A138145 -id:A138145 - cp's OEIS frontend

A147596 a(n) is the number whose binary representation is A138145(n).

1, 3, 7, 15, 31, 63, 119, 231, 455, 903, 1799, 3591, 7175, 14343, 28679, 57351, 114695, 229383, 458759, 917511, 1835015, 3670023, 7340039, 14680071, 29360135, 58720263, 117440519, 234881031, 469762055, 939524103, 1879048199, 3758096391

Offset: 1

Omar E. Pol, Nov 08 2008

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A138145, A145641, A147537, A147538, A147539.

Cf. A147540, A147590, A147595, A147597.

[1,3,7,15,31] cat [7*(1+2^(n-3)): n in [6..40]]; // G. C. Greubel, Oct 25 2022

Join[{1,3,7,15,31}, 7*(1+2^(Range[6, 40] -3))] (* G. C. Greubel, Oct 25 2022 *)

Vec(-x*(2*x^2-1)*(4*x^4+2*x^2+1)/((x-1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Sep 15 2013

def A147596(n): return 7*(1+2^(n-3)) -(1/8)*(63*int(n==0) +62*int(n==1) +60*int(n ==2)) -(7*int(n==3) +6*int(n==4) +4*int(n==5))
[A147596(n) for n in range(1,40)] # G. C. Greubel, Oct 25 2022

a(n) = 7*(2^(n-3) + 1) if n >= 6. - Hagen von Eitzen, Jun 02 2009
From Colin Barker, Sep 15 2013: (Start)
a(n) = 3*a(n-1) - 2*a(n-2), for n >= 8.
G.f.: x*(1-2*x^2)*(1+2*x^2+4*x^4) / ((1-x)*(1-2*x)). (End)
E.g.f.: (7/8)*(8*exp(x) + exp(2*x)) - (1/8)*(63 + 62*x + 30*x^2) - 7*x^3/6 - x^4/4 - x^5/30. - G. C. Greubel, Oct 25 2022

More terms from Hagen von Eitzen, Jun 02 2009

A000533 a(0)=1; a(n) = 10^n + 1, n >= 1.

Original entry on oeis.org

1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, 1000000000000001, 10000000000000001

Offset: 0

N. J. A. Sloane

Also, b^n+1 written in base b, for any base b >= 2.
Also, A083318 written in base 2. - Omar E. Pol, Feb 24 2008, Dec 30 2008
Also, palindromes formed from the reflected decimal expansion of the concatenation of 1 and infinite 0's. - Omar E. Pol, Dec 14 2008
It seems that the sequence gives 'all' positive integers m such that m^4 is a palindrome. Note that a(0)^4 = 1 is a palindrome and for n > 0, a(n)^4 = (10^n + 1)^4 = 10^(4n) + 4*10^(3n) + 6*10^(2n) + 4*10^(n) + 1 is a palindrome. - Farideh Firoozbakht, Oct 28 2014
a(n)^2 starts with a(n)+1 for n >= 1. - Dhilan Lahoti, Aug 31 2015
From Peter Bala, Sep 25 2015: (Start)
The simple continued fraction expansion of sqrt(a(2*n)) = [10^n; 2*10^n, 2*10^n, ...] has period 1.
The simple continued fraction expansion of sqrt(a(6*n))/a(2*n) = [10^n - 1; 1, 10^n - 1, 10^n - 1, 1, 2*(10^n - 1), ...] has period 5.
The simple continued fraction expansion of sqrt(a(10*n))/a(2*n) = [10^(3*n) - 10^n; 10^n, 10^n, 2*(10^(3*n) - 10^n), ...] has period 3.
As n increases, these expansions have large partial quotients.
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) begins [10^n; 3*10^(2*n), 10^n, 4.5*10^(2*n), 0.8*10^n, ( 9*10^(2*n + 2) - 144 + 24*(2^mod(n,3) - 1) )/168, ...]. As n increases, the expansion begins with 6 large partial quotients. An example is given below. Cf. A002283, A066138 and A168624.
(End)
a(1) and a(2) are the only prime terms up to n=100000. - Daniel Arribas, Jun 04 2016
Based on factors from A001271, the first abundant number in this sequence should occur in the first M terms, where M is the double factorial M=7607!!. Is any abundant number known in this sequence? - Sergio Pimentel, Oct 04 2019
The (3^5 * 5^2 * 7^2 * 11^2 * 13^2 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97 * 101 * 103 * 107 * 109 * 113 * 127 * 131 * 137 * 139 * 157 * 163 * 181 * 191 * 241 * 251 * 263)-th term of this sequence is an abundant number. - Jon E. Schoenfield, Nov 19 2019
The only perfect power (i.e., a perfect square, a perfect cube, and so forth) in the present sequence is a(0) by Mihăilescu's theorem (since 10^n is a perfect power not equal to 2^3 - see the links in A001597). - Marco Ripà, Feb 04 2025

