A000042 -id:A000042 - cp's OEIS frontend

A248323 Numbers n which appear at least twice in A037278(n), concatenation of their divisors written in base 10.

11, 12, 42, 84, 124, 135, 248, 325, 366, 510, 550, 555, 624, 650, 714, 1010, 1111, 1525, 1734, 2510, 3913, 4020, 5100, 5500, 5610, 5625, 8040, 11111, 13515, 16575, 21175, 24104, 25500, 28160, 34170, 35250, 35610, 36800, 37444, 44919, 50100, 51020, 51102, 51250, 52000

Offset: 1

Eric Angelini and M. F. Hasler, Oct 04 2014

Overlapping is allowed, so a(1) = 11 is in the sequence, with concatenated divisors A037278(11) = "111".
All repunits (10^k-1)/9 = A000042(k) = A002275(k) with even k = number of digits (as to be divisible by 11) but not multiples of 3, i.e., k in A047235, have divisors {1, 11, ..., 1010...101, 1111...111} and therefore are in this sequence.
Numbers n = floor(10^(8+3k)/7), k>=0, also belong to this sequence; for k>=2m, the number n appears (at least) m+2 times in A037278(n). [Found by extending results from Hans Havermann.]
The smallest terms that appear more than twice in the concatenation are 1111, 400200, 800400, 28571428, all 3 times, and 42857142, 4 times. - Hans Havermann, Oct 05 2014

			The divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42, and "42" appears twice in their concatenation A037278(42) = "12367142142".

Chai Wah Wu, Table of n, a(n) for n = 1..340
E. Angelini, Divisors showing a string, SeqFan list, Oct 03 2014

Cf. A037278.

is(n)={d=concat(apply(digits,divisors(n)));n=digits(n);for(j=0,#d-#n-1,for(i=1,#n,d[i+j]==n[i]||next(2));return(1))}

from sympy import divisors
import re
A248323_list = [n for n in range(1,10**7) if len(list(re.finditer('(?='+str(n)+')',''.join([str(d) for d in divisors(n)])))) > 1]
# Chai Wah Wu, Nov 01 2014

A332080 Irregular table in which row n = 1, 2, 3... lists the lexicographically first triangle of height and width n with minimal sum, distinct positive entries using only digits <= n and no diagonal having the same digit in two entries.

Original entry on oeis.org

1, 2, 1, 10, 3, 2, 20, 10, 1, 11, 4, 3, 30, 20, 2, 22, 1, 14, 10, 11, 5, 4, 44, 3, 33, 30, 2, 20, 25, 22, 1, 15, 14, 10, 11

Offset: 1

M. F. Hasler, Jul 02 2020

The triangles consist of n rows of length 1, 2, ..., n, respectively. Diagonals are considered in both directions, SE & SW. Alternatively, these are rows and columns if the triangle is the upper left half of an n X n matrix, entries read by rising antidiagonals.
To have digits up to n, we use base n+1 for the entries in row n. To make the data easier to read, we give the terms in the base-(n+1) representation. Up to row 9 or a(165) this is the same as if we used always base 10. For rows n >= 10^D, we can use D+1 decimal digits for each base-(n+1) digit.
There is always a "trivial" solution, non-minimal for n > 2, of the form:
            9
          8   88
       7   77   777
    ...           ...
  1   11  111 ...    11...1
Here, (9, 8, 7, ...) represent the digits (n, n-1, n-2, ...).
This solution, where the k-th row is filled with repdigits (A010785), can always be improved towards a smaller total sum of elements by replacing the largest terms in each row by smaller terms involving the additional digit '0' or a larger digit. In particular, one could replace 88 by 80 in the second row above; replace (7, 777) by (70, 7) in the 3rd row; (666, 6666) by (60, 666) in the 4th row. One can also avoid 666 by using 69 in the second position of the 4th row.
Then, no digit 0 would be possible in the 5th row, but one can still avoid 5555 and 55555 in favor of 58 and 59, etc. However, following this algorithm in a greedy manner will not always give a minimal solution, see examples for n >= 5.

