A033934 -id:A033934 - cp's OEIS frontend

A066138 a(n) = 10^(2*n) + 10^n + 1.

3, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001, 1000000000000001000000000000001

Offset: 0

Henry Bottomley, Dec 07 2001

Palindromes whose digit sum is 3.
Essentially the same as A135577. - R. J. Mathar Apr 29 2008
From Peter Bala, Sep 25 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(n)) = [10^n; 1, 1, 2/3*(10^n - 1), 1, 1, 2*10^n, ...] has period 6. As n increases, the expansion has the large partial quotients 2/3*(10^n - 1) and 2*10^n.
For n >= 1, the continued fraction expansion of sqrt(a(2*n))/a(n) = [0; 1, 10^n - 1, 1, 1, 1/3*(10^n - 4), 1, 4, 1, 1/3*(10^n - 4), 1, 1, 10^n - 1, 2, 10^n - 1, ...] has pre-period of length 3 and period 12 beginning 1, 1, 1/3*(10^n - 4), ....  As n increases, the expansion has the large partial quotients 10^n - 1 and 1/3*(10^n - 4).
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 exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n), for m >= 3. For example, it appears that the continued fraction expansion of a(3*n)^(1/3), for n >= 2, begins [10^(2*n); 3*10^n - 1, 1, 0.5*10^(2*n) - 1, 1.44*10^n - 1, 1, ...]. Cf. A000533, A002283 and A168624. (End)

			From _Peter Bala_, Sep 25 2015: (Start)
Simple continued fraction expansions showing large partial quotients:
a(9)^(1/3) =[1000000; 2999, 1, 499999, 1439, 1, 2582643, 1, 1, 1, 2, 3, 3, ...].
a(20)^(1/4) = [10000000000; 39999999999, 1, 3999999999, 16949152542, 2, 1, 2, 6, 1, 4872106, 3, 9, 2, 3, ...].
a(25)^(1/5) = [10000000000; 4999999999999999, 1, 3333333332, 2, 1, 217391304347825, 2, 2, 1, 1, 1, 2, 1, 23980814, 1, 1, 1, 1, 1, 7, ...]. (End)

Harry J. Smith, Table of n, a(n) for n=0..100
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000533, A002283, A033934, A062397, A074992, A135577, A168624.

[10^(2*n) + 10^n + 1: n in [0..20]]; // Vincenzo Librandi, Sep 27 2015

Table[10^(2 n) + 10^n + 1, {n, 0, 15}] (* Michael De Vlieger, Sep 27 2015 *)
CoefficientList[Series[(3 - 222 x + 1110 x^2)/((1 - 100 x) (1 - 10 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 27 2015 *)

a(n) = { 10^(2*n) + 10^n + 1 } \\ Harry J. Smith, Feb 02 2010

Vec(-3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

A168624(n) = a(2*n)/a(n). - Peter Bala, Sep 24 2015
G.f.: (3 - 222*x + 1110*x^2)/((1 - 100*x)*(1 - 10*x)*(1 - x)). - Vincenzo Librandi, Sep 27 2015
From Colin Barker, Sep 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
G.f.: -3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
From Elmo R. Oliveira, Aug 27 2024: (Start)
E.g.f.: exp(x)*(exp(99*x) + exp(9*x) + 1).
a(n) = 3*A074992(n). (End)

Offset changed from 1 to 0 by Harry J. Smith, Feb 02 2010

More terms from Michael De Vlieger, Sep 27 2015

A267769 Numbers whose base-9 representation is a square when read in base 10.

Original entry on oeis.org

0, 1, 4, 15, 23, 33, 58, 73, 81, 100, 121, 185, 213, 265, 298, 324, 361, 400, 474, 509, 555, 643, 685, 751, 861, 914, 1093, 1153, 1215, 1288, 1354, 1481, 1554, 1705, 1783, 1863, 1945, 2029, 2210, 2301, 2488, 2584, 2673, 2773, 2875, 3101, 3210, 3424, 3538, 3682, 3802, 4038, 4154, 4281, 4450

Offset: 1

M. F. Hasler, Jan 20 2016

Trivially includes powers of 81, since 81^k = 100..00_9 = 10^(2k) when read in base 10. Moreover, for any a(n) in the sequence, 81*a(n) is also in the sequence. One could call "primitive" the terms not of this form. These primitive terms include the subsequence 81^k + 2*9^k + 1 = (9^k+1)^2, k > 0, which yields A033934 when written in base 9.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A267763 - A267768 for bases 3 through 8. The base-2 analog is A000302 = powers of 4.