			The continued fraction expansion of a(9)^(1/3) begins [1000; 3000000, 1000, 4500000, 800, 5357142, 1, 6, 14, 6, 1, 5999999, 6, 1, 12, 7, 1, ...] with 5 large partial quotients immediately following the integer part of the number. - _Peter Bala_, Sep 25 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..100
John Rafael M. Antalan, A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle, arXiv:1908.06014 [math.HO], 2019.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002283, A066138, A083318, A138144, A138145, A152756, A168624.

[10^n + 1 - 0^n: n in [0..30]]; // Vincenzo Librandi, Jul 15 2011

Join[{1},LinearRecurrence[{11,-10},{11,101},20]] (* Harvey P. Dale, May 01 2014 *)

a(n)=if(n,10^n+1,1) \\ Charles R Greathouse IV, Oct 28 2014

a(n) = 10^n + 1 - 0^n. - Reinhard Zumkeller, Jun 10 2003
From Paul Barry, Feb 05 2005: (Start)
G.f.: (1-10*x^2)/((1-x)*(1-10*x));
a(n) = Sum_{k=0..n} binomial(n, k)*0^(k(n-k))*10^k. (End)
a(n) = A178500(n) + 1. - Reinhard Zumkeller, May 28 2010
E.g.f.: exp(x) + exp(10*x) - 1. - Ilya Gutkovskiy, Jun 03 2016

A138146 Palindromes with 2n-1 digits formed from the reflected decimal expansion of the concatenation of 1, 1, 1 and infinite 0's.

Original entry on oeis.org

1, 111, 11111, 1110111, 111000111, 11100000111, 1110000000111, 111000000000111, 11100000000000111, 1110000000000000111, 111000000000000000111, 11100000000000000000111

Offset: 1

Omar E. Pol, Mar 29 2008, May 18 2008

Bisection of A138145.

a(n) is also A147597(n) written in base 2. [Omar E. Pol, Nov 08 2008]

			n ............ a(n)
1 ............. 1
2 ............ 111
3 ........... 11111
4 .......... 1110111
5 ......... 111000111
6 ........ 11100000111
7 ....... 1110000000111
8 ...... 111000000000111
9 ..... 11100000000000111
10 ... 1110000000000000111

Index entries for linear recurrences with constant coefficients, signature (101,-100).

Cf. A138144, A138145.

Cf. A138118, A138119, A138120, A138721, A138826, A147597. [Omar E. Pol, Nov 08 2008]

Table[FromDigits@ If[n < 4, ConstantArray[1, 2 n - 1], Join[#, ConstantArray[0, 2 n - 7], #]] &@ ConstantArray[1, 3], {n, 14}] (* or *)
Rest@ CoefficientList[Series[-x (10 x - 1) (10 x + 1) (100 x^2 + 10 x + 1)/((x - 1) (100 x - 1)), {x, 0, 14}], x] (* Michael De Vlieger, Nov 25 2016 *)

Vec(-x*(10*x-1)*(10*x+1)*(100*x^2+10*x+1)/((x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

From Colin Barker, Sep 16 2013: (Start)
a(n) = 111 + 111*100^(n-2) for n>3.
a(n) = 101*a(n-1) - 100*a(n-2) for n>5.
G.f.: -x*(10*x-1)*(10*x+1)*(100*x^2+10*x+1) / ((x-1)*(100*x-1)). (End)

Better definition from Omar E. Pol, Nov 16 2008

A138120 Concatenation of n digits 1, 2n-1 digits 0 and n digits 1.

