A075415 -id:A075415 - cp's OEIS frontend

A178630 a(n) = 18*((10^n - 1)/9)^2.

18, 2178, 221778, 22217778, 2222177778, 222221777778, 22222217777778, 2222222177777778, 222222221777777778, 22222222217777777778, 2222222222177777777778, 222222222221777777777778, 22222222222217777777777778, 2222222222222177777777777778, 222222222222221777777777777778

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 18 = 9 * 2;
n=2: ................... 2178 = 99 * 22;
n=3: ................. 221778 = 999 * 222;
n=4: ............... 22217778 = 9999 * 2222;
n=5: ............. 2222177778 = 99999 * 22222;
n=6: ........... 222221777778 = 999999 * 222222;
n=7: ......... 22222217777778 = 9999999 * 2222222;
n=8: ....... 2222222177777778 = 99999999 * 22222222;
n=9: ..... 222222221777777778 = 999999999 * 222222222.

G. C. Greubel, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178631, A178632, A178633, A178634, A178635.

Cf. A002276, A002281, A002283, A002477.

List([1..20], n -> 18*((10^n-1)/9)^2); # G. C. Greubel, Jan 28 2019

[18*((10^n - 1)/9)^2: n in [1..20]]; // Vincenzo Librandi, Dec 28 2010

Table[18*((10^n-1)/9)^2, {n, 1, 20}] (* G. C. Greubel, Jan 28 2019 *)

a(n)=18*(10^n\9)^2 \\ Charles R Greathouse IV, Aug 21 2011

[18*((10^n-1)/9)^2 for n in (1..20)] # G. C. Greubel, Jan 28 2019

a(n) = 18*A002477(n) = A002283(n)*A002276(n).
a(n)=((A002276(n-1)*10 + 1)*10^(n-1) + A002281(n-1))*10 + 8.
G.f.: 18*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
From Elmo R. Oliveira, Jul 30 2025: (Start)
E.g.f.: 2*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) for n > 3. (End)

A178631 a(n) = 27*((10^n - 1)/9)^2.

Original entry on oeis.org

27, 3267, 332667, 33326667, 3333266667, 333332666667, 33333326666667, 3333333266666667, 333333332666666667, 33333333326666666667, 3333333333266666666667, 333333333332666666666667, 33333333333326666666666667, 3333333333333266666666666667, 333333333333332666666666666667

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 27 = 9 * 3;
n=2: ................... 3267 = 99 * 33;
n=3: ................. 332667 = 999 * 333;
n=4: ............... 33326667 = 9999 * 3333;
n=5: ............. 3333266667 = 99999 * 33333;
n=6: ........... 333332666667 = 999999 * 333333;
n=7: ......... 33333326666667 = 9999999 * 3333333;
n=8: ....... 3333333266666667 = 99999999 * 33333333;
n=9: ..... 333333332666666667 = 999999999 * 333333333.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178632, A178633, A178634, A178635.

Cf. A002277, A002280, A002283, A002477.

[27*((10^n-1)/9)^2: n in [1..50]]; // Vincenzo Librandi, Dec 28 2010

27*(FromDigits/@Table[PadRight[{},n,1],{n,20}])^2 (* or *) LinearRecurrence[ {111,-1110,1000},{27,3267,332667},20] (* Harvey P. Dale, Oct 11 2012 *)

A178631(n):=27*((10^n-1)/9)^2$ makelist(A178631(n),n,1,10); /* Martin Ettl, Nov 12 2012 */

a(n)=27*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = 27*A002477(n) = A002283(n)*A002277(n).
a(n) = ((A002277(n-1)*10 + 2)*10^(n-1) + A002280(n-1))*10 + 7.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>3, a(1)=27, a(2)=3267, a(3)=332667. - Harvey P. Dale, Oct 11 2012
G.f.: 27*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/3. - Elmo R. Oliveira, Aug 01 2025

A178632 a(n) = 45*((10^n - 1)/9)^2.