			For n = 1, the triangle is reduced to a single number, [ 1 ].
For n = 2, we have the triangle:       2
  so row 2 = [ 2 ; 1, 10 ].          1   10
  (Obviously the symmetric triangle [2; 10, 1] has the same minimal sum, but it comes later in lexicographical order.)
For n = 3, we have the triangle to the right:               3
  This gives row 3 = [3; 2, 20; 10, 1, 11]                2   20
  with minimal sum = 47 (using base 10).               10   1    11
  (Using base 4 the sum is 113[4] = 23[10].)
For n = 4, we have the triangle:                   4
  The entries yield row 4 =                      3   30
  [4; 3, 30; 20, 2, 22; 1, 14, 10, 11].       20   2   22
  See below for the sum.                    1   14   10  11
This is the lexicographically earliest triangle for n = 5 with minimal sum. Indeed:
- We have to start with 4 to avoid the larger number 40 elsewhere in the table; using 40 somewhere would make the sum of the entries larger by 10 or more.
- For a similar reason, we use leading digit 3 in the second row. If we used leading digit 2 here, we would need an entry >= 33 in the 3rd row.
- We can't for example put 2 in the first place of the second row, because in the second and third place, no digit 0 may appear, since there is already the one from "30" in the second "SW/NE" diagonal and in the rightmost "NW/SE" diagonal. Thus, if we don't start the row with 20, we'd have to use 222 later in that row (or in the next one if we use a digit 1 in this 2nd row).
- In the last row we can use "10" only in the 3rd place because of the digits 0 in 20 and 30. But we can use 14 in the 2nd place. This achieves the minimal sum of 117 if we compute 4 + 3 + 30 + 20 + 2 + 22 + 1 + 14 + 10 + 11 in base 10.
  If we consider these as numbers written in base 5, the sum is 232[5] = 67[10].
  In any case this is the smallest possible sum.
For n = 5, we have the triangle    5
                                 4  44
                               3  33  30
                             2  20  25  22
                           1  15  14  10  11
The sum of these base-6 numbers is 423[6] = 159[10].
Here we do not use 40 in place of 44, which would allow only solutions with larger total sum: more precisely, one would then need to use a term >= 55 in the last row (or larger terms in earlier rows).
One possibly minimal solution for n = 6 could be:
                         5
                       6  66
                    30   3  40
                   4  44  55  33
                 2  25  26  20  22
               1  10  13  15  14  11,  sum in base 7: 652[7] = 331[10].
and for n = 7:
                          5
                        6   7
                     77  55  44
                   40   4   3  30
                  3  36  50  47  66
                2  20  27  26  22  222
              1  11  16  15  13  10  111,  sum in base 8: 1353[8] = 747[10].
A proof of minimality of these, using, e.g., exhaustive search with backtracking, would be appreciated.

Eric Angelini, Du recuit simulé (Simulated annealing), personal blog, June 26 2020.

Cf. A010785 (repdigit numbers), A000042 and A002275 (repunits).

Row lengths are A000217 (triangular numbers).

A332080(n,r,c,d=n-r+1)={if(c==1, d*10^(r==3&&n<5), !r, n>5 && warning("non-minimal result for n>5!"); [[self()(n,r,c)|c<-[1..r]]|r<-[1..n]], c==r, d*((r!=(n>4)+2)+10), r<4, d*11^(n>4), 1<#d=setminus(concat([0,[d+1..n],d*10]), Set(concat([digits(self()(n, r-abs(c-.5-k)\/1, min(k,c))) | k<-[1..r-1]]))), d[1]+d[#d], c<3, d[1]\10*111, until(9M. F. Hasler, Aug 16 2020

A046414 Repunit of length a(n) has exactly 3 prime factors (counted with multiplicity).

Original entry on oeis.org

13, 31, 37, 38, 67, 73, 83, 97, 101, 103, 113, 127, 149, 163, 191, 227, 241, 263, 313, 331, 373, 379

Offset: 1

Patrick De Geest, Jul 15 1998

467 <= a(23) <= 857. 857, 1303, 1483, 2267 are terms of this sequence. - Chai Wah Wu, Nov 03 2019

			a(n)=13 so 1111111111111 = 53*79*265371653.

P. De Geest, Repunits prime factors
Makoto Kamada, Factorizations of 11...11 (Repunit)
Eric Weisstein's World of Mathematics, Repunit

Cf. A000042, A001222, A002275, A004022, A046053.

is(n)=bigomega(10^n\9)==3 \\ Charles R Greathouse IV, Mar 11 2014

More terms from Robert Gerbicz, Nov 22 2010

More terms from Bo Gyu Jeong, Jun 12 2012

A046430 Lengths of repunits with all prime factors ending with the digit 1.

Original entry on oeis.org

1, 2, 4, 5, 10, 19, 20, 23, 25, 38, 50, 59, 76, 95, 100, 115, 125, 190, 250, 295, 317, 380

Offset: 1

Patrick De Geest, Jul 15 1998

			a(n) = 10 -> 11*41*271*9091 -> 1(1)*4(1)*27(1)*909(1).

P. De Geest, Repunits prime factors
Eric Weisstein's World of Mathematics, Repunit

Cf. A000042, A004022, A046053.