For a "prime" analog see also A235265, A235266, A152079, A235461 - A235482, A065720 ⊂ A036952, A065721 - A065727, A235394, A235395, A089971 ⊂ A020449, A089981, A090707 - A091924.

Select[Range[0, 5000], IntegerQ@ Sqrt@ FromDigits@ IntegerDigits[#, 9] &] (* Michael De Vlieger, Jan 24 2016 *)

is(n,b=9,c=10)=issquare(subst(Pol(digits(n,b)),x,c))

A267769_list = [int(s, 9) for s in (str(i**2) for i in range(10**6)) if max(s) < '9'] # Chai Wah Wu, Jan 20 2016

A267763 Numbers whose base-3 representation is a square when read in base 10.

Original entry on oeis.org

0, 1, 9, 16, 81, 100, 144, 235, 729, 784, 900, 961, 1296, 1369, 2115, 6561, 6724, 7056, 7225, 8100, 8649, 11664, 11881, 12321, 15985, 19035, 59049, 59536, 60516, 61009, 63504, 64009, 65025, 72900, 73441, 77841, 104976, 105625, 106929, 110889, 143865, 171315, 182428, 531441, 532900, 535824, 537289, 544644, 546121

Offset: 1

M. F. Hasler, Jan 20 2016

Trivially includes powers of 9, since 9^k = 100..00_3 = 10^(2k) when read as a base-10 number. Moreover, for any a(n) in the sequence, 9*a(n) is also in the sequence. One could call "primitive" the terms not of this form; these would be 1, 16 = 121_3, 100 = 10201_3, 235 = 22201_3, 784 = 1002001_3, 961 = 1022121_3, ... These primitive terms include the subsequence 9^k + 2*3^k + 1, k > 0, which yields A033934 when written in base 3.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A267764 - A267769 for bases 4 through 9. The base-2 analog is A000302 = powers of 4.

For a "prime" analog see also A235265, A235266, A152079, A235461 - A235482, A065720 ⊂ A036952, A065721 - A065727, A235394, A235395, A089971 ⊂ A020449, A089981, A090707 - A091924.

[n: n in [0..10^6] | IsSquare(Seqint(Intseq(n, 3)))]; // Vincenzo Librandi, Dec 28 2016

Select[Range[0, 600000], IntegerQ@Sqrt@FromDigits@IntegerDigits[#, 3] &] (* Vincenzo Librandi Dec 28 2016 *)

is(n,b=3,c=10)=issquare(subst(Pol(digits(n,b)),x,c))

A267763_list = [int(d,3) for d in (str(i**2) for i in range(10**6)) if max(d) < '3'] # Chai Wah Wu, Mar 12 2016

A267764 Numbers whose base-4 representation is a square when read in base 10.

Original entry on oeis.org

0, 1, 16, 25, 256, 289, 400, 441, 673, 1761, 1849, 4096, 4225, 4624, 4761, 6400, 6561, 7056, 7713, 10768, 13401, 28176, 29584, 65536, 66049, 67600, 68121, 73984, 74529, 76176, 76729, 77985, 102400, 103041, 104976, 112896, 113569, 123408, 150081, 172288, 214416, 450816, 473344, 501433, 519873

Offset: 1

M. F. Hasler, Jan 20 2016

Trivially includes powers of 16, since 16^k = 100..00_4 = 10^(2k) when read as a base-10 number. Moreover, for any a(n) in the sequence, 16*a(n) is also in the sequence. One could call "primitive" the terms not of this form, these would be 1, 25 = 121_4, 289 = 10201_4, 441 = 12321_4, 673 = 22201_4, 1761 = 123201_4, ... These primitive terms include the subsequence 16^k + 2*4^k + 1 = (4^k+1)^2, k > 0, which yields A033934 when written in base 4.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A267763 - A267769 for bases 3 through 9. The base-2 analog is A000302 = powers of 4.