Original entry on oeis.org

45, 5445, 554445, 55544445, 5555444445, 555554444445, 55555544444445, 5555555444444445, 555555554444444445, 55555555544444444445, 5555555555444444444445, 555555555554444444444445, 55555555555544444444444445, 5555555555555444444444444445, 555555555555554444444444444445

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 45 = 9 * 5;
n=2: ................... 5445 = 99 * 55;
n=3: ................. 554445 = 999 * 555;
n=4: ............... 55544445 = 9999 * 5555;
n=5: ............. 5555444445 = 99999 * 55555;
n=6: ........... 555554444445 = 999999 * 555555;
n=7: ......... 55555544444445 = 9999999 * 5555555;
n=8: ....... 5555555444444445 = 99999999 * 55555555;
n=9: ..... 555555554444444445 = 999999999 * 555555555.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178631, A178633, A178634, A178635.

Cf. A002278, A002279, A002283, A002477.

[45*((10^n-1)/9)^2: n in [1..50]]; // Vincenzo Librandi, Dec 28 2010

45 (FromDigits/@Table[PadRight[{}, n, 1], {n, 20}])^2 (* Vincenzo Librandi, Mar 20 2014 *)
LinearRecurrence[{111,-1110,1000},{45,5445,554445},20] (* Harvey P. Dale, Jan 23 2019 *)

A178632(n):=45*((10^n-1)/9)^2$ makelist(A178632(n),n,1,12); /* Martin Ettl, Nov 08 2012 */

a(n)=45*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = 45*A002477(n) = A002283(n)*A002279(n).
a(n) = (A002279(n-1)*10^n + A002278(n))*10 + 5.
G.f.: 45*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
From Elmo R. Oliveira, Aug 01 2025: (Start)
E.g.f.: 5*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) for n > 3. (End)

A178633 a(n) = 54*((10^n - 1)/9)^2.

Original entry on oeis.org

54, 6534, 665334, 66653334, 6666533334, 666665333334, 66666653333334, 6666666533333334, 666666665333333334, 66666666653333333334, 6666666666533333333334, 666666666665333333333334, 66666666666653333333333334, 6666666666666533333333333334, 666666666666665333333333333334

Offset: 1

Reinhard Zumkeller, May 31 2010

			n = 1:                   54 = 9 * 6;
n = 2:                 6534 = 99 * 66;
n = 3:               665334 = 999 * 666;
n = 4:             66653334 = 9999 * 6666;
n = 5:           6666533334 = 99999 * 66666;
n = 6:         666665333334 = 999999 * 666666;
n = 7:       66666653333334 = 9999999 * 6666666;
n = 8:     6666666533333334 = 99999999 * 66666666;
n = 9:   666666665333333334 = 999999999 * 666666666.

Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.

Colin Barker, Table of n, a(n) for n = 1..499
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178631, A178632, A178634, A178635.

Cf. A002277, A002280, A002283, A002477.

[54*((10^n-1)/9)^2: n in [1..50]]; // Vincenzo Librandi, Dec 28 2010

54 ((10^Range[20] - 1)/9)^2 (* Vincenzo Librandi, Dec 07 2015 *)

a(n)=54*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

Vec(54*x*(1+10*x)/((1-x)*(1-10*x)*(1-100*x)) + O(x^20)) \\ Colin Barker, Dec 07 2015

a(n) = 54*A002477(n) = A002283(n)*A002280(n).
a(n) = ((A002280(n-1)*10 + 5)*10^(n-1) + A002277(n-1))*10 + 4 = (2/3)*(10^n - 1)^2.
From Colin Barker, Dec 07 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>3.
G.f.: 54*x*(1+10*x)/((1-x)*(1-10*x)*(1-100*x)). (End)
E.g.f.: 2*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/3. - Elmo R. Oliveira, Aug 01 2025

A178634 a(n) = 63*((10^n - 1)/9)^2.

