cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A075412 Squares of A002277.

Original entry on oeis.org

0, 9, 1089, 110889, 11108889, 1111088889, 111110888889, 11111108888889, 1111111088888889, 111111110888888889, 11111111108888888889, 1111111111088888888889, 111111111110888888888889, 11111111111108888888888889, 1111111111111088888888888889, 111111111111110888888888888889
Offset: 0

Views

Author

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

Keywords

Comments

A transformation of the Wonderful Demlo numbers (A002477).

Examples

			a(2) = 33^2 = 1089.
Contribution from _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ...................... 9 = 9 * 1;
n=2: ................... 1089 = 99 * 11;
n=3: ................. 110889 = 999 * 111;
n=4: ............... 11108889 = 9999 * 1111;
n=5: ............. 1111088889 = 99999 * 11111;
n=6: ........... 111110888889 = 999999 * 111111;
n=7: ......... 11111108888889 = 9999999 * 1111111;
n=8: ....... 1111111088888889 = 99999999 * 11111111;
n=9: ..... 111111110888888889 = 999999999 * 111111111. (End)
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{11, -10}, {0, 3}, 20]^2 (* Vincenzo Librandi, Mar 20 2014 *)
    Table[FromDigits[PadRight[{},n,9]]FromDigits[PadRight[{},n,1]],{n,0,15}] (* Harvey P. Dale, Feb 12 2023 *)

Formula

a(n) = A002277(n)^2 = (3*A002275(n))^2 = 9*A002275(n)^2.
a(n) = {111111... (2n times)} - 2*{ 111... (n times)} a(n) = A000042(2*n) - 2*A000042(n). - Amarnath Murthy, Jul 21 2003
a(n) = {333... (n times)}^2 = {111...(n times)}{000... (n times)} - {111... (n times)}. For example, 333^2 = 111000 - 111 = 110889. - Kyle D. Balliet, Mar 07 2009
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = A002283(n)*A002275(n).
For n>0, a(n) = (A002275(n-1)*10^n + A002282(n-1))*10 + 9. (End)
a(n) = (10^(n+1)-10)^2/900. - José de Jesús Camacho Medina, Apr 01 2016
From Elmo R. Oliveira, Jul 27 2025: (Start)
G.f.: 9*x*(1+10*x)/((1-x)*(1-10*x)*(1-100*x)).
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 9*A002477(n). (End)

A034048 Numbers with multiplicative digital root value 0.

Original entry on oeis.org

0, 10, 20, 25, 30, 40, 45, 50, 52, 54, 55, 56, 58, 59, 60, 65, 69, 70, 78, 80, 85, 87, 90, 95, 96, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 120, 125, 130, 140, 145, 150, 152, 154, 155, 156, 158, 159, 160, 165, 169, 170, 178, 180, 185, 187, 190, 195
Offset: 1

Views

Author

Patrick De Geest, Sep 15 1998

Keywords

Comments

Numbers with root value 1 are the 'repunits' (see A000042).
This sequence has density 1. - Franklin T. Adams-Watters, Apr 01 2009
Any integer with at least one 0 among its base 10 digits is in this sequence. - Alonso del Arte, Aug 29 2014
This sequence is 10-automatic: it contains numbers with a 0, or with a 5 and any even digit. - Charles R Greathouse IV, Feb 13 2017

Examples

			20 is in the sequence because 2 * 0 = 0.
25 is in the sequence because 2 * 5 = 10 and 1 * 0 = 0.
		

Crossrefs

Cf. A031347, A034048, A002275, A034049, A034050, A034051, A034052, A034053, A034054, A034055, A034056 (numbers having multiplicative digital roots 0-9).
Cf. the subsets A011540 and A008592.