Select[Range[60],Length[fi=FactorInteger[(10^#-1)/9]]==Length[Select[First/@fi,Mod[#,10]==1&]] &](* Ray Chandler, Apr 24 2017 *)

a(16)-a(21) from Ray Chandler, Apr 24 2017
a(22) from Max Alekseyev, May 14 2022
Definition corrected by Max Alekseyev, Aug 19 2024

A060002 Digitized partition numbers: numbers with (weakly) increasing digits ordered by sum of their digits then by the numbers themselves.

Original entry on oeis.org

0, 1, 2, 11, 3, 12, 111, 4, 13, 22, 112, 1111, 5, 14, 23, 113, 122, 1112, 11111, 6, 15, 24, 33, 114, 123, 222, 1113, 1122, 11112, 111111, 7, 16, 25, 34, 115, 124, 133, 223, 1114, 1123, 1222, 11113, 11122, 111112, 1111111, 8, 17, 26, 35, 44, 116, 125, 134, 224

Offset: 0

Henry Bottomley, Mar 14 2001

a(97) cannot be written in decimal since it requires ten to be written as a single digit.

Number of values of a(n) with a digit sum of k is A000041, number of values of a(n) with a digit sum of k and m digits is A008284, a(A000070(n))=n+1 written as a single digit, a(A026905(n))=A000042(n).

A063431 Square array read by antidiagonals of n written in base k (n,k>0).

Original entry on oeis.org

1, 1, 11, 1, 10, 111, 1, 2, 11, 1111, 1, 2, 10, 100, 11111, 1, 2, 3, 11, 101, 111111, 1, 2, 3, 10, 12, 110, 1111111, 1, 2, 3, 4, 11, 20, 111, 11111111, 1, 2, 3, 4, 10, 12, 21, 1000, 111111111, 1, 2, 3, 4, 5, 11, 13, 22, 1001, 1111111111, 1, 2, 3, 4, 5, 10, 12, 20, 100

Offset: 1

Henry Bottomley, Jul 20 2001

It is difficult to write ten in base 11 using decimal digits.

			Rows start (1, 1, 1, 1, 1,...), (11, 10, 2, 2, 2,...), (111, 11, 10, 3, 3,...), (1111, 100, 11, 10, 4,...), etc.

Cf. A063432. Rows start effectively as the reverse of A001731, A001732, A001733, A001734, A001735, A001736, A008707, A008708, A008709, A008710, A008711, A008712, A008713, A008714, A008715, A008716, A008717 etc. Columns are A000042, A007088, A007089, A007090, A007091, A007092, A007093, A007094, A007095, A000027 and perhaps A055649 etc.

A063750 Number of divisors (A000005) of the Wonderful Demlo numbers A002477.

Original entry on oeis.org

1, 3, 9, 9, 9, 243, 9, 81, 45, 81, 9, 2187, 27, 81, 729, 729, 9, 10935, 3, 2187, 2187, 1215, 3, 59049, 243, 729, 567, 6561, 243, 1594323, 27, 177147, 729, 729, 2187, 295245, 27, 27, 729, 177147, 81, 7971615, 81, 98415, 32805, 729

Offset: 1

Jason Earls, Aug 11 2001

nonn

Amiram Eldar, Table of n, a(n) for n = 1..352 (terms 1..80 from Harry J. Smith)

Cf. A000005, A000042, A002477, A048691.

a:=n->((10^n-1)/9)^2;;  A063750:=List([1..50],n->Tau(a(n))); # Muniru A Asiru, Feb 17 2018

a:=n->((10^n-1)/9)^2: A063750:=[seq(numtheory[tau](a(n)),n=1..50)]; # Muniru A Asiru, Feb 17 2018

Array[DivisorSigma[0, FromDigits[PadRight[{}, #, 1]]^2] &, 46] (* Michael De Vlieger, Nov 18 2017, after Harvey P. Dale at A002477 *)

vector(45,n, numdiv((10^n\9)^2) ) \\ Edited by M. F. Hasler, Nov 18 2017

apply( A063750=n->numdiv((10^n\9)^2), [1..80]) \\ M. F. Hasler, Nov 18 2017

a(n) = A000005(A002477(n)) = d(((10^n-1)/9)^2). - M. F. Hasler, Nov 18 2017

a(n) = A048691(A000042(n)). - Amiram Eldar, Nov 19 2024

Edited and offset corrected by M. F. Hasler, Nov 18 2017

A063751 a(n) = phi(A002477(n)).