Original entry on oeis.org

63, 7623, 776223, 77762223, 7777622223, 777776222223, 77777762222223, 7777777622222223, 777777776222222223, 77777777762222222223, 7777777777622222222223, 777777777776222222222223, 77777777777762222222222223, 7777777777777622222222222223, 777777777777776222222222222223

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 63 = 9 * 7;
n=2: ................... 7623 = 99 * 77;
n=3: ................. 776223 = 999 * 777;
n=4: ............... 77762223 = 9999 * 7777;
n=5: ............. 7777622223 = 99999 * 77777;
n=6: ........... 777776222223 = 999999 * 777777;
n=7: ......... 77777762222223 = 9999999 * 7777777;
n=8: ....... 7777777622222223 = 99999999 * 77777777;
n=9: ..... 777777776222222223 = 999999999 * 777777777.

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 33 at p. 62.
Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.

G. C. Greubel, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178631, A178632, A178633, A178635.

Cf. A002276, A002281, A002283, A002477.

List([1..20], n -> 63*((10^n - 1)/9)^2); # G. C. Greubel, Jan 28 2019

[63*((10^n - 1)/9)^2: n in [1..20]]; // Vincenzo Librandi, Dec 28 2010

63((10^Range[15]-1)/9)^2 (* or *) Table[FromDigits[Join[PadRight[{},n,7],{6},PadRight[{},n,2],{3}]],{n,0,15}] (* Harvey P. Dale, Apr 23 2012 *)

a(n)=63*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

[63*((10^n - 1)/9)^2 for n in (1..20)] # G. C. Greubel, Jan 28 2019

a(n) = 63*A002477(n) = A002283(n)*A002281(n).
a(n) = ((A002281(n-1)*10 + 6)*10^(n-1) + A002276(n-1))*10 + 3.
G.f.: 63*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 7*exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 3. - Elmo R. Oliveira, Aug 01 2025

A178635 a(n) = 72*((10^n - 1)/9)^2.

Original entry on oeis.org

72, 8712, 887112, 88871112, 8888711112, 888887111112, 88888871111112, 8888888711111112, 888888887111111112, 88888888871111111112, 8888888888711111111112, 888888888887111111111112, 88888888888871111111111112, 8888888888888711111111111112, 888888888888887111111111111112

Offset: 1

Reinhard Zumkeller, May 31 2010

			n=1: ..................... 72 = 9 * 8;
n=2: ................... 8712 = 99 * 88;
n=3: ................. 887112 = 999 * 888;
n=4: ............... 88871112 = 9999 * 8888;
n=5: ............. 8888711112 = 99999 * 88888;
n=6: ........... 888887111112 = 999999 * 888888;
n=7: ......... 88888871111112 = 9999999 * 8888888;
n=8: ....... 8888888711111112 = 99999999 * 88888888;
n=9: ..... 888888887111111112 = 999999999 * 888888888.

Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A059988, A075412, A075415, A178630, A178631, A178632, A178633, A178634.

Cf. A002275, A002282, A002283, A002477.

[72*((10^n-1)/9)^2: n in [1..50]]; // Vincenzo Librandi, Dec 28 2010

72 (FromDigits/@Table[PadRight[{}, n, 1], {n, 40}])^2 (* Vincenzo Librandi, Mar 21 2014 *)

a(n)=72*(10^n\9)^2 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = 72*A002477(n) = A002283(n)*A002282(n).
a(n) = ((A002282(n-1)*10 + 7)*10^(n-1) + A002275(n-1))*10 + 2.
G.f.: 72*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Ilya Gutkovskiy, Feb 24 2017
From Elmo R. Oliveira, Aug 01 2025: (Start)
E.g.f.: 8*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) for n > 3. (End)

A102807 a(n) is the square of one plus the number consisting of n 3's.

