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.

Showing 1-10 of 26 results. Next

A102794 Duplicate of A075415.

Original entry on oeis.org

36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556
Offset: 1

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A002275 Repunits: (10^n - 1)/9. Often denoted by R_n.

Original entry on oeis.org

0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111, 11111111111, 111111111111, 1111111111111, 11111111111111, 111111111111111, 1111111111111111, 11111111111111111, 111111111111111111, 1111111111111111111, 11111111111111111111
Offset: 0

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Comments

R_n is a string of n 1's.
Base-4 representation of Jacobsthal bisection sequence A002450. E.g., a(4)= 1111 because A002450(4)= 85 (in base 10) = 64 + 16 + 4 + 1 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 1. - Paul Barry, Mar 12 2004
Except for the first two terms, these numbers cannot be perfect squares, because x^2 != 11 (mod 100). - Zak Seidov, Dec 05 2008
For n >= 0: a(n) = (A000225(n) written in base 2). - Jaroslav Krizek, Jul 27 2009, edited by M. F. Hasler, Jul 03 2020
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Except 0, 1 and 11, all these integers are Brazilian numbers, A125134. - Bernard Schott, Dec 24 2012
Numbers n such that 11...111 = R_n = (10^n - 1)/9 is prime are in A004023. - Bernard Schott, Dec 24 2012
The terms 0 and 1 are the only squares in this sequence, as a(n) == 3 (mod 4) for n>=2. - Nehul Yadav, Sep 26 2013
For n>=2 the multiplicative order of 10 modulo the a(n) is n. - Robert G. Wilson v, Aug 20 2014
The above is a special case of the statement that the order of z modulo (z^n-1)/(z-1) is n, here for z=10. - Joerg Arndt, Aug 21 2014
From Peter Bala, Sep 20 2015: (Start)
Let d be a divisor of a(n). Let m*d be any multiple of d. Split the decimal expansion of m*d into 2 blocks of contiguous digits a and b, so we have m*d = 10^k*a + b for some k, where 0 <= k < number of decimal digits of m*d. Then d divides a^n - (-b)^n (see McGough). For example, 271 divides a(5) and we find 2^5 + 71^5 = 11*73*271*8291 and 27^5 + 1^5 = 2^2*7*31*61*271 are both divisible by 271. Similarly, 4*271 = 1084 and 10^5 + 84^5 = 2^5*31*47*271*331 while 108^5 + 4^5 = 2^12*7*31*61*271 are again both divisible by 271. (End)
Starting with the second term this sequence is the binary representation of the n-th iteration of the Rule 220 and 252 elementary cellular automaton starting with a single ON (black) cell. - Robert Price, Feb 21 2016
If p > 5 is a prime, then p divides a(p-1). - Thomas Ordowski, Apr 10 2016
0, 1 and 11 are only terms that are of the form x^2 + y^2 + z^2 where x, y, z are integers. In other words, a(n) is a member of A004215 for all n > 2. - Altug Alkan, May 08 2016
Except for the initial terms, the binary representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Mar 17 2017
The term "repunit" was coined by Albert H. Beiler in 1964. - Amiram Eldar, Nov 13 2020
q-integers for q = 10. - John Keith, Apr 12 2021
Binomial transform of A001019 with leading zero. - Jules Beauchamp, Jan 04 2022

References

  • Albert H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, New York: Dover Publications, 1964, chapter XI, p. 83.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 235-237.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 197-198.
  • Samuel Yates, Peculiar Properties of Repunits, J. Recr. Math. 2, 139-146, 1969.
  • Samuel Yates, Prime Divisors of Repunits, J. Recr. Math. 8, 33-38, 1975.

Crossrefs

Programs

  • Haskell
    a002275 = (`div` 9) . subtract 1 . (10 ^)
    a002275_list = iterate ((+ 1) . (* 10)) 0
    -- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012
    
  • Magma
    [(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014
    
  • Maple
    seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013
  • Mathematica
    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 *)
  • Maxima
    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 */
    
  • PARI
    a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009
    
  • PARI
    my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016
    
  • Python
    print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022
  • Sage
    [lucas_number1(n, 11, 10) for n in range(21)]  # Zerinvary Lajos, Apr 27 2009
    

Formula

a(n) = 10*a(n-1) + 1, a(0)=0.
a(n) = A000042(n) for n >= 1.
Second binomial transform of Jacobsthal trisection A001045(3n)/3 (A015565). - Paul Barry, Mar 24 2004
G.f.: x/((1-10*x)*(1-x)). Regarded as base b numbers, g.f. x/((1-b*x)*(1-x)). - Franklin T. Adams-Watters, Jun 15 2006
a(n) = 11*a(n-1) - 10*a(n-2), a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
a(n) = A125118(n,9) for n>8. - Reinhard Zumkeller, Nov 21 2006
a(n) = A075412(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = A242614(n,A242622(n)). - Reinhard Zumkeller, Jul 17 2014
E.g.f.: (exp(9*x) - 1)*exp(x)/9. - Ilya Gutkovskiy, May 11 2016
a(n) = Sum_{k=0..n-1} 10^k. - Torlach Rush, Nov 03 2020
Sum_{n>=1} 1/a(n) = A065444. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Aug 02 2025: (Start)
a(n) = A002283(n)/9 = A105279(n)/10.
a(n) = A010785(A017173(n-1)) for n >= 1. (End)

A002283 a(n) = 10^n - 1.