Programs

  • Mathematica
    mdr0Q[n_]:=NestWhile[Times@@IntegerDigits[#]&,n,#>9&]==0; Select[Range[ 0,200],mdr0Q] (* Harvey P. Dale, Jul 21 2020 *)
  • PARI
    is(n)=factorback(digits(n))==0 \\ Charles R Greathouse IV, Feb 13 2017

A023705 Numbers with no 0's in base-4 expansion.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 21, 22, 23, 25, 26, 27, 29, 30, 31, 37, 38, 39, 41, 42, 43, 45, 46, 47, 53, 54, 55, 57, 58, 59, 61, 62, 63, 85, 86, 87, 89, 90, 91, 93, 94, 95, 101, 102, 103, 105, 106, 107, 109, 110, 111, 117, 118, 119, 121, 122, 123
Offset: 1

Views

Author

Keywords

Comments

A032925 is the intersection of this sequence and A023717; cf. A179888. - Reinhard Zumkeller, Jul 31 2010

Crossrefs

Zeroless numbers in some other bases <= 10: A000042 (base 2), A032924 (base 3), A248910 (base 6), A255805 (base 8), A255808 (base 9), A052382 (base 10).
Cf. A100968 (subsequence).

Programs

  • C
    #include 
    uint32_t a_next(uint32_t a_n) { return (a_n + 1) | ((a_n & (a_n + 0xaaaaaaab)) >> 1); } /* Falk Hüffner, Jan 22 2022 */
    
  • Haskell
    a023705 n = a023705_list !! (n-1)
    a023705_list = iterate f 1 where
       f x = 1 + if r < 3 then x else 4 * f x'
             where (x', r) = divMod x 4
    -- Reinhard Zumkeller, Mar 06 2015, Oct 19 2011
    
  • Magma
    [n: n in [1..130] | not 0 in Intseq(n,4)]; // Vincenzo Librandi, Oct 04 2018
    
  • Maple
    R:= [1,2,3]: A:= 1,2,3:
    for i from 1 to 4 do
      R:= map(t -> (4*t+1,4*t+2,4*t+3), R);
      A:= A, op(R);
    od:
    A; # Robert Israel, Oct 04 2018
  • Mathematica
    Select[ Range[ 120 ], (Count[ IntegerDigits[ #, 4 ], 0 ]==0)& ]
    Select[Range[200],DigitCount[#,4,0]==0&] (* Harvey P. Dale, Dec 23 2015 *)
  • PARI
    isok(n) = vecmin(digits(n, 4)); \\ Michel Marcus, Jul 04 2015
    
  • Python
    from sympy import integer_log
    def A023705(n):
        m = integer_log(k:=(n<<1)+1,3)[0]
        return sum(1+(k-3**m)//(3**j<<1)%3<<(j<<1) for j in range(m)) # Chai Wah Wu, Jun 27 2025

Formula

G.f. g(x) satisfies g(x) = (x+2*x^2+3*x^3)/(1-x^3) + 4*(x+x^2+x^3)*g(x^3). - Robert Israel, Oct 04 2018

A031973 a(n) = Sum_{k=0..n} n^k.

Original entry on oeis.org

1, 2, 7, 40, 341, 3906, 55987, 960800, 19173961, 435848050, 11111111111, 313842837672, 9726655034461, 328114698808274, 11966776581370171, 469172025408063616, 19676527011956855057, 878942778254232811938, 41660902667961039785743, 2088331858752553232964200
Offset: 0

Views

Author

Keywords

Comments

These are the generalized repunits of length n+1 in base n for all n >= 1: a(n) expressed in base n is 111...111 (n+1 1's): a(1) = 1^0 + 1^1 = 2 = A000042(2), a(2) = 2^0 + 2^1 + 2^2 = 7 = A000225(3), a(3) = 3^0 + 3^1 + 3^2 + 3^3 = 40 = A003462(4), etc., a(10) = 10^0 + 10^1 + 10^2 + ... + 10^9 + 10^10 = 11111111111 = A002275(11), etc. - Rick L. Shepherd, Aug 26 2004
a(n)=the total number of ordered selections of up to n objects from n types with repetitions allowed. Thus for 2 objects a,b there are 7 possible selections: aa,bb,ab,ba,a,b, and the null set. - J. M. Bergot, Mar 26 2014
a(n)=the total number of ordered arrangements of 0,1,2..n objects, with repetitions allowed, selected from n types of objects. - J. M. Bergot, Apr 11 2014

Examples

			a(3) = 3^0 + 3^1 + 3^2 + 3^3 = 40.
		