For a "prime" analog see also A235265, A235266, A152079, A235461 - A235482, A065720 ⊂ A036952, A065721 - A065727, A235394, A235395, A089971 ⊂ A020449, A089981, A090707 - A091924.

Select[Range[1000], IntegerQ[Sqrt[FromDigits[IntegerDigits[#, 4]]]] &] (* Alonso del Arte, Jan 23 2016 *)

is(n,b=4,c=10)=issquare(subst(Pol(digits(n,b)),x,c))

A267764_list = [int(d,4) for d in (str(i**2) for i in range(10**6)) if max(d) < '4'] # Chai Wah Wu, Feb 23 2016

A267768 Numbers whose base-8 representation is a square when read in base 10.

Original entry on oeis.org

0, 1, 4, 14, 21, 30, 52, 64, 81, 100, 149, 174, 212, 241, 256, 289, 382, 405, 446, 532, 622, 661, 804, 849, 896, 1012, 1045, 1102, 1220, 1281, 1344, 1409, 1476, 1557, 1630, 1780, 1920, 2001, 2197, 2286, 2452, 2545, 2593, 2878, 2965, 3070, 3233, 3328, 3441, 3540, 3630, 3733, 4068, 4096

Offset: 1

M. F. Hasler, Jan 20 2016

Trivially includes powers of 64, since 64^k = 100..00_8 = 10^(2k) when read in base 10. Moreover, for any a(n) in the sequence, 64*a(n) is also in the sequence. One could call "primitive" the terms not of this form. These primitive terms include the subsequence 64^k + 2*8^k + 1 = (8^k+1)^2, k > 0, which yields A033934 when written in base 8.

Motivated by the subsequence A267490 which lists the primes in this sequence.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A267763 - A267769 for bases 3 through 9. The base-2 analog is A000302 = powers of 4.

For a "prime" analog see also A235265, A235266, A152079, A235461 - A235482, A065720 ⊂ A036952, A065721 - A065727, A235394, A235395, A089971 ⊂ A020449, A089981, A090707 - A091924.

[n: n in [0..10^4] | IsSquare(Seqint(Intseq(n, 8)))]; // Vincenzo Librandi, Dec 28 2016

Select[Range[0, 2 10^4], IntegerQ@Sqrt@FromDigits@IntegerDigits[#, 8] &] (* Vincenzo Librandi, Dec 28 2016 *)

is(n,b=8,c=10)=issquare(subst(Pol(digits(n,b)),x,c))

A267768_list = [int(s, 8) for s in (str(i**2) for i in range(10**6)) if max(s) < '8'] # Chai Wah Wu, Jan 20 2016

A249784 Number of divisors of n^(n^n).

Original entry on oeis.org

1, 5, 28, 513, 3126, 2176875649, 823544, 50331649, 774840979, 100000000020000000001, 285311670612, 158993694406808436568227841, 302875106592254, 123476695691247958050243432972289, 191751059232884087544279144287109376, 73786976294838206465

Offset: 1

Jon E. Schoenfield, Nov 05 2014

nonn

An infinite number of squares are terms of this sequence.
Proof: for any n of the form (p*q)^k (with p and q distinct primes), a(n) = (k * n^n + 1)^2.
It seems likely that the only nontrivial palindromes in this sequence comprise a subset of these squares and occur at n = 10^(10^M) for M>=0; at such values of n, a(n) = (10^(10^(10^M + M) + M) + 1)^2 = A033934(10^(10^M + M) + M). The actual decimal expansion of each of these numbers is of the form 1000...0002000...0001, where the total number of zero digits on each side of the 2 is 10^(10^M + M) + M - 1.

			12 = 2^2 * 3^1 (two distinct prime factors, with multiplicities e_1=2 and e_2=1), so a(12) = (2*k+1)*(1*k+1) = 2*k^2 + 3*k + 1 where k = 12^12, so a(12) = 158993694406808436568227841.

Jon E. Schoenfield, Table of n, a(n) for n = 1..155

Cf. A000005, A000312, A002488, A014566, A033934, A225739.

