A000533 -id:A000533 - cp's OEIS frontend

A344897 a(n) is the number of divisors of 10^n + 1.

2, 2, 2, 8, 4, 4, 4, 4, 4, 32, 8, 24, 8, 8, 16, 128, 32, 16, 8, 4, 16, 192, 16, 32, 8, 32, 8, 128, 16, 8, 128, 4, 16, 384, 16, 32, 64, 16, 8, 768, 16, 8, 128, 16, 16, 4096, 16, 16, 512, 16, 128, 256, 16, 4, 64, 768, 32, 64, 32, 16, 64, 8, 8, 3072, 8, 64, 256, 4, 16, 1024, 2048, 8, 32, 16, 128, 2048, 64, 3072, 128, 16

Offset: 0

Seiichi Manyama, Jun 01 2021

nonn

a(n) is even because 10^n + 1 is not a square number.

Max Alekseyev, Table of n, a(n) for n = 0..331

Cf. A000005, A000533, A003021, A038371, A046798, A057934, A070528, A119704, A344859, A087021, A087022,

a[0] = 2; a[n_] := DivisorSigma[0, 10^n + 1]; Array[a, 60, 0] (* Amiram Eldar, Jun 01 2021 *)

PARI
```
a(n) = numdiv(10^n+1);
```

a(n) = A000005(A000533(n)).

A066138 a(n) = 10^(2*n) + 10^n + 1.

Original entry on oeis.org

3, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001, 1000000000000001000000000000001

Offset: 0

Henry Bottomley, Dec 07 2001

Palindromes whose digit sum is 3.
Essentially the same as A135577. - R. J. Mathar Apr 29 2008
From Peter Bala, Sep 25 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(n)) = [10^n; 1, 1, 2/3*(10^n - 1), 1, 1, 2*10^n, ...] has period 6. As n increases, the expansion has the large partial quotients 2/3*(10^n - 1) and 2*10^n.
For n >= 1, the continued fraction expansion of sqrt(a(2*n))/a(n) = [0; 1, 10^n - 1, 1, 1, 1/3*(10^n - 4), 1, 4, 1, 1/3*(10^n - 4), 1, 1, 10^n - 1, 2, 10^n - 1, ...] has pre-period of length 3 and period 12 beginning 1, 1, 1/3*(10^n - 4), ....  As n increases, the expansion has the large partial quotients 10^n - 1 and 1/3*(10^n - 4).
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n), for m >= 3. For example, it appears that the continued fraction expansion of a(3*n)^(1/3), for n >= 2, begins [10^(2*n); 3*10^n - 1, 1, 0.5*10^(2*n) - 1, 1.44*10^n - 1, 1, ...]. Cf. A000533, A002283 and A168624. (End)

			From _Peter Bala_, Sep 25 2015: (Start)
Simple continued fraction expansions showing large partial quotients:
a(9)^(1/3) =[1000000; 2999, 1, 499999, 1439, 1, 2582643, 1, 1, 1, 2, 3, 3, ...].
a(20)^(1/4) = [10000000000; 39999999999, 1, 3999999999, 16949152542, 2, 1, 2, 6, 1, 4872106, 3, 9, 2, 3, ...].
a(25)^(1/5) = [10000000000; 4999999999999999, 1, 3333333332, 2, 1, 217391304347825, 2, 2, 1, 1, 1, 2, 1, 23980814, 1, 1, 1, 1, 1, 7, ...]. (End)

Harry J. Smith, Table of n, a(n) for n=0..100
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000533, A002283, A033934, A062397, A074992, A135577, A168624.

[10^(2*n) + 10^n + 1: n in [0..20]]; // Vincenzo Librandi, Sep 27 2015

Table[10^(2 n) + 10^n + 1, {n, 0, 15}] (* Michael De Vlieger, Sep 27 2015 *)
CoefficientList[Series[(3 - 222 x + 1110 x^2)/((1 - 100 x) (1 - 10 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 27 2015 *)

a(n) = { 10^(2*n) + 10^n + 1 } \\ Harry J. Smith, Feb 02 2010

Vec(-3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

A168624(n) = a(2*n)/a(n). - Peter Bala, Sep 24 2015
G.f.: (3 - 222*x + 1110*x^2)/((1 - 100*x)*(1 - 10*x)*(1 - x)). - Vincenzo Librandi, Sep 27 2015
From Colin Barker, Sep 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
G.f.: -3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
From Elmo R. Oliveira, Aug 27 2024: (Start)
E.g.f.: exp(x)*(exp(99*x) + exp(9*x) + 1).
a(n) = 3*A074992(n). (End)

Offset changed from 1 to 0 by Harry J. Smith, Feb 02 2010

More terms from Michael De Vlieger, Sep 27 2015

A138144 Palindromes formed from the reflected decimal expansion of the concatenation of 1, 1 and infinite 0's.