Original entry on oeis.org

101, 1100011, 11100000111, 111100000001111, 1111100000000011111, 11111100000000000111111, 111111100000000000001111111, 1111111100000000000000011111111, 11111111100000000000000000111111111, 111111111100000000000000000001111111111

Offset: 1

Omar E. Pol, Apr 06 2008

a(n) has 4n-1 digits.

a(n) is also A147539(n) written in base 2. [Omar E. Pol, Nov 08 2008]

			n ........... a(n)
1 ........... 101
2 ......... 1100011
3 ....... 11100000111
4 ..... 111100000001111
5 ... 1111100000000011111

Index entries for linear recurrences with constant coefficients, signature (11011,-10121010,110110000,-100000000).

Cf. A138144, A138145, A138146, A138148, A138720, A138721, A138722, A138826.

Cf. A147539. [Omar E. Pol, Nov 08 2008]

a:= n-> parse(cat(1$n,0$(2*n-1),1$n)):
seq(a(n), n=1..11);  # Alois P. Heinz, Mar 03 2022

Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},2n-1,0], PadRight[ {},n,1]]],{n,10}] (* or *) LinearRecurrence[{11011,-10121010,110110000,-100000000},{101,1100011,11100000111,111100000001111},10] (* Harvey P. Dale, Mar 19 2016 *)

Vec(x*(10001000*x^2-12100*x+101)/((x-1)*(10*x-1)*(1000*x-1)*(10000*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

def a(n): return int("1"*n + "0"*(2*n-1) + "1"*n)
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, Mar 03 2022

G.f.: x*(10001000*x^2-12100*x+101) / ((x-1)*(10*x-1)*(1000*x-1)*(10000*x-1)). [Colin Barker, Sep 16 2013]

A138721 Concatenation of n digits 1, n digits 0 and n digits 1.

Original entry on oeis.org

101, 110011, 111000111, 111100001111, 111110000011111, 111111000000111111, 111111100000001111111, 111111110000000011111111, 111111111000000000111111111, 111111111100000000001111111111, 111111111110000000000011111111111, 111111111111000000000000111111111111

Offset: 1

Omar E. Pol, Mar 29 2008

a(n) is also A145641(n) written in base 2. - Omar E. Pol, Oct 15 2008

a(n) has 3n digits. - Omar E. Pol, Nov 12 2008

			From _Omar E. Pol_, Nov 12 2008: (Start)
n         Successive digits of a(n)
1                 ( 1 0 1 )
2              ( 1 1 0 0 1 1 )
3           ( 1 1 1 0 0 0 1 1 1 )
4        ( 1 1 1 1 0 0 0 0 1 1 1 1 )
5     ( 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 )
(End)

Paolo Xausa, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

Cf. A000533, A135577, A138120, A138144, A138145, A138146, A138826, A145641, A147757, A147759.

a:= n-> parse(cat(1$n,0$n,1$n)):
seq(a(n), n=1..14);  # Alois P. Heinz, Mar 03 2022

Table[(100^n + 1)*(10^n - 1)/9, {n, 15}] (* Paolo Xausa, Aug 02 2024 *)

Vec(x*(101000*x^2-2200*x+101)/((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

G.f.: x*(101000*x^2 - 2200*x + 101) / ((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)). - Colin Barker, Sep 16 2013

a(n) = (100^n+1)*(10^n-1)/9. - Paolo Xausa, Aug 02 2024

A138144 Palindromes formed from the reflected decimal expansion of the concatenation of 1, 1 and infinite 0's.

Original entry on oeis.org

1, 11, 111, 1111, 11011, 110011, 1100011, 11000011, 110000011, 1100000011, 11000000011, 110000000011, 1100000000011, 11000000000011, 110000000000011, 1100000000000011, 11000000000000011, 110000000000000011

Offset: 1

Omar E. Pol, Mar 29 2008

a(n) is also A147595(n) written in base 2. [From Omar E. Pol, Nov 08 2008]

			n .... a(n)
1 .... 1
2 .... 11
3 .... 111
4 .... 1111
5 .... 11011
6 .... 110011
7 .... 1100011
8 .... 11000011
9 .... 110000011
10 ... 1100000011

Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000533.