Original entry on oeis.org

0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999
Offset: 0

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Comments

A friend from Germany remarks that the sequence 9, 99, 999, 9999, 99999, 999999, ... might be called the grumpy German sequence: nein!, nein! nein!, nein! nein! nein!, ...
The Regan link shows that integers of the form 10^n -1 have binary representations with exactly n trailing 1 bits. Also those integers have quinary expressions with exactly n trailing 4's. For example, 10^4 -1 = (304444)5. The first digits in quinary correspond to the number 2^n -1, in our example (30)5 = 2^4 -1. A similar pattern occurs in the binary case. Consider 9 = (1001)2. - Washington Bomfim Dec 23 2010
a(n) is the number of positive integers with less than n+1 digits. - Bui Quang Tuan, Mar 09 2015
From Peter Bala, Sep 27 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(2*n)) = [10^n - 1; 1, 2*(10^n - 1), 1, 2*(10^n - 1), ...] has period 2. The simple continued fraction expansion of sqrt(a(2*n))/a(n) = [1; 10^n - 1, 2, 10^n - 1, 2, ...] also has period 2. Note the occurrence of large partial quotients in both expansions.
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 the presence of unexpectedly large partial quotients early in the continued fraction expansions of the m-th roots of the numbers a(m*n) for m >= 3. Some typical examples are given below. (End)
For n > 0, numbers whose smallest decimal digit is 9. - Stefano Spezia, Nov 16 2023

Examples

			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)
		

Crossrefs

Programs

Formula

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-10*x)-1/(1-x).
E.g.f.: e^(10*x)-e^x. (End)
a(n) = A075412(n)/A002275(n) = A178630(n)/A002276(n) = A178631(n)/A002277(n) = A075415(n)/A002278(n) = A178632(n)/A002279(n) = A178633(n)/A002280(n) = A178634(n)/A002281(n) = A178635(n)/A002282(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 9*10^(n-1) with a(0)=0; Also: a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=9. - Vincenzo Librandi, Jul 22 2010
For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - Reinhard Zumkeller, Aug 06 2010
A048379(a(n)) = 0. - Reinhard Zumkeller, Feb 21 2014
a(n) = Sum_{k=1..n} 9*10^k. - Carauleanu Marc, Sep 03 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 9*A002275(n).
a(n) = A010785(A008591(n)). (End)

Extensions

More terms from Michael De Vlieger, Sep 27 2015

A002279 a(n) = 5*(10^n - 1)/9.

Original entry on oeis.org

0, 5, 55, 555, 5555, 55555, 555555, 5555555, 55555555, 555555555, 5555555555, 55555555555, 555555555555, 5555555555555, 55555555555555, 555555555555555, 5555555555555555, 55555555555555555, 555555555555555555, 5555555555555555555, 55555555555555555555, 555555555555555555555
Offset: 0

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Comments

Arithmetic mean of all n-digit odd numbers. E.g., a(2) = arithmetic mean of {11,13,15,...,97,99} = (11+99)/2 = 55. - Amarnath Murthy, Aug 02 2005

Crossrefs

Programs

Formula

a(n) = A178632(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 5*10^(n-1) with a(0)=0.
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=5. (End)
G.f.: 5*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 5*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
From Karl-Heinz Hofmann, Nov 28 2023: (Start)
a(n) = A010785(9*n-4) for n > 0.
a(n) = 5 * A002275(n).
a(n) = 5 * A002283(n) / 9. (End)

A002280 a(n) = 6*(10^n - 1)/9.

Original entry on oeis.org

0, 6, 66, 666, 6666, 66666, 666666, 6666666, 66666666, 666666666, 6666666666, 66666666666, 666666666666, 6666666666666, 66666666666666, 666666666666666, 6666666666666666, 66666666666666666, 666666666666666666, 6666666666666666666, 66666666666666666666, 666666666666666666666
Offset: 0

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a(n-1) = number of Fibonacci numbers F(k), k <= 10^n, which end in 0. a(1)=6 because there are 6 Fibonacci numbers up to 10^2 which end in 0. - Shyam Sunder Gupta and Benoit Cloitre, Aug 15 2002
a(n) is the total number of holes in a certain triangle fractal (start with 10 triangles, 6 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015

Crossrefs

Programs

Formula

a(n) = 6*A002275(n).
From Jaume Oliver Lafont, Feb 03 2009: (Start)
G.f.: 6*x/((1-x)*(1-10*x)).
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=6. (End)
a(n) = A178633(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 6*10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
E.g.f.: 2*exp(x)*(exp(9*x) - 1)/3. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 21 2025: (Start)
a(n) = A073551(n+1)/2 for n >= 1.
a(n) = A010785(A017233(n-1)) for n >= 1. (End)

A002477 Wonderful Demlo numbers: a(n) = ((10^n - 1)/9)^2.