Crossrefs

Cf. A000042 (unary representations), A000225 (2^n-1: binary repunits shown in decimal), A003462 ((3^n-1)/2: ternary repunits shown in decimal), A002275 ((10^n-1)/9: decimal repunits).
Cf. A104878.

Programs

  • Magma
    [&+[n^k: k in [0..n]]: n in [0..30]]; // Vincenzo Librandi, Apr 18 2011
    
  • Maple
    a:= proc(n) local c, i; c:=1; for i to n do c:= c*n+1 od; c end:
    seq(a(n), n=0..20); # Alois P. Heinz, Aug 15 2013
  • Mathematica
    Join[{1},Table[Total[n^Range[0,n]],{n,20}]] (* Harvey P. Dale, Nov 13 2011 *)
  • PARI
    a(n)=(n^(n+1)-1)/(n-1) \\ Charles R Greathouse IV, Mar 26 2014
  • Sage
    [lucas_number1(n,n,n-1) for n in range(1, 19)] # Zerinvary Lajos, May 16 2009
    

Formula

a(n) = (n^(n+1)-1)/(n-1) = (A007778(n)-1)/(n-1) = A023037(n)+A000312(n) = A031972(n)+1. - Henry Bottomley, Apr 04 2003
a(n) = A125118(n,n-2) for n>2. - Reinhard Zumkeller, Nov 21 2006
a(n) = [x^n] 1/((1 - x)*(1 - n*x)). - Ilya Gutkovskiy, Oct 04 2017
a(n) = A104878(2n,n). - Alois P. Heinz, May 04 2021

A105992 Near-repunit primes.

Original entry on oeis.org

101, 113, 131, 151, 181, 191, 211, 311, 811, 911, 1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111, 10111, 11113, 11117, 11119, 11131, 11161, 11171, 11311, 11411, 16111, 101111, 111119, 111121, 111191, 111211, 111611, 112111, 113111, 131111, 311111, 511111
Offset: 1

Views

Author

Shyam Sunder Gupta, Apr 29 2005

Keywords

Comments

According to the prime glossary "a near-repunit prime is a prime all but one of whose digits are 1." This would also include {2, 3, 5, 7, 13, 17, 19, 31, 41, 61 and 71}, but this sequence only lists terms with more than two digits. - M. F. Hasler, Feb 10 2020

Examples

			a(2)=113 is a term because 113 is a prime and all digits are 1 except one.
		

References

  • C. Caldwell and H. Dubner, "The near repunit primes 1(n-k-1)01(1k)," J. Recreational Math., 27 (1995) 35-41.
  • Heleen, J. P., "More near-repunit primes 1(n-k-1)D(1)1(k), D=2,3, ..., 9," J. Recreational Math., 29:3 (1998) 190-195.

Crossrefs

Programs

  • Mathematica
    lst = {}; Do[r = (10^n - 1)/9; Do[AppendTo[lst, DeleteCases[Select[FromDigits[Permutations[Append[IntegerDigits[r], d]]], PrimeQ], r]], {d, 0, 9}], {n, 2, 14}]; Sort[Flatten[lst]] (* Arkadiusz Wesolowski, Sep 20 2011 *)

A214676 A(n,k) is n represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 1, 11, 1, 2, 111, 1, 2, 11, 1111, 1, 2, 3, 12, 11111, 1, 2, 3, 11, 21, 111111, 1, 2, 3, 4, 12, 22, 1111111, 1, 2, 3, 4, 11, 13, 111, 11111111, 1, 2, 3, 4, 5, 12, 21, 112, 111111111, 1, 2, 3, 4, 5, 11, 13, 22, 121, 1111111111
Offset: 1

Views

Author

Alois P. Heinz, Jul 25 2012

Keywords

Comments

The digit set for bijective base-k numeration is {1, 2, ..., k}.