Cf. A138118, A138119, A138120, A138145, A138146, A138721, A138826, A147595. [From Omar E. Pol, Nov 08 2008]

LinearRecurrence[{11,-10},{1,11,111,1111,11011},20] (* Harvey P. Dale, Aug 21 2016 *)

Vec(-x*(10*x^2-1)*(10*x^2+1)/((x-1)*(10*x-1)) + O(x^100)) \\ Colin Barker, Sep 15 2013

a(n) = 11+11*10^(n-2) for n>3. a(n) = 11*a(n-1)-10*a(n-2). G.f.: -x*(10*x^2-1)*(10*x^2+1) / ((x-1)*(10*x-1)). - Colin Barker, Sep 15 2013

Better definition. - Omar E. Pol, Nov 15 2008

A138826 Concatenation of 2n-1 digits 1, n digits 0 and 2n-1 digits 1.

Original entry on oeis.org

101, 11100111, 1111100011111, 111111100001111111, 11111111100000111111111, 1111111111100000011111111111, 111111111111100000001111111111111, 11111111111111100000000111111111111111

Offset: 1

Omar E. Pol, Apr 06 2008

a(n) has 5n-2 digits.

a(n) is also A147540(n) written in base 2. [Omar E. Pol, Nov 08 2008]

			n ........... a(n)
1 ........... 101
2 ......... 11100111
3 ....... 1111100011111
4 ..... 111111100001111111
5 ... 11111111100000111111111

Paolo Xausa, Table of n, a(n) for n = 1..195
Index entries for linear recurrences with constant coefficients, signature (101101,-110201100,10110100000,-10000000000).

Cf. A138120, A138144, A138145, A138146, A138148, A138720, A138721, A138722, A147540.

Table[(1000^n + 10)*(100^n - 10)/900, {n, 10}] (* Paolo Xausa, Aug 08 2024 *)

Vec(x*(1100000000*x^3-2000000*x^2+888910*x+101)/((x-1)*(100*x-1)*(1000*x-1)*(100000*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

a(n) = (10^(2n-1)-1+10^(5n-2)-10^(3n-1))/9. [R. J. Mathar, Nov 07 2008, corrected Nov 09 2008]

G.f.: x*(1100000000*x^3-2000000*x^2+888910*x+101) / ((x-1)*(100*x-1)*(1000*x-1)*(100000*x-1)). - Colin Barker, Sep 16 2013

A138119 Concatenation of n digits 1 and 2*n-1 digits 0.

Original entry on oeis.org

10, 11000, 11100000, 11110000000, 11111000000000, 11111100000000000, 11111110000000000000, 11111111000000000000000, 11111111100000000000000000, 11111111110000000000000000000, 11111111111000000000000000000000, 11111111111100000000000000000000000

Offset: 1

Omar E. Pol, Apr 03 2008

a(n) has 3*n-1 digits.

a(n) is also A147538(n) written in base 2. - Omar E. Pol, Nov 08 2008.

			n ...... a(n)
1 ....... 10
2 ...... 11000
3 ..... 11100000
4 .... 11110000000
5 ... 11111000000000

Index entries for linear recurrences with constant coefficients, signature (1100,-100000).

Cf. A138118, A138145, A138146, A138147, A138719, A138720, A138721, A138722, A147538.

LinearRecurrence[{1100, -100000}, {10, 11000}, 15] (* Paolo Xausa, Feb 06 2024 *)

Vec(10*x/((100*x-1)*(1000*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

From Colin Barker, Sep 16 2013: (Start)
a(n) = 1100*a(n-1) - 100000*a(n-2).
G.f.: 10*x / ((100*x-1)*(1000*x-1)). (End)

A138147 Concatenation of n digits 1 and n digits 0.

Original entry on oeis.org

10, 1100, 111000, 11110000, 1111100000, 111111000000, 11111110000000, 1111111100000000, 111111111000000000, 11111111110000000000, 1111111111100000000000, 111111111111000000000000, 11111111111110000000000000, 1111111111111100000000000000, 111111111111111000000000000000

Offset: 1

Omar E. Pol, Mar 29 2008

Also, a(n) = binary representation of A020522(n), for n>0 (see example).

			n ... A020522(n) ..... a(n)
1 ....... 2 ........... 10
2 ...... 12 .......... 1100
3 ...... 56 ......... 111000
4 ..... 240 ........ 11110000
5 ..... 992 ....... 1111100000
6 .... 4032 ...... 111111000000
7 ... 16256 ..... 11111110000000

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 136, Ex. 4.2.2. - N. J. A. Sloane, Jul 27 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..95
Index entries for linear recurrences with constant coefficients, signature (110,-1000).