// program to generate b-file
for n in [1..155] do
   k:=n^n;
   F:=Factorization(n);
   prod:=1;
   for j in [1..#F] do
      prod*:=F[j,2]*k + 1;
   end for;
   n, prod;
end for;

a(n)=my(v=factor(n)[,2]*n^n); prod(i=1,#v,v[i]+1) \\ Charles R Greathouse IV, Jul 21 2015

def A249784(n):
   n_exp_n = n^n
   return mul(exp[1]*n_exp_n + 1 for exp in factor(n))
[A249784(n) for n in (1..16)] # Peter Luschny, Nov 08 2014

a(n) = A000005(A002488(n)).
a(n) = Product_{j=1..m} (e_j * n^n + 1)
   where m = number of distinct prime factors of n
   and e_j = multiplicity of the j-th prime factor.
If n is a prime p, then m=1 and e_1=1, so
   a(p) = p^p + 1 = A000312(p) + 1 = A014566(p).
If n=10^L, then m=2 and e_1=e_2=L, so
   a(10^L) = (L * 10^(L * 10^L) + 1)^2.

A267765 Numbers whose base-5 representation is a square when read in base 10.

Original entry on oeis.org

0, 1, 4, 25, 36, 49, 89, 100, 121, 139, 249, 329, 351, 625, 676, 729, 900, 961, 1225, 1551, 1654, 2146, 2225, 2289, 2500, 2601, 3025, 3289, 3475, 3521, 3814, 4324, 4529, 4801, 5086, 5149, 6225, 6726, 6829, 7374, 8225, 8464, 8775, 9454, 9601, 13926, 15625, 15876, 16129, 16900, 17161

Offset: 1

M. F. Hasler, Jan 20 2016

Trivially includes powers of 25, since 25^k = 100..00_5 = 10^(2k) when read in base 10. Moreover, for any a(n) in the sequence, 25*a(n) is also in the sequence. One could call "primitive" the terms not of this form, these would be 1, 4, 36 = 121_5, 49 = 144_5, 89 = 324_5, ... These primitive terms include the subsequence 25^k + 2*5^k + 1 = (5^k+1)^2, k > 0, which yields A033934 when written in base 5.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A267763 - A267769 for bases 3 through 9. The base-2 analog is A000302 = powers of 4.

For a "prime" analog see also A235265, A235266, A152079, A235461 - A235482, A065720 ⊂ A036952, A065721 - A065727, A235394, A235395, A089971 ⊂ A020449, A089981, A090707 - A091924.

Select[Range[0, 17200], IntegerQ@ Sqrt@ FromDigits@ IntegerDigits[#, 5] &] (* Michael De Vlieger, Jan 24 2016 *)

is(n,b=5,c=10)=issquare(subst(Pol(digits(n,b)),x,c))

A267765_list = [int(d,5) for d in (str(i**2) for i in range(10**6)) if max(d) < '5'] # Chai Wah Wu, Mar 12 2016

A267766 Numbers whose base-6 representation is a square when read in base 10.

Original entry on oeis.org

0, 1, 4, 17, 36, 49, 64, 89, 124, 144, 169, 232, 305, 388, 409, 449, 544, 577, 612, 665, 953, 1105, 1296, 1369, 1444, 1529, 1764, 1849, 1936, 2033, 2304, 2825, 3097, 3204, 3280, 3473, 4345, 4464, 4588, 4841, 5104, 5184, 5329, 5633, 6084, 6241, 7081, 7649, 8044, 8352, 8449, 9160, 9593

Offset: 1

M. F. Hasler, Jan 20 2016

Trivially includes powers of 36, since 36^k = 100..00_6 = 10^(2k) when read in base 10. Moreover, for any a(n) in the sequence, 36*a(n) is also in the sequence. One could call "primitive" the terms not of this form. These primitive terms include the subsequence 36^k + 2*6^k + 1 = (6^k+1)^2, k > 0, which yields A033934 when written in base 6.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A267763 - A267769 for bases 3 through 9. The base-2 analog is A000302 = powers of 4.