Examples

			Square array A(n,k) begins:
:         1,   1,  1,  1,  1,  1,  1,  1, ...
:        11,   2,  2,  2,  2,  2,  2,  2, ...
:       111,  11,  3,  3,  3,  3,  3,  3, ...
:      1111,  12, 11,  4,  4,  4,  4,  4, ...
:     11111,  21, 12, 11,  5,  5,  5,  5, ...
:    111111,  22, 13, 12, 11,  6,  6,  6, ...
:   1111111, 111, 21, 13, 12, 11,  7,  7, ...
:  11111111, 112, 22, 14, 13, 12, 11,  8, ...
		

Crossrefs

A(n+1,n) gives A010850.

Programs

  • Maple
    A:= proc(n, b) local d, l, m; m:= n; l:= NULL;
          while m>0 do  d:= irem(m, b, 'm');
            if d=0 then d:=b; m:=m-1 fi;
            l:= d, l
          od; parse(cat(l))
        end:
    seq(seq(A(n, 1+d-n), n=1..d), d=1..12);
  • Mathematica
    A[n_, b_] := Module[{d, l, m}, m = n; l = Nothing; While[m > 0, {m, d} = QuotientRemainder[m, b]; If[d == 0, d = b; m--]; l = {d, l}]; FromDigits @ Flatten @ l];
    Table[A[n, d-n+1], {d, 1, 12}, {n, 1, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A260851 a(n) in base n is the concatenation of the base n expansions of (1, 2, 3, ..., n-1, n, n-1, ..., 3, 2, 1).

Original entry on oeis.org

1, 13, 439, 27961, 3034961, 522134761, 131870760799, 45954960939217, 21107054541321649, 12345678910987654321, 8954302429379707945271, 7883984846509322664831433, 8281481197999449959084458465, 10228838696316240496325238416281, 14674825961700306151086890240104831
Offset: 1

Views

Author

M. F. Hasler, Aug 01 2015

Keywords

Comments

Sequences A173427, A260853 - A260859, A173426, A260861 - A260866, A260860 list the numbers A_b(n) whose base b expansion is the concatenation of the base b expansions of (1, 2, ..., n, n-1, ..., 1). For n < b these are the squares of the repdigits of length n in base b, so the first candidate for a prime is the term with n = b. These are the numbers listed here. Sequence A260343 gives the bases b for which this is indeed a prime, the corresponding primes a(A260343(n)) are listed in A260852.
The initial term a(1) = 1 refers to the unary or "tally mark" representation of the numbers, cf. A000042. It can be considered as purely conventional.

Examples

			a(1) = 1 is the "concatenation" of (1) which is the unary representation of 1, cf A000042.
a(2) = 13 = 1101[2] = concatenation of (1, 10, 1), where 10 is the base 2 representation of 2.
a(3) = 439 = 121021[3] = concatenation of (1, 2, 10, 2, 1), where 10 is the base 3 representation of 3.
a(10) = 12345678910987654321 is the concatenation of (1, 2, 3, ..., 9, 10, 9, 8, ..., 2, 1); it is also a prime.
		

Crossrefs

For primes in this sequence see A260343, A260852.

