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 1-8 of 8 results.

A075415 Squares of A002280 or numbers (666...6)^2.

Original entry on oeis.org

0, 36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556, 444444444443555555555556, 44444444444435555555555556, 4444444444444355555555555556, 444444444444443555555555555556
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) = 66^2 = 4356.
From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 36 = 9 * 4;
n=2: ................... 4356 = 99 * 44;
n=3: ................. 443556 = 999 * 444;
n=4: ............... 44435556 = 9999 * 4444;
n=5: ............. 4444355556 = 99999 * 44444;
n=6: ........... 444443555556 = 999999 * 444444;
n=7: ......... 44444435555556 = 9999999 * 4444444;
n=8: ....... 4444444355555556 = 99999999 * 44444444;
n=9: ..... 444444443555555556 = 999999999 * 444444444. (End)
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits[PadRight[{},n,6]]^2,{n,0,20}] (* or *) LinearRecurrence[ {111,-1110,1000},{0,36,4356},20] (* Harvey P. Dale, May 20 2021 *)

Formula

a(n) = A002280(n)^2 = (6*A002275(n))^2 = 36*A002275(n)^2.
a(n) = (6*(10^n-1)/9)^2 = (4/9)*(10^(2*n) - 2*10^n + 1), which is n-1 4's, followed by a 3, n-1 5's and a 6. - Ignacio Larrosa Cañestro, Feb 26 2005
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = ((A002278(n-1)*10 + 3)*10^(n-1) + A002279(n-1))*10 + 6 for n>0.
a(n) = A002283(n)*A002278(n). (End)
G.f.: 36*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Arkadiusz Wesolowski, Dec 26 2011
From Elmo R. Oliveira, Jul 27 2025: (Start)
E.g.f.: 4*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) = 36*A002477(n). (End)

Extensions

Edited by Alois P. Heinz, Aug 21 2019 (merged with A102794, submitted by Richard C. Schroeppel, Feb 26 2005)

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

Views

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)

A075411 Squares of A002276.

Original entry on oeis.org

0, 4, 484, 49284, 4937284, 493817284, 49382617284, 4938270617284, 493827150617284, 49382715950617284, 4938271603950617284, 493827160483950617284, 49382716049283950617284, 4938271604937283950617284, 493827160493817283950617284, 49382716049382617283950617284
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) = 22^2 = 484.
		

Crossrefs

Programs

  • Magma
    I:=[0,4,484]; [n le 3 select I[n] else 111*Self(n-1)-1110*Self(n-2)+1000*Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 25 2017
  • Mathematica
    LinearRecurrence[{111, -1110, 1000}, {0, 4, 484}, 30] (* Vincenzo Librandi, Apr 25 2017 *)

Formula

a(n) = A002276(n)^2 = (2*A002275(n))^2 = 4*A002275(n)^2.
From Chai Wah Wu, Apr 24 2017: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
G.f.: 4*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). (End)
a(n) = 4*(10^n - 1)^2/81. - Colin Barker, Apr 25 2017
From Elmo R. Oliveira, Jul 28 2025: (Start)
E.g.f.: 4*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 4*A002477(n). (End)

A075413 Squares of A002278.

Original entry on oeis.org

0, 16, 1936, 197136, 19749136, 1975269136, 197530469136, 19753082469136, 1975308602469136, 197530863802469136, 19753086415802469136, 1975308641935802469136, 197530864197135802469136, 19753086419749135802469136, 1975308641975269135802469136, 197530864197530469135802469136
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) = 44^2 = 1936.
		

Crossrefs

Programs

  • PARI
    concat(0, Vec(16*x*(1 + 10*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20))) \\ Colin Barker, Jul 17 2019

Formula

a(n) = A002278(n)^2 = (4*A002275(n))^2 = 16*A002275(n)^2.
From Colin Barker, Jul 17 2019: (Start)
G.f.: 16*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2.
a(n) = 16*(10^n-1)^2/81. (End)
From Elmo R. Oliveira, Jul 29 2025: (Start)
E.g.f.: 16*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 16*A002477(n). (End)

A075414 Squares of A002279: a(n) = (5*(10^n - 1)/9)^2.

Original entry on oeis.org

0, 25, 3025, 308025, 30858025, 3086358025, 308641358025, 30864191358025, 3086419691358025, 308641974691358025, 30864197524691358025, 3086419753024691358025, 308641975308024691358025, 30864197530858024691358025, 3086419753086358024691358025, 308641975308641358024691358025
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) = 55^2 = 3025.
		