Original entry on oeis.org

1, 11, 111, 1111, 11011, 110011, 1100011, 11000011, 110000011, 1100000011, 11000000011, 110000000011, 1100000000011, 11000000000011, 110000000000011, 1100000000000011, 11000000000000011, 110000000000000011

Offset: 1

Omar E. Pol, Mar 29 2008

a(n) is also A147595(n) written in base 2. [From Omar E. Pol, Nov 08 2008]

			n .... a(n)
1 .... 1
2 .... 11
3 .... 111
4 .... 1111
5 .... 11011
6 .... 110011
7 .... 1100011
8 .... 11000011
9 .... 110000011
10 ... 1100000011

Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000533.

Cf. A138118, A138119, A138120, A138145, A138146, A138721, A138826, A147595. [From Omar E. Pol, Nov 08 2008]

LinearRecurrence[{11,-10},{1,11,111,1111,11011},20] (* Harvey P. Dale, Aug 21 2016 *)

Vec(-x*(10*x^2-1)*(10*x^2+1)/((x-1)*(10*x-1)) + O(x^100)) \\ Colin Barker, Sep 15 2013

a(n) = 11+11*10^(n-2) for n>3. a(n) = 11*a(n-1)-10*a(n-2). G.f.: -x*(10*x^2-1)*(10*x^2+1) / ((x-1)*(10*x-1)). - Colin Barker, Sep 15 2013

Better definition. - Omar E. Pol, Nov 15 2008

A329914 Numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k.

Original entry on oeis.org

21, 23, 27, 29, 33, 39, 57, 59, 69, 71, 83, 87, 99, 101, 107

Offset: 1

Bernard Schott, Nov 24 2019

The idea of this sequence comes from the 21-digit integer 112359550561797732809 in Penguin dictionary (see reference) with this property: "The smallest number which, when 1 is placed at both ends, the number is multiplied by 99". The terms of this sequence are the other numbers k that have the same property than 99 and the corresponding smallest numbers M in each set {M_k} are in A329915 (see link).
The Diophantine equation to solve is 1M1 = k * M  with M that has q digits, this is equivalent to 10^(q+1) + 1 = (k-10) * M, with number of zeros in 10^(q+1) + 1 = q also.
Some results coming from this Diophantine equation:
q >= 2 and 21 <= k <= 110, so this sequence is finite. The integers (k-10) end with 1, 3, 7 or 9, hence k also.
Integer (k-10) must be a divisor of 10^(q+1)+1 = A000533(q+1).
For k = 21, there is 21 * 91 = 1[91]1 but also 21 * 9091 = 1[9091]1; hence, 91 and 9091 are terms of M_21.
Since 10^(q+1)+1 mod (k-10) is periodic and the period length cannot exceed k-10, it is easy to check that the sequence is indeed full. - Giovanni Resta, Nov 26 2019

			23 * 77 = 1[77]1, so k = 23 is a term and 13 * 77 = 1001; remark: number M = 77 has 2 digits and 10^3+1 has 2 zeros also.
29 * 52631579 = 1[52631579]1, so 29 is a term et 19 * 52631579 = 10^9 + 1 = 1000000001.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Bernard Schott, Array with values of (q,k,M)

Cf. A000533, A329915 (corresponding numbers M).

Select[Range[21, 110], GCD[10, # - 10] == 1 && MemberQ[Mod[10^Range[#] + 1, # - 10], 0] &] (* Giovanni Resta, Nov 26 2019 *)

A349194 a(n) is the product of the sum of the first i digits of n, as i goes from 1 to the total number of digits of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 49, 56, 63, 70, 77

Offset: 1

Malo David, Nov 10 2021

The only primes in the sequence are 2, 3, 5 and 7. - Bernard Schott, Nov 23 2021

			For n=256, a(256) = 2*(2+5)*(2+5+6) = 182.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Malo David, About the Product Sum Digit Sequence

Cf. A055642, A284001 (binary analog), A349190 (fixed points).
Cf. A007953 (sum of digits), A059995 (floor(n/10)).
Cf. A349278 (similar, with the last digits).
Cf. A349898, A349899.

