A102794 Duplicate of A075415.
36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556
Offset: 1
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.
a002275 = (`div` 9) . subtract 1 . (10 ^) a002275_list = iterate ((+ 1) . (* 10)) 0 -- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012
[(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014
seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013
Table[(10^n - 1)/9, {n, 0, 19}] (* Alonso del Arte, Nov 15 2011 *) Join[{0},Table[FromDigits[PadRight[{},n,1]],{n,20}]] (* Harvey P. Dale, Mar 04 2012 *)
a[0]:0$ a[1]:1$ a[n]:=11*a[n-1]-10*a[n-2]$ A002275(n):=a[n]$ makelist(A002275(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009
my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016
print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022
[lucas_number1(n, 11, 10) for n in range(21)] # Zerinvary Lajos, Apr 27 2009
From _Peter Bala_, Sep 27 2015: (Start) Continued fraction expansions showing large partial quotients: a(12)^(1/3) = [9999; 1, 299999998, 1, 9998, 1, 449999998, 1, 7998, 1, 535714284, 1, 2, 2, 142, 2, 2, 1, 599999999, 3, 1, 1,...]. Compare with a(30)^(1/3) = [9999999999; 1, 299999999999999999998, 1, 9999999998, 1, 449999999999999999998, 1, 7999999998, 1, 535714285714285714284, 1, 2, 2, 142857142, 2, 2, 1, 599999999999999999999, 3, 1, 1,...]. a(24)^(1/4) = [999999; 1, 3999999999999999998, 1, 666665, 1, 1, 1, 799999999999999999, 3, 476190, 7, 190476190476190476, 21, 43289, 1, 229, 1, 1864801864801863, 1, 4, 6,...]. Compare with a(48)^(1/4) = [999999999999; 1, 3999999999999999999999999999999999998, 1, 666666666665, 1, 1, 1, 799999999999999999999999999999999999, 3, 476190476190, 7, 190476190476190476190476190476190476, 21, 43290043289, 1, 229, 1, 1864801864801864801864801864801863, 1, 4, 6,...]. a(25)^(1/5) = [99999, 1, 499999999999999999998, 1, 49998, 1, 999999999999999999998, 1, 33332, 3, 151515151515151515151, 5, 1, 1, 1947, 1, 1, 38, 3787878787878787878, 1, 3, 5,...]. (End)
a002283 = subtract 1 . (10 ^) -- Reinhard Zumkeller, Feb 21 2014
[(10^n-1): n in [0..20]]; // Vincenzo Librandi, Apr 26 2011
Table[10^n - 1, {n, 0, 22}] (* Michael De Vlieger, Sep 27 2015 *)
A002283(n):=10^n-1$ makelist(A002283(n),n,0,20); /* Martin Ettl, Nov 08 2012 */
a(n)=10^n-1; \\ Charles R Greathouse IV, Jan 30 2012
def a(n): return 10**n-1 # Michael S. Branicky, Feb 25 2023
Table[FromDigits[PadRight[{},n,5]],{n,0,20}] (* Harvey P. Dale, Oct 05 2013 *)
A002279(n):=5*(10^n-1)/9$ makelist(A002279(n),n,0,30); /* Martin Ettl, Nov 08 2012 */
a(n)=5*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015
def A002279(n): return 5*(10**n-1)//9 # Karl-Heinz Hofmann, Nov 28 2023
LinearRecurrence[{11,-10},{0,6},20] (* Harvey P. Dale, Dec 20 2012 *)
a(n)=6*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015
From _José de Jesús Camacho Medina_, Apr 01 2016: (Start) n=1: ....................... 1 = 9 / 9; n=2: ..................... 121 = 1089 / 9; n=3: ................... 12321 = 110889 / 9; n=4: ................. 1234321 = 11108889 / 9; n=5: ............... 123454321 = 1111088889 / 9; n=6: ............. 12345654321 = 111110888889 / 9; n=7: ........... 1234567654321 = 11111108888889 / 9; n=8: ......... 123456787654321 = 1111111088888889 / 9; n=9: ....... 12345678987654321 = 111111110888888889 / 9. (End) a(11) = concat(L = 1234567901, R = 20987654321), with L + R = 22222222222 = 2*(10^11-1)/9, of same length as R. - _M. F. Hasler_, Nov 23 2017
[((10^n - 1)/9)^2: n in [1..20]]; // Vincenzo Librandi, Jul 26 2011
A002477 := proc(n) (10^n-1)^2/81 ; end proc: seq(A002477(n),n=1..12) ; # R. J. Mathar, Aug 06 2019
Table[FromDigits[PadRight[{},n,1]]^2,{n,15}] (* Harvey P. Dale, Oct 16 2012 *) (10^Range[20] - 1)^2/81 (* Paolo Xausa, Aug 03 2025 *)
A002477(n):=((10^n - 1)/9)^2$ makelist(A002477(n),n,1,10); /* Martin Ettl, Nov 12 2012 */
a(n) = (10^n\9)^2 \\ Charles R Greathouse IV, Jul 25 2011
LinearRecurrence[{11, -10}, {0, 4}, 20] (* Robert G. Wilson v, Jul 06 2013 *)
a(n)=4*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015
Select[Range[66]^2, FreeQ[IntegerDigits[#],0]==True &] (* Jayanta Basu, May 25 2013 *)
From _Reinhard Zumkeller_, May 31 2010: (Start) n=1: ..................... 81 = 9^2; n=2: ................... 9801 = 99^2; n=3: ................. 998001 = 999^2; n=4: ............... 99980001 = 9999^2; n=5: ............. 9999800001 = 99999^2; n=6: ........... 999998000001 = 999999^2; n=7: ......... 99999980000001 = 9999999^2; n=8: ....... 9999999800000001 = 99999999^2; n=9: ..... 999999998000000001 = 999999999^2. (End)
A059988:=n->(10^n-1)^2; seq(A059988(n), n=0..20); # Wesley Ivan Hurt, Nov 21 2013
Table[(10^n-1)^2, {n,0,20}] (* Wesley Ivan Hurt, Nov 21 2013 *)
a(n) = { (10^n - 1)^2 } \\ Harry J. Smith, Jul 01 2009
def a(n): return (10**n - 1)**2 # Martin Gergov, Sep 10 2022
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)
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 *)
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