Programs

  • Magma
    [1] cat [((n^n-1)/(n-1) - n + 1)*(1 + n*(n^n-1)/(n-1)) - 1: n in [2..15]]; // Vincenzo Librandi, Aug 02 2015
    
  • Maple
    f:=proc(b) local i;
    add((i+1)*b^i, i=0..b-2) + b^b + add(i*b^(2*b-i),i=1..b-1); end;
    [seq(f(b),b=1..25)]; # N. J. A. Sloane, Sep 26 2015
  • Mathematica
    Join[{1}, Table[((n^n - 1)/(n - 1) - n + 1) (1 + n (n^n - 1)/(n - 1)) - 1, {n, 2, 30}]] (* Vincenzo Librandi, Aug 02 2015 *)
  • PARI
    A260851(n)=(1+n*r=if(n>2,n^n\(n-1),n*2-1))*(r-n+1)-1
    
  • Python
    def A260851(n): return sum(i*(n**(2*n-i)+n**(i-1)) for i in range(1, n)) + n**n # Ya-Ping Lu, Dec 23 2021

Formula

a(n) = n*r + (r - n)*(1 + n*r) = (r - n + 1)*(1 + n*r) - 1, where r = (n^n-1)/(n-1) is the base n repunit of length n, r = 1 for n = 1.
Another closed-form expression for the series is a(n) = (n^(2*n+1) + (-n^3 + 2*n^2 - 2*n - 1)*n^n + 1)/(n - 1)^2. - Serge Batalov, Aug 02 2015

A046413 Numbers k such that the repunit of length k (11...11, with k 1's) has exactly 2 prime factors.

Original entry on oeis.org

3, 4, 5, 7, 11, 17, 47, 59, 71, 139, 211, 251, 311, 347, 457, 461
Offset: 1

Views

Author

Patrick De Geest, Jul 15 1998

Keywords

Comments

347, 457, 461 and 701 are also terms. The only other possible terms up to 1000 are 263, 311, 509, 557, 617, 647 and 991; repunits of these lengths are known to be composite but the linked sources do not provide their factors. - Rick L. Shepherd, Mar 11 2003
The Yousuke Koide reference now shows the repunit of length 263 partially factored; 263 is no longer a possible candidate for this sequence. - Ray Chandler, Sep 06 2005
The repunit of length 263 has 3 prime factors; the repunit of length 617 has one known prime factor and a large composite. Possible terms > 1000 are 1117, 1213, 1259, 1291, 1373, 1447, 1607, 1637, 1663, 1669, 1759, 1823, 1949, 1987, 2063 & 2087. - Robert G. Wilson v, Apr 26 2010
All terms are either primes or squares of primes in A004023. In particular, the only composite below a million is 4. - Charles R Greathouse IV, Nov 21 2014
a(17) >= 509. The only confirmed term below 2500 is 701. As of July 2019, no factorization is known for the potential terms 509, 557, 647, 991, 1117, 1259, 1447, 1607, 1637, 1663, 1669, 1759, 1823, 1949, 1987, 2063, 2087, 2111, 2203, 2269, 2309, 2341, 2467, 2503, 2521, ... Unless the least prime factors of the respective composites have fewer than ~80 decimal digits and are thus accessible by massive ECM computations, there is no chance for an extension using current publicly available factorization methods. See links to factordb.com for the status of the factorization of the smallest unknown terms. - Hugo Pfoertner, Jul 30 2019

Examples

			7 is a term because 1111111 = 239*4649.
		

References

  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.

Crossrefs

Cf. A000042, A004022 (repunit primes), A046053, A102782.

Programs

  • Mathematica
    Select[Range[60],PrimeOmega[FromDigits[PadRight[{},#,1]]]==2&] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Aug 26 2024 *)

Extensions

More terms from Rick L. Shepherd, Mar 11 2003
a(13)-a(16) from Robert G. Wilson v, Apr 26 2010

A083956 a(n) = sum 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, 33, 666, 11110, 166665, 2333331, 31111108, 399999996, 4999999995, 509876543215, 52098641976336, 5331076296399558, 546238942849832881, 56038035699304276305, 5755318721445859729830, 591693488306202516193456
Offset: 1

Views

Author

Amarnath Murthy, May 10 2003

Keywords

Comments

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

Examples

			a(1) = 1, a(2) = 12 + 21, a(3) = 123 + 231 + 312 = 666.
a(11) = 1234567891011 + 2345678910111 + ... + 1011123456789 + 1112345678910.
		

