A000042 -id:A000042 - cp's OEIS frontend

A102782 Repunit semiprimes.

111, 1111, 11111, 1111111, 11111111111, 11111111111111111, 11111111111111111111111111111111111111111111111, 11111111111111111111111111111111111111111111111111111111111

Offset: 1

Shyam Sunder Gupta, Feb 11 2005

			a(2)=1111 because 1111=11*101, so 1111 is semiprime as well as a repunit number.

Cf. A046413 the repunit of length n has exactly 2 prime factors.

Select[Table[FromDigits[PadRight[{},n,1]],{n,60}],PrimeOmega[#]==2&] (* Harvey P. Dale, Aug 28 2013 *)

a(n) = A000042(A046413(n-1)). - Ray Chandler, Sep 06 2005

A108047 Concatenation of the previous pair of numbers, with the first two terms both 1.

Original entry on oeis.org

1, 1, 11, 111, 11111, 11111111, 1111111111111, 111111111111111111111, 1111111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111, 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

Offset: 1

Parthasarathy Nambi, Jun 01 2005

The Fibonacci numbers, A000045, represented in base 1 (see A000042).

			The third term is 11 which is the concatenation of the first two terms 1 and 1.

Column b=1 of A214326.
Column k=1 of A214679.
Cf. A037842, A131293.

a(n) = (10^A000045(n)-1)/9.

a(n) = A000042(A000045(n)).

Edited by Jason Kimberley, Dec 15 2012

A046412 Lengths of nonsquarefree repunits.

Original entry on oeis.org

9, 18, 22, 27, 36, 42, 44, 45, 54, 63, 66, 72, 78, 81, 84, 88, 90, 99, 108, 110, 111, 117, 126, 132, 135, 144, 153, 154, 156, 162, 168, 171, 176, 180, 189, 198, 205, 207, 210, 216, 220, 222, 225, 234, 242, 243, 252, 261, 264, 270, 272, 279, 286, 288, 294, 297, 306, 308, 312, 315, 324, 330, 333, 336, 342, 351, 352

Offset: 1

Patrick De Geest, Jul 15 1998

This is the set of all positive multiples of all positive members of A087094. What is the asymptotic density of this set? - Jeppe Stig Nielsen, Dec 28 2015

Patrick De Geest, Repunits prime factors
Shyam Sunder Gupta, Repunit Numbers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 11, 327-352.
Eric Weisstein's World of Mathematics, Repunit

Cf. A000042, A004022, A046053, A084006, A087094.

remove(t -> numtheory:-issqrfree((10^t-1)/9), [$1..90]); # Robert Israel, Dec 30 2015

Select[Range[300],!SquareFreeQ[(10^#-1)/9]&] (* Harvey P. Dale, Jun 26 2013 *)

isok(n) = ! issquarefree((10^n-1)/9); \\ Michel Marcus, Dec 31 2015

a(n)=k where (10^k-1)/9 is not squarefree. - Ray Chandler, Aug 10 2003

Terms to a(60) from Ray Chandler, Aug 10 2003

a(61)-a(67) from Max Alekseyev, Apr 29 2022

A052983 Least multiple of n consisting of a succession of 1's followed by a succession of 0's.

Original entry on oeis.org

10, 10, 1110, 100, 10, 1110, 1111110, 1000, 1111111110, 10, 110, 11100, 1111110, 1111110, 1110, 10000, 11111111111111110, 1111111110, 1111111111111111110, 100, 1111110, 110, 11111111111111111111110, 111000, 100, 1111110, 1111111111111111111111111110

Offset: 1

Lekraj Beedassy, Jun 26 2003

All entries are differences of two terms of A000042. Since the pigeonhole principle guarantees that, for any m, two among the first m+1 entries of A000042 are congruent modulo m, their difference (i.e. belonging to this sequence) is therefore divisible by m, so that such numbers exist for all m. This sequence is thus infinite.

For n>1, a(n) consists of s 1's and t 0's, where s=A084681(X) and t is the greater of p or q (s=1 for X=1, t=1 for p=q=0), when we write n=X*Y with (X,Y)=1 and Y=2^p*5^q.

			We have a(6)=1110 because 6 divides 1110=6*185, the smallest such one with a string of 1's followed by that of 0's

Cf. A000042, A002371, A007732.

Cf. A084681.

f[n_] := Select[ Map[ FromDigits, IntegerDigits[ Table[ Sum[2^i, {i, k, j, -1}], {j, k, 1, -1}], 2]]/n, IntegerQ[ # ] & ]; g[n_] := Block[{k = 1}, While[ f[n] == {}, k++ ]; n*Min[ f[n]]]; Table[ g[n], {n, 1, 27}]
nn=30;With[{nos=Sort[Flatten[Table[FromDigits[Join[Table[1,{n}], Table[ 0,{i}]]],{n,nn},{i,5}]]]},Flatten[Table[Select[nos,Divisible[#,n]&,1],{n,nn}]]] (* Harvey P. Dale, Mar 09 2014 *)

a(n) = A276348(n) * n; A227362(a(n)) = 10. - Jaroslav Krizek, Aug 30 2016

Edited, corrected and extended by Robert G. Wilson v, Jun 26 2003

A083811 Numbers n such that 2n+1 is the digit reversal of n+1.