Original entry on oeis.org

1, 16, 1156, 111556, 11115556, 1111155556, 111111555556, 11111115555556, 1111111155555556, 111111111555555556, 11111111115555555556, 1111111111155555555556, 111111111111555555555556, 11111111111115555555555556, 1111111111111155555555555556, 111111111111111555555555555556

Offset: 0

Ron Knott, Feb 27 2005

Old name was: The number (333...334)^2.
An infinite sequence of squares with no zeros in base 10.
a(n) = A104265(2n) for n > 0. - Chai Wah Wu, Mar 24 2020

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 31 at p. 61.
Italo Ghersi, Matematica dilettevole e curiosa, pp. 111-112, Hoepli, Milano, 1967. [Vincenzo Librandi, Dec 31 2008]

Seiichi Manyama, Table of n, a(n) for n = 0..500
Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A052041, A093137, A075415, A109344, A104265.

a:= n-> (1+parse(cat(0, 3$n)))^2:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 03 2018

Table[(10^n + 2)^2/9, {n, 0, 20}] (* Paolo Xausa, Jun 26 2024 *)

From R. J. Mathar, Jan 06 2009: (Start)
a(n) = (100^n + 4*10^n + 4)/9.
G.f.: (1 - 95*x + 490*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)
E.g.f.: exp(x)*(4 + 4*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024

New name from Alois P. Heinz, Sep 03 2018

A104264 Number of n-digit squares with no zero digits.

Original entry on oeis.org

3, 6, 19, 44, 136, 376, 1061, 2985, 8431, 24009, 67983, 193359, 549697, 1563545, 4446173, 12650545, 35999714, 102439796, 291532841, 829634988, 2360947327, 6719171580, 19122499510, 54423038535, 154888366195

Offset: 1

Reinhard Zumkeller and Ron Knott, Feb 26 2005

Comments from David W. Wilson, Feb 26 2005: (Start)
"There are approximately s(d) = (10^d)^(1/2) - (10^(d-1))^(1/2) d-digit squares. A random d-digit number has the probability p(d) = (9/10)^(d-1) of being zeroless (exponent d-1 as opposed to d because the first digit is not zero). So we expect p(d)s(d) zeroless d-digit squares.
"For d = 1 through 12, we get (truncating): 1, 5, 15, 44, 127, 363, 1034, 2943, 8377, 23841, 67854, 193117, ... The elements grow approximately geometrically with limit ratio (9/10)*10^(1/2) = 2.846+.
"The same naive estimate can easily be generalize to k-th powers, giving the estimate s(d) = (10^d)^(1/k) - (10^(d-1))^(1/k) for d-digit k-th powers. p(d) remains the same. The resulting estimates have ratio (9/10)*10^(1/k).
"We should expect an infinite number of zeroless k-th powers when this ratio is >= 1, which it is for k <= 21. For k >= 22, the ratio is < 1 and we should expect a finite number of zeroless k-th powers." (End)

			a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.

Cf. A052041, A104265, A104266, A075415, A102807.

def aupton(terms):
  c, k, kk = [0 for i in range(terms)], 1, 1
  while kk < 10**terms:
    s = str(kk)
    c[len(s)-1], k, kk = c[len(s)-1] + (s.count('0')==0), k+1, kk + 2*k + 1
  return c
print(aupton(14)) # Michael S. Branicky, Mar 06 2021

a(14)-a(18) from Donovan Johnson, Nov 05 2009
a(19)-a(21) from Donovan Johnson, Mar 23 2011
a(22)-a(25) from Donovan Johnson, Jan 29 2013

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

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

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 22^2 = 484.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417.

Cf. A002275, A002276, A002283, A002477.

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

LinearRecurrence[{111, -1110, 1000}, {0, 4, 484}, 30] (* Vincenzo Librandi, Apr 25 2017 *)

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

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

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 44^2 = 1936.

Colin Barker, Table of n, a(n) for n = 0..100
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417.

Cf. A002275, A002278, A002283, A002477.

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

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)

cp's OEIS Frontend

A178630 a(n) = 18*((10^n - 1)/9)^2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

A178631 a(n) = 27*((10^n - 1)/9)^2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A178632 a(n) = 45*((10^n - 1)/9)^2.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Formula

A178633 a(n) = 54*((10^n - 1)/9)^2.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

A178634 a(n) = 63*((10^n - 1)/9)^2.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

A178635 a(n) = 72*((10^n - 1)/9)^2.

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Magma