Crossrefs

Cf. A083957.

Programs

  • Maple
    # count digits in positive integer digs := proc(inp::integer) local resul,shiftinp : resul := 1 : shiftinp := iquo(inp,10) : while shiftinp > 0 do resul := resul+1 : shiftinp := iquo(shiftinp,10) : od : RETURN(resul) : end: # provide number of concatenation up to lst, permuted by cycl newnum := proc(lst::integer,cycl::integer) local resul,i,insrt : resul := 0 : for i from 1 to lst do insrt := ((i+cycl-1) mod lst) +1 : resul := resul*10^digs(insrt)+insrt : od : RETURN(resul) ; end : n := 2 : while n < 13 do su := 0 : for cycl from 0 to n-1 do # print(n," add ",newnum(n,cycl)) ; su := su + newnum(n,cycl) : od : printf("%a,",su) : n := n+1 : od : # R. J. Mathar, Mar 13 2006
    A083956 := n -> add( convert( cat( 'modp(j+i,n)+1' $ j=1..n ),decimal,10), i=1..n ); # M. F. Hasler, Nov 08 2006

Extensions

More terms from R. J. Mathar, Mar 13 2006
Further terms from M. F. Hasler, Nov 08 2006

A246057 a(n) = (5*10^n - 2)/3.

Original entry on oeis.org

1, 16, 166, 1666, 16666, 166666, 1666666, 16666666, 166666666, 1666666666, 16666666666, 166666666666, 1666666666666, 16666666666666, 166666666666666, 1666666666666666, 16666666666666666, 166666666666666666, 1666666666666666666, 16666666666666666666, 166666666666666666666
Offset: 0

Views

Author

Vincenzo Librandi, Aug 13 2014

Keywords

Comments

a(k-1) = (10^k - 4)/6, together with b(k) = 3*a(k-1) + 2 = A093143(k) and c(k) = 2*a(k-1) + 1 = A002277(k) are k-digit numbers for k >= 1 satisfying the so-called curious cubic identity a(k-1)^3 + b(k)^3 + c(k)^3 = a(k)*10^(2*k) + b(k)*10^k + c(k) (concatenated a(k)b(k)c(k)). This k-family and the proof of the identity has been given in the introduction of the van der Poorten reference. Thanks go to S. Heinemeyer for bringing these identities to my attention. - Wolfdieter Lang, Feb 07 2017

Examples

			Curious cubic identities (see a comment and reference above): 1^3 + 5^3 + 3^3 = 153, 16^3 + 50^3 + 33^3 = 165033, 166^3 + 500^3 + 333^3 = 166500333, ... - _Wolfdieter Lang_, Feb 07 2017
		

Crossrefs

Cf. sequences with terms of the form 1k..k where the digit k is repeated n times: A000042 (k=1), A090843 (k=2), A097166 (k=3), A099914 (k=4), A099915 (k=5), this sequence (k=6), A246058 (k=7), A246059 (k=8), A067272 (k=9).

Programs

  • Magma
    [(5*10^n-2)/3: n in [0..20]];
    
  • Mathematica
    Table[(5 10^n - 2)/3, {n, 0, 20}]
  • PARI
    vector(50, n, (5*10^(n-1)-2)/3) \\ Derek Orr, Aug 13 2014

Formula

G.f.: (1 + 5*x)/((1 - x)*(1 - 10*x)).
a(n) = 11*a(n-1) - 10*a(n-2).
E.g.f.: exp(x)*(5*exp(9*x) - 2)/3. - Stefano Spezia, May 02 2025
a(n) = A323639(n+1)/2 = A086948(n+1)/12. - Elmo R. Oliveira, May 07 2025
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