Original entry on oeis.org

36, 396, 3996, 39996, 399996, 3999996, 39999996, 399999996, 3999999996, 39999999996, 399999999996, 3999999999996

Offset: 1

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 08 2003

1. a(n) = 36 + 360 + 3600+ ...+ up to n terms. a(n) = sum of n terms of the geometric progression with the first term 36 and common ratio 10. 2. a(n) = 36*A000042(n).( the unary sequence).

			36 + 1 = 37, 2*36 + 1 = 73.

Cf. A000042.

a(n) = 4*(10^n - 1).

A111066 Numbers with digits 1 and 2 and at least one of each.

Original entry on oeis.org

12, 21, 112, 121, 122, 211, 212, 221, 1112, 1121, 1122, 1211, 1212, 1221, 1222, 2111, 2112, 2121, 2122, 2211, 2212, 2221, 11112, 11121, 11122, 11211, 11212, 11221, 11222, 12111, 12112, 12121, 12122, 12211, 12212, 12221, 12222, 21111, 21112, 21121, 21122, 21211

Offset: 1

Alexandre Wajnberg & Youri Mora, Oct 08 2005

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Equals A007931 minus A000042 and A002276. Supersequence of A214218.

Cf. A043494, A037415, A062289, A000225.

FromDigits /@ Select[ IntegerDigits[ Range[210], 3], Union[ # ] == {1, 2} &] (* Robert G. Wilson v, Oct 09 2005 *)
Union[FromDigits/@Select[Flatten[Table[Tuples[{1,2},n],{n,2,5}],1], Union[#] == {1,2}&]] (* Harvey P. Dale, Sep 05 2013 *)

from itertools import count, islice
def agen():
    for i in count(1):
        s = bin(i+1)[3:].replace('1', '2').replace('0', '1')
        if 0 < s.count('1') < len(s):
            yield int(s)
print(list(islice(agen(), 42))) # Michael S. Branicky, Dec 21 2021

More terms from Robert G. Wilson v, Oct 09 2005

Crossrefs from Charles R Greathouse IV, Aug 03 2010

A130835 Sum of all numbers having n or fewer digits and having the sum of their digits equal to n.

Original entry on oeis.org

1, 33, 1110, 38885, 1399986, 51333282, 1906666476, 71499999285, 2701111108410, 102631111100848, 3917722222183045, 150126888888738762, 5771538888888311735, 222499777777775552780, 8598259999999991401740, 332968856666666633369781, 12918171566666666537484951

Offset: 1

J. M. Bergot, Jul 18 2007

			Take n = 3. The numbers to be summed are 111, 3, 30, 300, 210, 201, 120, 102, 21 and 12, which add to 1110.

Alois P. Heinz, Table of n, a(n) for n = 1..626

Cf. A000042, A002275, A007953, A167403, A211072, A331672, A331673.

A007953 := proc(n) add(i,i=convert(n,base,10)) ; end: A130835 := proc(n) local a,i; a := 0 ; for i from 1 to 10^n-1 do if A007953(i) = n then a := a+i ; fi ; od ; RETURN(a) ; end: seq(A130835(n),n=1..4) ; # R. J. Mathar, Aug 01 2007
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1,
      `if`(i=0, 0, add(b(n-j, i-1), j=0..min(n, 9))))
    end:
a:= n-> b(n, n)*(10^n-1)/9:
seq(a(n), n=1..20); # Alois P. Heinz, Nov 02 2009

a(n) = (10^n-1)/9 * [x^n] ((x^10-1)/(x-1))^n. - Alois P. Heinz, Feb 07 2012

a(n) = A000042(n) * A167403(n) = A002275(n) * A167403(n). - Alois P. Heinz, Aug 16 2018

a(4)-a(6) from R. J. Mathar, Aug 01 2007
a(7)-a(12) from Donovan Johnson, Jul 02 2009
More terms from Alois P. Heinz, Nov 02 2009

A136308 a(n) = (10^2^n - 1)/9.

Original entry on oeis.org

1, 11, 1111, 11111111, 1111111111111111, 11111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111111111111

Offset: 0

Ctibor O. Zizka, Mar 22 2008

More generally, reading in base B >= 2: a(n) = (B^2^n - 1)/(B-1).
Recurrence: a(n) = a(n-1)*(B^K + 1) and a(0)=1 where K = floor(log_B a(n-1)) + 1.
B = 2 gives A051179; B = 3 gives A059918.