For a "prime" analog see also A235265, A235266, A152079, A235461 - A235482, A065720 ⊂ A036952, A065721 - A065727, A235394, A235395, A089971 ⊂ A020449, A089981, A090707 - A091924.

[n: n in [0..10^4] | IsSquare(Seqint(Intseq(n,6)))]; // Bruno Berselli, Jan 20 2016

Select[Range[0, 10^4], IntegerQ@ Sqrt@ FromDigits@ IntegerDigits[#, 6] &] (* Michael De Vlieger, Jan 24 2016 *)

is(n,b=6,c=10)=issquare(subst(Pol(digits(n,b)),x,c))

A267766_list = [int(d,6) for d in (str(i**2) for i in range(10**6)) if max(d) < '6'] # Chai Wah Wu, Mar 12 2016

A225810 a(n) = (10^n)^2 + 4*(10^n) + 1.

Original entry on oeis.org

6, 141, 10401, 1004001, 100040001, 10000400001, 1000004000001, 100000040000001, 10000000400000001, 1000000004000000001, 100000000040000000001, 10000000000400000000001, 1000000000004000000000001, 100000000000040000000000001, 10000000000000400000000000001

Offset: 0

Lance J. Weingartz, Jul 29 2013

This is an instance of (10^n)^2 + x(10^n) + 1 which umbrellas A066138, A033934, A171375, A171410, A171461, A171513 and A171553 which all produce palindromes of the form 1...n...1 when n <> 0.

Colin Barker, Table of n, a(n) for n = 0..499
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A033934, A066138, A171375, A171410, A171461, A171513, A171553.

Table[(10^n)^2 + 4*(10^n) + 1, {n, 0, 20}] (* T. D. Noe, Aug 12 2013 *)
LinearRecurrence[{111,-1110,1000},{6,141,10401},20] (* Harvey P. Dale, Oct 28 2017 *)

Vec(-3*(470*x^2-175*x+2)/((x-1)*(10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Apr 27 2015

From Colin Barker, Apr 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
G.f.: -3*(470*x^2-175*x+2)/((x-1)*(10*x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(1 + 4*exp(9*x) + exp(99*x)). - Elmo R. Oliveira, Jul 04 2025

A225813 a(n) = (10^n)^2 + 7*(10^n) + 1.

Original entry on oeis.org

9, 171, 10701, 1007001, 100070001, 10000700001, 1000007000001, 100000070000001, 10000000700000001, 1000000007000000001, 100000000070000000001, 10000000000700000000001, 1000000000007000000000001, 100000000000070000000000001, 10000000000000700000000000001

Offset: 0

Lance J. Weingartz, Jul 29 2013

This is an instance of (10^n)^2 + x(10^n) + 1 which umbrellas A066138, A033934, A171375, A171410, A171461, A171513 and A171553 which all produce palindromes of the form 1...x...1 when n <> 0.

Colin Barker, Table of n, a(n) for n = 0..499
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A033934, A066138, A171375, A171410, A171461, A171513, A171553.

A225813:=n->(10^n)^2 + 7*(10^n) + 1: seq(A225813(n), n=0..20); # Wesley Ivan Hurt, Apr 08 2017

Table[(10^n)^2 + 7*(10^n) + 1, {n, 0, 20}] (* T. D. Noe, Aug 12 2013 *)
LinearRecurrence[{111,-1110,1000},{9,171,10701},20] (* Harvey P. Dale, Apr 12 2020 *)

Vec(-9*(190*x^2-92*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Apr 27 2015

From Colin Barker, Apr 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
G.f.: -9*(190*x^2-92*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
E.g.f.: exp(x)*(1 + 7*exp(9*x) + exp(99*x)). - Elmo R. Oliveira, Jul 04 2025

cp's OEIS Frontend

A066138 a(n) = 10^(2*n) + 10^n + 1.

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A267769 Numbers whose base-9 representation is a square when read in base 10.

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A267763 Numbers whose base-3 representation is a square when read in base 10.

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A267764 Numbers whose base-4 representation is a square when read in base 10.

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A267768 Numbers whose base-8 representation is a square when read in base 10.

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A249784 Number of divisors of n^(n^n).

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A267765 Numbers whose base-5 representation is a square when read in base 10.

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