=sum(k=N+1,N=eval(Str(N,n)),#select(d->d==7,digits(k))))) \\ For illustration; more efficient code below. - M. F. Hasler, Oct 31 2016

A277635(n, m=7)=if(n>m,A277635(n, m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ Valid only for n <= 9. - M. F. Hasler, Nov 02 2016

A277838 Number of '8' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713306, 8367628, 96021959, 1083676380, 12071331701, 133058996022, 1454046750343, 15775035404664, 170096033058985, 1824417120713306, 19478739108367627, 207133070096021958, 2194787491083676380, 23182442812071331701

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '8' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '8' in { 18, 28, ..., 78, 80, ..., 89, 98, 108, 118 }, where 88 accounts for two '8's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277837, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==8,digits(k)))))

A277838(n,m=8)=if(n>m,A277838(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 8, a(8) = A277849(8) + 1 = A277837(8) - 9.

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

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

A019566 The differences 1-1, 21-12, 321-123, ..., 10987654321-12345678910, 1110987654321-1234567891011, etc.

Original entry on oeis.org

0, 9, 198, 3087, 41976, 530865, 6419754, 75308643, 864197532, -1358024589, -123580236690, -2345801446791, 775432077543108, 178553219976533007, 27956332009875522906, 3805734210999774512805, 481583522109989673502704, 58259362312008979572492603

Offset: 1

Robert Dickau

All terms are divisible by 9, cf. A083449. There is an increasingly longer subsequence of negative terms starting at each power of 10, namely for indices n = 10..12, 100..123, 1000..1234, etc. - M. F. Hasler, Nov 02 2016

Gupta (1988) calls these "unique numbers".

S. S. Gupta, Unique Numbers, Science Today, Jan 01 1988, India.

Alois P. Heinz, Table of n, a(n) for n = 1..350
Shyam Sunder Gupta, Unique Numbers
Shyam Sunder Gupta, On Some Marvellous Numbers of Kaprekar, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 9, 275-315. See Section 9.4.

Cf. A000422, A007908.

u:= proc(n) u(n):= `if`(n=1, 1, parse(cat(u(n-1), n))) end:
d:= proc(n) d(n):= `if`(n=1, 1, parse(cat(n, d(n-1)))) end:
a:= n-> d(n)-u(n):
seq(a(n), n=1..20);  # Alois P. Heinz, Dec 06 2014

f[n_] := Block[ {a = "", k = 1}, While[k < n + 1, a = StringJoin[ ToString[k], a]; k++ ]; Return[ ToExpression[a] - ToExpression[ StringReverse[a]]]]; Table[ f[n], {n, 1, 17} ]

A = vector(25); c = 1; f = 1; for (i = 2, 9, c = 10*c + i; f = f + i*10^(i - 1); A[i] = (f - c)); for (i = 10, 25, c = 100*c + i; f = f + i*10^(2*i - 11);; A[i] = (f - c)); A \\ David Wasserman, Nov 09 2004

apply( {A019566(n)=A000422(n)-A007908(n)}, [1..22]) \\ Replacing code from Jan 13 2013, following a comment from Nov 02 2016. - M. F. Hasler, Nov 07 2020

a(n) = A000422(n) - A007908(n) = 9*A083449(n).

More terms from Robert G. Wilson v, Jan 11 2002
More terms from David Wasserman, Nov 09 2004
Edited by N. J. A. Sloane, Nov 22 2020

A277837 Number of '7' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713307, 8367637, 96022049, 1083677281, 12071340713, 133059086145, 1454047651577, 15775044417009, 170096123182441, 1824418021947873, 19478748120713314, 207133160219478837, 2194788392318245180, 23182451824417019723

Offset: 0

M. F. Hasler, Nov 01 2016

This sequence is very similar (actually equal, for 1 <= n <= 9) to A277635, which was the original motivation for considering the family A277830 - A277838 and A277849. The main difference is that A277635 is based on A007908 (where 123456789 is followed by 12345678910) while this family is based on A014824, starts as the latter at offset 0, and therefore has a strongly different growth for n > 9.

			For n=2 there is only one digit '7' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '7' in { 17, 27, ..., 67, 70, ..., 79, 87, 97, 107, 117 }, where 77 accounts for two '7's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==7,digits(k)))))

A277837(n,m=7)=if(n>16,error("n>16 not yet implemented"), n>m,A277837(n,m+1)+(m+2)*10^(n-m-1),(9*n-11)*(10^n+1)\729+2-(m>n)) \\ Edited by M. F. Hasler, Dec 29 2020

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 7,
a(n) = A277836(n) - 8*10^(n-7) [for n >= 7] = A277838(n) + 9*10^(n-8) [for n >= 8].
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277831 Number of '1' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 1, 5, 57, 689, 8121, 93553, 1058985, 11824417, 130589849, 1429355281, 15528120716, 167626886179, 1799725651922, 19231824420465, 204663923217008, 2170096022293551, 22935528124170094, 241700960254046637, 2540466392663923180, 26639231827873799724

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 are counted the same '1' as for n=1, plus the 4 additional digits '1' in 10, 11 and 12.

David A. Corneth, Table of n, a(n) for n = 0..998
M. F. Hasler, "Digits d in 0 through 123...n", OEIS Wiki, Nov. 2016.

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==1,digits(k)))))

A277831(n)=if(n<2,n, n<11, A277832(n)+3*10^(n-2), error("n > 10 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

a(n) = A277832(n) + 3*10^(n-2), for 2 <= n <= 10.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277832 Number of '2' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 2, 27, 389, 5121, 63553, 758985, 8824417, 100589849, 1129355281, 12528120713, 137626886149, 1499725651622, 16231824417465, 174663923187008, 1870096021993551, 19935528121170094, 211700960224046637, 2240466392363923180, 23639231824873799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 are counted the two '2's in { 2, 12 }.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A102685, A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Array[Total@ DigitCount[Range[Sum[10^i - 1, {i, #}]/9], 10, 2] &, 7] (* Michael De Vlieger, Dec 31 2020 *)

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==2,digits(k)))))

A277832(n)=if(n<3,(n==2)*2, n<13,A277833(n)+4*10^(n-3), error("n > 12 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

a(n) = {if(n == 0, return(0)); n = (10^(n+1)\9-n)\9; f(n, 2) }
f(n, {c = 2}) = { my(d = digits(n), res = 0); for(i = 1, #d - 1, res += d[i] * (#d - i)*10^(#d - i - 1); if(d[i]==c, res+=(n % (10^(#d - i)) + 1); ); if(d[i] > c, res+=(10^(#d - i)) ); ); if(d[#d] >= c, res++); res } \\ David A. Corneth, Dec 31 2020

a(n) = A277831(n) - 3*10^(n-2) [for n >= 2] = A277833(n) + 4*10^(n-3) [n >= 3].

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

cp's OEIS Frontend

A272525 Convolution of nonzero repunits (A002275) with themselves.

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A277830 Number of digits '0' in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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A277849 Number of digits '9' in the set of all numbers from 0 to A014824(n) = sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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A277635 Number of 7's appearing in the sequence of consecutive natural numbers from 1 to A007908(n), where A007908 = (1, 12, 123, 1234, ...).

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A277838 Number of '8' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

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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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A019566 The differences 1-1, 21-12, 321-123, ..., 10987654321-12345678910, 1110987654321-1234567891011, etc.

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