Cf. A000042 (repunits).
In other bases: A051179, A059918.
Cf. A000079, A007088, A007089, A007953.

A136308 := func; [A136308(n):n in[0..7]];

(10^2^Range[0, 10] - 1)/9 (* G. C. Greubel, Apr 19 2021 *)

[(10^2^n -1)/9 for n in (0..10)] # G. C. Greubel, Apr 19 2021

a(n) = a(n-1)*(10^K + 1) and a(0)=1 where K=floor(log_10 a(n-1)) + 1 = 2^n + 1.
a(n) = A000042(A000079(n)) = A007088(A051179(n)) = A007089(A059918(n)).
A007953(a(n)) = 2^n. - Stefano Spezia, Mar 27 2025

Edited by Jason Kimberley, Dec 18 2012

A309761 Numbers that are sums of consecutive powers of 10.

Original entry on oeis.org

1, 10, 11, 100, 110, 111, 1000, 1100, 1110, 1111, 10000, 11000, 11100, 11110, 11111, 100000, 110000, 111000, 111100, 111110, 111111, 1000000, 1100000, 1110000, 1111000, 1111100, 1111110, 1111111, 10000000, 11000000, 11100000, 11110000, 11111000, 11111100

Offset: 1

Ilya Gutkovskiy, Aug 15 2019

nonn

Numbers of the form (10^i - 10^j)/9 with i > j.

			11100 = 10^2 + 10^3 + 10^4, so 11100 is in the sequence.

Robert Israel, Table of n, a(n) for n = 1..10011

Cf. A000042, A007088, A011557, A023758, A038444, A276349, A309758, A309759.

seq(seq((10^i-10^(i-j))/9, j=1..i),i=1..10); # Robert Israel, Aug 16 2019

from math import isqrt
def A309761(n): return (10**(m:=isqrt(n<<3)+1>>1)-10**(m*(m+1)-(n<<1)>>1))//9 # Chai Wah Wu, Apr 04 2025

a(n) = A007088(A023758(n+1)).
a(i*(i-1)/2 + j) = (10^i - 10^(i-j))/9 for 1<=j<=i. - Robert Israel, Aug 16 2019
a(n) = A276349(n)/10. - Chai Wah Wu, Jun 16 2025

A048612 Find smallest pair (x,y) such that x^2-y^2 = 11...1 (n times) = (10^n-1)/9; sequence gives value of y.

Original entry on oeis.org

0, 5, 17, 45, 115, 67, 2205, 2933, 166667, 44445, 245795, 6667, 132683733, 4444445, 2012917, 23767083, 2680575317, 666667, 555555555555555555, 83053525, 3263104267, 12488376483, 5555555555555555555555, 66666667, 2952525627555

Offset: 1

Felice Russo

Least solutions for 'Difference between two squares is a repunit of length n'.

			For n=2, 6^2 - 5^2 = 11.

David Wells, "Curious and Interesting Numbers", Revised Ed. 1997, Penguin Books, p. 119. ISBN 0-14-026149-4.

H. Havermann, Repunit Square Differences (gives many more terms)

Cf. A048611, A000042, A002275, A033676, A033677.

s = Flatten[Table[r = (10^i - 1)/9; d = Divisors[r]; p = d[[Length[d]/2]]; Solve[{x - y == p, x + y == r/p}, {y, x}], {i, 2, 56}]]; Prepend[Cases[s, Rule[y, n_] -> n], 0]
Join[{0},Table[y/.Solve[{x>0,y>0,x^2-y^2==FromDigits[PadRight[{},n,1]]},{x,y},Integers][[1]],{n,2,30}]](* Harvey P. Dale, Jun 12 2018 *)

from sympy import divisors
def A048612(n):
    d = divisors((10**n-1)//9)
    l = len(d)
    return (d[l//2]-d[(l-1)//2])//2 # Chai Wah Wu, Apr 05 2021

a(n) = (A033677((10^n-1)/9)-A033676((10^n-1)/9))/2. - Chai Wah Wu, Apr 05 2021

Corrected and extended by Patrick De Geest, Jun 15 1999
More terms from Hans Havermann, Jul 02 2000
Offset corrected by Chai Wah Wu, Apr 05 2021

cp's OEIS Frontend

A102782 Repunit semiprimes.

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A108047 Concatenation of the previous pair of numbers, with the first two terms both 1.

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A046412 Lengths of nonsquarefree repunits.

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PARI

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A052983 Least multiple of n consisting of a succession of 1's followed by a succession of 0's.

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Mathematica

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A083811 Numbers n such that 2n+1 is the digit reversal of n+1.

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A111066 Numbers with digits 1 and 2 and at least one of each.

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A130835 Sum of all numbers having n or fewer digits and having the sum of their digits equal to n.

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Maple

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A136308 a(n) = (10^2^n - 1)/9.

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Magma