f:=func; [f(n):n in [1..100]]; // Marius A. Burtea, Nov 23 2021

Table[Product[Sum[Part[IntegerDigits[n],j],{j,i}],{i,Length[IntegerDigits[n]]}],{n,74}] (* Stefano Spezia, Nov 10 2021 *)

a(n) = my(d=digits(n)); prod(i=1, #d, sum(j=1, i, d[j])); \\ Michel Marcus, Nov 10 2021

first(n)=if(n<9,return([1..n])); my(v=vector(n)); for(i=1,9,v[i]=i); for(i=10,n, v[i]=sumdigits(i)*v[i\10]); v \\ Charles R Greathouse IV, Dec 04 2021

from math import prod
from itertools import accumulate
def a(n): return prod(accumulate(map(int, str(n))))
print([a(n) for n in range(1, 100)]) # Michael S. Branicky, Nov 10 2021

For n>10: a(n) = a(A059995(n))*A007953(n) where A059995(n) = floor(n/10).
In particular, for n<100: a(n) = floor(n/10)*A007953(n)
From Bernard Schott, Nov 23 2021: (Start)
a(n) = 1 iff n = 10^k, k >= 0 (A011557).
a(n) = 2 iff n = 10^k + 1, k >= 0 (A000533 \ {1}).
a(n) = 3 iff n = 10^k + 2, k >= 0 (A133384).
a(n) = 5 iff n = 10^k + 4, k >= 0.
a(n) = 7 iff n = 10^k + 6, k >= 0. (End)
From Marius A. Burtea, Nov 23 2021: (Start)
a(A002275(n)) = n! = A000142(n), n >= 1.
a(A090843(n - 1)) = (2*n - 1)!! = A001147(n), n >= 1.
a(A097166(n)) = (3*n - 2)!!! = A007559(n).
a(A093136(n)) = 2^n = A000079(n).
a(A093138(n)) = 3^n = A000244(n). (End)

A037156 a(n) = 10^n*(10^n+1)/2.

Original entry on oeis.org

1, 55, 5050, 500500, 50005000, 5000050000, 500000500000, 50000005000000, 5000000050000000, 500000000500000000, 50000000005000000000, 5000000000050000000000, 500000000000500000000000, 50000000000005000000000000, 5000000000000050000000000000

Offset: 0

Marvin Ray Burns

Sum of first 10^n positive integers. - Omar E. Pol, May 03 2015

			From _Omar E. Pol_, May 03 2015: (Start)
For n = 0; a(0) = 1                       =    1 * 1   = 1
For n = 1; a(1) = 1 + 2 + ...... + 9 + 10 =   11 * 5   = 55
For n = 2; a(2) = 1 + 2 + .... + 99 + 100 =  101 * 50  = 5050
For n = 3; a(3) = 1 + 2 + .. + 999 + 1000 = 1001 * 500 = 500500
...
(End)

C. Pickover, Wonders of Numbers, Oxford University Press, NY, 2001, p. 328.

Brian Hayes, Gauss's Day of Reckoning, American Scientist
Bill Johnson, The great Gauss summation trick
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
Index entries for linear recurrences with constant coefficients, signature (110, -1000).

A subsequence of the triangular numbers A000217.

Cf. A038544.

LinearRecurrence[{110,-1000},{1,55},20] (* Harvey P. Dale, Oct 11 2023 *)

a(n) = A000533(n) * A093143(n). - Omar E. Pol, May 03 2015
From Chai Wah Wu, May 28 2016: (Start)
a(n) = 110*a(n-1) - 1000*a(n-2).
G.f.: (1 - 55*x)/((10*x - 1)*(100*x - 1)).
(End)
a(n) = sqrt(A038544(n)). - Bernard Schott, Jan 20 2022

Corrected by T. D. Noe, Nov 07 2006

A046851 Numbers n such that n^2 can be obtained from n by inserting internal (but not necessarily contiguous) digits.

Original entry on oeis.org

0, 1, 10, 11, 95, 96, 100, 101, 105, 110, 125, 950, 960, 976, 995, 996, 1000, 1001, 1005, 1006, 1010, 1011, 1021, 1025, 1026, 1036, 1046, 1050, 1100, 1101, 1105, 1201, 1205, 1250, 1276, 1305, 1316, 1376, 1405, 9500, 9505, 9511, 9525, 9600, 9605, 9625

Offset: 1

David W. Wilson

Contains A038444.  In particular, the sequence is infinite. - Robert Israel, Oct 20 2016
If n is any positive term, then b_n(k) := n*10^k (k >= 0) is an infinite subsequence. - Rick L. Shepherd, Nov 01 2016
From Robert Israel's comment it follows that the subsequence of terms with no trailing zeros is also infinite (contains A000533). - Rick L. Shepherd, Nov 01 2016

			110^2 = 12100 (insert "2" and "0" into "1_1_0").