Crossrefs

Programs

  • PARI
    concat(0, Vec(25*x*(1 + 10*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20))) \\ Colin Barker, Jul 17 2019

Formula

a(n) = A002279(n)^2 = (5*A002275(n))^2 = 25*A002275(n)^2.
From Colin Barker, Jul 17 2019: (Start)
G.f.: 25*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2.
a(n) = 25*(10^n-1)^2/81. (End)
From Elmo R. Oliveira, Jul 29 2025: (Start)
E.g.f.: 25*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 25*A002477(n). (End)

A075416 Squares of A002281.

Original entry on oeis.org

0, 49, 5929, 603729, 60481729, 6049261729, 604937061729, 60493815061729, 6049382595061729, 604938270395061729, 60493827148395061729, 6049382715928395061729, 604938271603728395061729, 60493827160481728395061729, 6049382716049261728395061729, 604938271604937061728395061729
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) = 77^2 = 5929.
		

Crossrefs

Programs

  • PARI
    concat(0, Vec(49*x*(1 + 10*x) / ((1 - x)*(1 - 10*x)*(1 - 100*x)) + O(x^20))) \\ Colin Barker, Jul 17 2019

Formula

a(n) = A002281(n)^2 = (7*A002275(n))^2 = 49*A002275(n)^2.
From Colin Barker, Jul 17 2019: (Start)
G.f.: 49*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2.
a(n) = 49*(10^n-1)^2/81. (End)
From Elmo R. Oliveira, Jul 29 2025: (Start)
E.g.f.: 49*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/81.
a(n) = 49*A002477(n). (End)

A271528 a(n) = 2*(10^n - 1)^2/27.

Original entry on oeis.org

0, 6, 726, 73926, 7405926, 740725926, 74073925926, 7407405925926, 740740725925926, 74074073925925926, 7407407405925925926, 740740740725925925926, 74074074073925925925926, 7407407407405925925925926, 740740740740725925925925926, 74074074074073925925925925926
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 09 2016

Keywords

Comments

All terms are multiple of 6.
Converges in a 10-adic sense to ...925925925926.
A transformation of the Wonderful Demlo numbers (A002477).
More generally, the ordinary generating function for the transformation of the Wonderful Demlo numbers, is k*x*(1 + 10*x)/(1 - 111*x + 1110*x^2 - 1000*x^3).

Examples

			n=1:                  6 = 2 * 3;
n=2:                726 = 22 * 33;
n=3:              73926 = 222 * 333;
n=4:            7405926 = 2222 * 3333;
n=5:          740725926 = 22222 * 33333;
n=6:        74073925926 = 222222 * 333333;
n=7:      7407405925926 = 2222222 * 3333333;
n=8:    740740725925926 = 22222222 * 33333333;
n=9:  74074073925925926 = 222222222 * 333333333, etc.
		

Crossrefs

Cf. similar sequences of the form k*((10^n - 1)/9)^2: A075411 (k=4), this sequence (k=6), A075412 (k=9), A075413 (k=16), A178630 (k=18), A075414 (k=25), A178631 (k=27), A075415 (k=36), A178632 (k=45), A075416 (k=49), A178633 (k=54), A178634 (k=63), A075417 (k=64), A178635 (k=72), A059988 (k=81).

Programs

  • Mathematica
    Table[2 ((10^n - 1)^2/27), {n, 0, 15}]
    LinearRecurrence[{111, -1110, 1000}, {0, 6, 726}, 16]
  • PARI
    x='x+O('x^99); concat(0, Vec(6*x*(1+10*x)/(1-111*x+1110*x^2-1000*x^3))) \\ Altug Alkan, Apr 09 2016
    
  • Python
    for n in range(0,10**1):print((int)((2*(10**n-1)**2)/27))
    # Soumil Mandal, Apr 10 2016

Formula

O.g.f.: 6*x*(1 + 10*x)/(1 - 111*x + 1110*x^2 - 1000*x^3).
E.g.f.: 2 (exp(x) - 2*exp(10*x) + exp(100*x))/27.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 6*A002477(n) = 6*A002275(n)^2 = A002276(n)*A002277(n) = sqrt(A075411(n)*A075412(n)).
Sum_{n>=1} 1/a(n) = 0.1680577405662077350849154881928636039793563...
Lim_{n -> infinity} a(n + 1)/a(n) = 100.
Showing 1-8 of 8 results.