Cf. A002275, A020522, A138144, A138145, A138720, A276352.

Cf. A109241, A138118, A138119, A138120, A138146, A138721, A138826. - Omar E. Pol, Nov 08 2008

[(10^(2*n) - 10^n)/9: n in [1..30]]; // Vincenzo Librandi, Apr 26 2011

Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},n,0]]],{n,15}] (* Harvey P. Dale, Nov 20 2011 *)

Vec(10*x/((10*x-1)*(100*x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2013

a(n) = (10^(2*n) - 10^n)/9 = A002275(n)*10^n. - Omar E. Pol, Apr 16 2008
a(n) = 10*A109241(n-1). - Omar E. Pol, Nov 08 2008
From Colin Barker, Sep 16 2013: (Start)
a(n) = 110*a(n-1) - 1000*a(n-2).
G.f.: 10*x/((10*x-1)*(100*x-1)). (End)
From Elmo R. Oliveira, Jun 13 2025: (Start)
E.g.f.: exp(10*x)*(exp(90*x) - 1)/9.
a(n) = A276352(n)/9. (End)

A147759 Palindromes formed from the reflected decimal expansion of the infinite concatenation of 1's and 0's.

Original entry on oeis.org

1, 11, 101, 1001, 10101, 101101, 1010101, 10100101, 101010101, 1010110101, 10101010101, 101010010101, 1010101010101, 10101011010101, 101010101010101, 1010101001010101, 10101010101010101, 101010101101010101

Offset: 1

Omar E. Pol, Nov 11 2008

a(k(n)) is divisible by 3 iff k(n) is defined by k(1) = 5 and k(n+1) - k(n) = A100285(n+2). - Altug Alkan, Dec 05 2015

			n .... Successive digits of a(n)
1 ............. ( 1 )
2 ............ ( 1 1 )
3 ........... ( 1 0 1 )
4 .......... ( 1 0 0 1 )
5 ......... ( 1 0 1 0 1 )
6 ........ ( 1 0 1 1 0 1 )
7 ....... ( 1 0 1 0 1 0 1 )
8 ...... ( 1 0 1 0 0 1 0 1 )
9 ..... ( 1 0 1 0 1 0 1 0 1 )
10 ... ( 1 0 1 0 1 1 0 1 0 1 )

Matthew Schulz, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (11,-20,110,-100).

Cf. A000533, A094028, A135577, A138120, A138144, A138145, A138146, A138721, A138826, A147757.

I:=[1,11,101,1001]; [n le 4 select I[n] else 11*Self(n-1)-20*Self(n-2)+110*Self(n-3)-100*Self(n-4): n in [1..30]]; // Vincenzo Librandi, Dec 05 2015

CoefficientList[Series[x/((1 - x) (1 - 10 x) (1 + 10 x^2)),{x, 0, 20}], x] (* Vincenzo Librandi, Dec 05 2015 *)
LinearRecurrence[{11,-20,110,-100},{1,11,101,1001},30] (* Harvey P. Dale, Apr 10 2022 *)

Vec(x/((1-x)*(1-10*x)*(1+10*x^2)) + O(x^30)) \\ Michel Marcus, Dec 05 2015

From R. J. Mathar, Feb 20 2009: (Start)
a(n) = 11*a(n-1)-20*a(n-2)+110*a(n-3)-100*a(n-4).
G.f.: x/((1-x)*(1-10*x)*(1+10*x^2)). (End)
E.g.f.: (exp(x)*(10*exp(9*x) - 1) - 9*cos(sqrt(10)*x))/99. - Stefano Spezia, Oct 12 2024

cp's OEIS Frontend

A147596 a(n) is the number whose binary representation is A138145(n).

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A000533 a(0)=1; a(n) = 10^n + 1, n >= 1.

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A138146 Palindromes with 2n-1 digits formed from the reflected decimal expansion of the concatenation of 1, 1, 1 and infinite 0's.

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A138120 Concatenation of n digits 1, 2n-1 digits 0 and n digits 1.

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A138721 Concatenation of n digits 1, n digits 0 and n digits 1.

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A138144 Palindromes formed from the reflected decimal expansion of the concatenation of 1, 1 and infinite 0's.

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