Original entry on oeis.org

1, 110, 7992, 1111000, 119998800, 5759994240, 1229137654864, 108799998912000, 8007983991992016, 1090799999890920000, 123450846932098824864, 5702399999994297600000, 1195941578346547072508832, 111739686239998882603137600, 7532766719999992467233280000

Offset: 1

Jason Earls, Aug 11 2001

nonn

Amiram Eldar, Table of n, a(n) for n = 1..352 (terms 1..81 from Harry J. Smith)

Cf. A000010, A000042, A002477.

EulerPhi[((10^Range[20] - 1)/9)^2] (* Paolo Xausa, Nov 19 2024 *)

j=[]; for(n=0,25,j=concat(j,eulerphi( ((10^(n+1)-1)/9)^2) )); j

{ for (n=0, 80, a=eulerphi(((10^(n + 1) - 1)/9)^2); write("b063751.txt", n, " ", a) ) } \\ Harry J. Smith, Aug 29 2009

a(n) = A000042(n) * phi(A000042(n)). - Amiram Eldar, Nov 19 2024

Offset corrected by Amiram Eldar, Nov 19 2024

A083957 a(n) = product of all cyclic permutation of concatenation of first n numbers. In each case the digits of a number are kept together for n>9.

Original entry on oeis.org

1, 252, 8864856, 40638618022744, 23098236228791472696480, 1568841035670687222903484318855680, 12397389650957442295025809189993580551236444160

Offset: 1

Amarnath Murthy, May 10 2003

Initial terms are {n(n+1)/2}*{A000042(n)}.

			a(1) = 1, a(2) = 12*21 = 252, a(3) = 123*231*312 = 8864856.

Cf. A000042, A083956.

More terms from David Wasserman, Dec 03 2004

A088330 Sum of the remainders when n is divided by nonzero numbers obtained by deleting one digit. The sum ranges over all the digits.

Original entry on oeis.org

0, 0, 0, 1, 2, 0, 4, 3, 2, 1, 0, 1, 0, 3, 0, 1, 2, 7, 4, 3, 0, 1, 2, 0, 3, 2, 0, 3, 8, 3, 0, 1, 2, 4, 0, 1, 6, 8, 0, 5, 0, 1, 2, 5, 6, 0, 3, 3, 5, 9, 0, 1, 2, 3, 4, 5, 0, 5, 6, 9, 0, 1, 2, 4, 6, 5, 10, 0, 7, 9, 0, 1, 2, 5, 4, 5, 8, 10, 0, 9, 0, 1, 2, 3, 6, 5, 6, 13, 10, 0, 0, 3, 8, 16, 10, 5, 20, 14, 12, 24, 0

Offset: 10

Amarnath Murthy, Oct 01 2003

Differs from A067458 first at n=101, where A067458(101)=0 and a(101) = (101 mod 1) + (101 mod 11) + (101 mod 10) = 0+2+1 = 3. - R. J. Mathar, Sep 11 2008

			a(1234) = Rem[1234/123] + Rem[1234/124]+ Rem[1234/134] + Rem[1234/234] = 4+ 118 + 28 + 64 = 214 where Rem [a/b] = the remainder when a is divided by b.

Robert Israel, Table of n, a(n) for n = 10..10000

Cf. A000042.

f:= proc(n) local L,d,i,j,x,t;
     L:= convert(n,base,10); d:= nops(L); t:= 0;
     for i from 1 to d do
       x:= add(L[j]*10^(j-1),j=1..i-1) + add(L[j]*10^(j-2),j=i+1..d);
       if x <> 0 then t:= t + (n mod x) fi;
     od;
     t
end proc:
map(f, [$10 .. 200]); # Robert Israel, Dec 05 2024

a((10^n - 1)/9) = n. for n > 2. a(1111111 n times ) = a(A000042(n)) = n, n > 2.

a(10 * n) = 10 * a(n). - Robert Israel, Dec 05 2024

More terms from Ray Chandler, Oct 06 2003

cp's OEIS Frontend

A248323 Numbers n which appear at least twice in A037278(n), concatenation of their divisors written in base 10.

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A332080 Irregular table in which row n = 1, 2, 3... lists the lexicographically first triangle of height and width n with minimal sum, distinct positive entries using only digits <= n and no diagonal having the same digit in two entries.

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A046414 Repunit of length a(n) has exactly 3 prime factors (counted with multiplicity).

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A046430 Lengths of repunits with all prime factors ending with the digit 1.

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A060002 Digitized partition numbers: numbers with (weakly) increasing digits ordered by sum of their digits then by the numbers themselves.

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A063431 Square array read by antidiagonals of n written in base k (n,k>0).

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A063750 Number of divisors (A000005) of the Wonderful Demlo numbers A002477.

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A063751 a(n) = phi(A002477(n)).

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