Original entry on oeis.org

1, 121, 12321, 1234321, 123454321, 12345654321, 1234567654321, 123456787654321, 12345678987654321, 1234567900987654321, 123456790120987654321, 12345679012320987654321, 1234567901234320987654321
Offset: 1

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Comments

Only the first nine terms of this sequence are palindromes. - Bui Quang Tuan, Mar 30 2015
Not all of the terms are Demlo numbers as defined by Kaprekar, i.e., concat(L,M,R) with M and L+R repdigits using the same digit. For example, a(10), a(19), a(28) are not, but a(k) for k = 11, 12, ..., 18 are. - M. F. Hasler, Nov 18 2017

Examples

			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
		

References

  • D. R. Kaprekar, On Wonderful Demlo numbers, Math. Stud., 6 (1938), 68.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 29.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

G.f.: x*(1+10*x) / ((1-x)*(1-10*x)*(1-100*x)). - Simon Plouffe in his 1992 dissertation
a(n+1) = 100*a(n) + 20*A000042(n) + 1; a(1) = 1. - Reinhard Zumkeller, May 31 2010
a(n) = A000042(n)^2.
a(n) = A075412(n)/9 = A178630(n)/18 = A178631(n)/27 = A075415(n)/36 = A178632(n)/45 = A178633(n)/54 = A178634(n)/63 = A178635(n)/72 = A059988(n)/81. - Reinhard Zumkeller, May 31 2010
a(n+2) = -1000*a(n)+110*a(n+1)+11. - Alexander R. Povolotsky, Jun 06 2014
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81. - Stefano Spezia, May 23 2025

Extensions

Minor edits from N. J. A. Sloane, Aug 18 2009
Further edits from Reinhard Zumkeller, May 12 2010

A002278 a(n) = 4*(10^n - 1)/9.

Original entry on oeis.org

0, 4, 44, 444, 4444, 44444, 444444, 4444444, 44444444, 444444444, 4444444444, 44444444444, 444444444444, 4444444444444, 44444444444444, 444444444444444, 4444444444444444, 44444444444444444, 444444444444444444, 4444444444444444444, 44444444444444444444, 444444444444444444444
Offset: 0

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Crossrefs

Programs

Formula

a(n) = A075415(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 4*10^(n-1) with a(0)=0;
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=4. (End)
G.f.: 4*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 4*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
a(n) = A007091(A024049(n)). - Michel Marcus, Jun 16 2024
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 4*A002275(n).
a(n) = A010785(A017209(n-1)) for n >= 1. (End)

A052041 Squares lacking the digit zero in their decimal expansion.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 81, 121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961, 1156, 1225, 1296, 1369, 1444, 1521, 1681, 1764, 1849, 1936, 2116, 2916, 3136, 3249, 3364, 3481, 3721, 3844, 3969, 4225, 4356
Offset: 1

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Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

This sequence is infinite: see A075415 or A102807 for a constructive proof.
Intersection of A052382 and A000290; A168046(a(n))*A010052(a(n))=1. - Reinhard Zumkeller, Dec 01 2009

Crossrefs

Programs

  • Mathematica
    Select[Range[66]^2, FreeQ[IntegerDigits[#],0]==True &] (* Jayanta Basu, May 25 2013 *)

Formula

a(n) = A052040(n)^2. - R. J. Mathar, Jul 23 2025

A059988 a(n) = (10^n - 1)^2.

Original entry on oeis.org

0, 81, 9801, 998001, 99980001, 9999800001, 999998000001, 99999980000001, 9999999800000001, 999999998000000001, 99999999980000000001, 9999999999800000000001, 999999999998000000000001, 99999999999980000000000001, 9999999999999800000000000001, 999999999999998000000000000001
Offset: 0

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Author

Henry Bottomley, Mar 07 2001

Keywords

Comments

From James D. Klein, Feb 05 2012: (Start)
The periods of the reciprocals of a(n) are the consecutive integers from 0 to 10^n-1, omitting the one integer 10^n-2, right-justified in field widths of size n.
E.g.:
1/81 = 0.012345679...
1/9801 = 0.000102030405060708091011...9799000102...
1/998001 = 0.000001002003004005...997999000001002... (End)
Sum of first 10^n - 1 odd numbers. - Arkadiusz Wesolowski, Jun 12 2013

Examples

			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)
		

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.
  • Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 34.

Crossrefs

Programs

Formula

a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.
a(n) = {999... (n times)}^2 = {999... (n times), 000... (n times)} - {999... (n times)}. For example, 999^2 = 999000 - 999 = 998001. - Kyle D. Balliet, Mar 07 2009
a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - Reinhard Zumkeller, May 31 2010
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)
Sum_{n>=1} 1/a(n) = (log(10)*(QPolyGamma(0, 1, 1/10) - log(10/9)) + QPolyGamma(1, 1, 1/10))/log(10)^2 = 0.012448721523422795191... . - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Elmo R. Oliveira, Aug 02 2025

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)
Showing 1-10 of 26 results. Next