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A045953, A008851, A018834, A038444, A086457 (subsequence).

import Data.List (isInfixOf)
a046851 n = a046851_list !! (n-1)
a046851_list = filter chi a008851_list where
   chi n = (x == y && xs `isSub` ys) where
      x:xs = show $ div n 10
      y:ys = show $ div (n^2) 10
   isSub [] ys       = True
   isSub _  []       = False
   isSub us'@(u:us) (v:vs)
         | u == v    = isSub us vs
         | otherwise = isSub us' vs
-- Reinhard Zumkeller, Jul 27 2011

IsSublist:= proc(a, b)
  local i,bp,j;
  bp:= b;
  for i from 1 to nops(a) do
    j:= ListTools:-Search(a[i],bp);
    if j = 0 then return false fi;
    bp:= bp[j+1..-1];
  od;
  true
end proc:
filter:= proc(n) local A,B;
  A:= convert(n,base,10);
  B:= convert(n^2,base,10);
  if not(A[1] = B[1] and A[-1] = B[-1]) then return false fi;
  if nops(A) <= 2 then return true fi;
  IsSublist(A[2..-2],B[2..-2])
end proc:
select(filter, [$0..10^4]); # Robert Israel, Oct 20 2016

id[n_]:=IntegerDigits[n];
insQ[n_]:=First[id[n]]==First[id[n^2]]&&Last[id[n]]==Last[id[n^2]];
sort[n_]:=Flatten/@Table[Position[id[n^2],id[n][[i]]],{i,1,Length[id[n]]}];
takeQ[n_]:=Module[{lst={First[sort[n][[1]]]}},
   Do[
    Do[
     If[Last[lst]Ivan N. Ianakiev, Oct 19 2016 *)

A147759 Palindromes formed from the reflected decimal expansion of the infinite concatenation of 1's and 0's.

Original entry on oeis.org

1, 11, 101, 1001, 10101, 101101, 1010101, 10100101, 101010101, 1010110101, 10101010101, 101010010101, 1010101010101, 10101011010101, 101010101010101, 1010101001010101, 10101010101010101, 101010101101010101

Offset: 1

Omar E. Pol, Nov 11 2008

a(k(n)) is divisible by 3 iff k(n) is defined by k(1) = 5 and k(n+1) - k(n) = A100285(n+2). - Altug Alkan, Dec 05 2015

			n .... Successive digits of a(n)
1 ............. ( 1 )
2 ............ ( 1 1 )
3 ........... ( 1 0 1 )
4 .......... ( 1 0 0 1 )
5 ......... ( 1 0 1 0 1 )
6 ........ ( 1 0 1 1 0 1 )
7 ....... ( 1 0 1 0 1 0 1 )
8 ...... ( 1 0 1 0 0 1 0 1 )
9 ..... ( 1 0 1 0 1 0 1 0 1 )
10 ... ( 1 0 1 0 1 1 0 1 0 1 )

Matthew Schulz, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (11,-20,110,-100).

Cf. A000533, A094028, A135577, A138120, A138144, A138145, A138146, A138721, A138826, A147757.

I:=[1,11,101,1001]; [n le 4 select I[n] else 11*Self(n-1)-20*Self(n-2)+110*Self(n-3)-100*Self(n-4): n in [1..30]]; // Vincenzo Librandi, Dec 05 2015

CoefficientList[Series[x/((1 - x) (1 - 10 x) (1 + 10 x^2)),{x, 0, 20}], x] (* Vincenzo Librandi, Dec 05 2015 *)
LinearRecurrence[{11,-20,110,-100},{1,11,101,1001},30] (* Harvey P. Dale, Apr 10 2022 *)

Vec(x/((1-x)*(1-10*x)*(1+10*x^2)) + O(x^30)) \\ Michel Marcus, Dec 05 2015

From R. J. Mathar, Feb 20 2009: (Start)
a(n) = 11*a(n-1)-20*a(n-2)+110*a(n-3)-100*a(n-4).
G.f.: x/((1-x)*(1-10*x)*(1+10*x^2)). (End)
E.g.f.: (exp(x)*(10*exp(9*x) - 1) - 9*cos(sqrt(10)*x))/99. - Stefano Spezia, Oct 12 2024

A168624 a(n) = 1 - 10^n + 100^n.

Original entry on oeis.org

1, 91, 9901, 999001, 99990001, 9999900001, 999999000001, 99999990000001, 9999999900000001, 999999999000000001, 99999999990000000001, 9999999999900000000001, 999999999999000000000001, 99999999999990000000000001, 9999999999999900000000000001, 999999999999999000000000000001

Offset: 0

Jason Earls, Dec 01 2009

Prime values for n = 2,4,6,8, with no others up to n = 3400. Beiler mentions this pattern in the reference.
From Peter Bala, Sep 27 2015: (Start)
Calculation suggests the continued fraction expansion of sqrt(a(n)), for n >= 1, begins [10^n - 1, 1, 1, 1/3*(2*10^n - 5), 1, 5, 1/9*(2*10^n - 11), 1, 17, (2*10^n - 20 - 9*(1 - MOD(n, 3)))/27, ...]. Note the large partial quotients early in the expansion.
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions. Empirically, we also see exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n) for m = 2, 3, 4,... as n increases. Some examples are given below. Cf. A000533, A002283, A066138. (End)

			Simple continued fraction expansions showing large partial quotients:
sqrt(a(10)) = [9999999999; 1, 1, 6666666665, 1, 5, 2222222221, 1, 17, 740740740, 1, 1, 1, 5, 2, 1, 246913579, 1, 1, 4, 1, 1, 3, 1, 1, ...].
a(18)^(1/3) = [999999999999; 1, 2999999, 499999999999, 1, 1439999, 2582644628099, 5, 1, 3, 4, 1, 58, 1, 1, 1, 8, ...].
a(30)^(1/5) = [999999999999; 1, 4999999999999999999, 333333333333, 3, 217391304347826086, 1, 1, 1, 1, 1, 8, 2398081534, 1, 1, 1, 9, 1, 98, 1, 125052522059263, 1, 9, 7, 1, ...]. - _Peter Bala_, Sep 27 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 85.

Colin Barker, Table of n, a(n) for n = 0..499
P. Bala, A168624 and some empirical continued fraction expansions.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000533, A002283, A066138, A187868.

Table[1-10^n+100^n,{n,0,20}] (* Harvey P. Dale, Dec 01 2013 *)

Vec(-(910*x^2-20*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

From Colin Barker, Sep 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
G.f.: -(910*x^2-20*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(exp(99*x) - exp(9*x) + 1). - Elmo R. Oliveira, Sep 12 2024

A329915 a(n) is the least M such that A329914(n) * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

Original entry on oeis.org

91, 77, 5882353, 52631579, 4347826087, 3448275862069, 2127659574468085106383, 20408163265306122449, 1694915254237288135593220339, 16393442622950819672131147541, 137, 13, 112359550561797732809, 11, 10309278350515463917525773195876288659793814433

Offset: 1

Bernard Schott, Nov 24 2019

When M is a q-digit term, then M is a divisor of 10^(q+1) + 1.
For each term k in A329914, there exist a set of numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k. This sequence gives the smallest integer M(k) = M of each set {M_k}.
See A329914 for further information about these numbers.

			A329914(1) = 21 and 21 * 91 = 1[91]1, and there is no integer < 91 that satisfies this relation, so a(1) = 91.
A329914(2) = 23 and 23 * 77 = 1[77]1, and there is no integer < 77 that satisfies this relation, so a(2) = 77.
A329914(5) = 33 and 33 * 4347826087 = 1[4347826087]1, and there is no integer < 4347826087 that satisfies this relation, so a(5) = 4347826087.

D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

Cf. A000533, A329914 (corresponding numbers k).

Some corresponding sets {M_k} : A095372 \ {1} = {M_21}, A331630 = {M_23}, A351237 = {M_83}, A351238 = {M_87}, A351239 = {M_101}.

cp's OEIS Frontend

A344897 a(n) is the number of divisors of 10^n + 1.

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A066138 a(n) = 10^(2*n) + 10^n + 1.

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A138144 Palindromes formed from the reflected decimal expansion of the concatenation of 1, 1 and infinite 0's.

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A329914 Numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k.

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A349194 a(n) is the product of the sum of the first i digits of n, as i goes from 1 to the total number of digits of n.

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A037156 a(n) = 10^n*(10^n+1)/2.

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A046851 Numbers n such that n^2 can be obtained from n by inserting internal (but not necessarily contiguous) digits.