A269025 -id:A269025 - cp's OEIS frontend

A094028 Expansion of 1/((1-x)(1-100x)).

1, 101, 10101, 1010101, 101010101, 10101010101, 1010101010101, 101010101010101, 10101010101010101, 1010101010101010101, 101010101010101010101, 10101010101010101010101, 1010101010101010101010101, 101010101010101010101010101, 10101010101010101010101010101

Offset: 0

Paul Barry, Apr 22 2004

Regarded as binary numbers and converted to decimal, these become 1,5,21,85,... the partial sums of 4^n (see A002450).
Partial sums of 100^n.
Odd terms of A056830. - Alexandre Wajnberg, May 31 2005
101 is the only term that is prime, since (100^k-1)/99 = (10^k+1)/11 * (10^k-1)/9. When k is odd and not 1, (10^k+1)/11 is an integer > 1 and thus (100^k-1)/99 is nonprime. When k is even and greater than 2, (100^k-1)/99 has the prime factor 101 and is nonprime. - Felix Fröhlich, Oct 17 2015
Previous comment is the answer to the problem A1 proposed during the 50th Putnam Competition in 1989 (link). - Bernard Schott, Mar 24 2023

			From _Omar E. Pol_, Dec 13 2008: (Start)
=======================
n ....... a(n)
0 ........ 1
1 ....... 101
2 ...... 10101
3 ..... 1010101
4 .... 101010101
5 ... 10101010101
======================
(End)

Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Robert Price, Table of n, a(n) for n = 0..999
Kiran S. Kedlaya, The 50th William Lowell Putnam Mathematical Competition, Problem A1, Dec 02 1989.
J. V. Leyendekkers and A.G. Shannon, Modular Rings and the Integer 3, Notes on Number Theory & Discrete Mathematics, 17 (2011), pp. 47-51.
Robert Price, Comments on A094028 concerning Elementary Cellular Automata, Feb 21 2016.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
Stephen Wolfram, A New Kind of Science.
Index to Elementary Cellular Automata.
Index entries for sequences related to cellular automata.
Index entries for linear recurrences with constant coefficients, signature (101,-100).
Index to sequences related to Olympiads and other Mathematical Competitions.

Bisection of A147759. [Omar E. Pol, Nov 13 2008]
Cf. A002450, A056830, A094027.
Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[1+100*(100^n-1)/99 : n in [0..15]]; // Wesley Ivan Hurt, Oct 17 2015

A094028:=n->1+100*(100^n-1)/99: seq(A094028(n), n=0..15); # Wesley Ivan Hurt, Oct 17 2015

CoefficientList[Series[1/((1-x)(1-100x)),{x,0,20}],x] (* or *) Table[ FromDigits[ PadRight[{},2n-1,{1,0}]],{n,20}] (* or *) LinearRecurrence[ {101,-100},{1,101},20] (* or *) NestList[100#+1&,1,20] (* Harvey P. Dale, Apr 27 2015 *)

A094028(n):=1+100*(100^n-1)/99$
makelist(A094028(n),n,0,30); /* Martin Ettl, Nov 06 2012 */

a(n) = 1+100*(100^n-1)/99 \\ Felix Fröhlich, Oct 17 2015

Vec(1/((1-x)*(1-100*x)) + O(x^100)) \\ Altug Alkan, Oct 17 2015

G.f.: 1/((1-x)*(1-100*x)).
a(n) = 1 + 100*(100^n-1)/99. - N. J. A. Sloane, Apr 20 2008
a(n) = 100^(n+1)/99 - 1/99.
a(n) = A094027(2*n+1).
a(n) = 100*a(n-1) + 1, a(0) = 1. - Philippe Deléham, Feb 22 2014
a(n) = 101*a(n-1) - 100*a(n-2) for n > 1. - Wesley Ivan Hurt, Oct 17 2015
a(n) = (100^(n+1) - 1)/99. - Bernard Schott, Apr 15 2021
E.g.f.: exp(x)*(100*exp(99*x) - 1)/99. - Elmo R. Oliveira, Mar 06 2025

A133853 a(n) = (64^n - 1)/63.

Original entry on oeis.org

0, 1, 65, 4161, 266305, 17043521, 1090785345, 69810262081, 4467856773185, 285942833483841, 18300341342965825, 1171221845949812801, 74958198140788019265, 4797324681010433232961, 307028779584667726909505, 19649841893418734522208321, 1257589881178799009421332545

Offset: 0

Paul Curtz, Jan 07 2008

Partial sums of powers of 64 (A089357), a.k.a. q-numbers for q=64.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. See Table 1.
Index entries related to partial sums
Index entries related to q-numbers
Index entries for linear recurrences with constant coefficients, signature (65,-64).

Cf. A000364.

Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[(64^n-1)/63: n in [0..20]]; // Vincenzo Librandi, Aug 10 2011

LinearRecurrence[{65,-64},{0,1},20] (* Harvey P. Dale, Aug 20 2017 *)

makelist((64^n-1)/63, n, 0, 20); /* Martin Ettl, Nov 12 2012 */

PARI
```
A133853(n)=64^n\63
```

From Wolfdieter Lang, Apr 08 2022: (Start)
a(n) = Sum_{j=0..n-1} 2^(6*j). See the comment.
G.f.: x/((1 - 64*x)*(1 - x)).
E.g.f.: exp(x)*(exp(63*x) - 1)/63. (End)

a(6)-a(15) from Vincenzo Librandi, Aug 10 2011

A218723 a(n) = (256^n - 1)/255.

Original entry on oeis.org

0, 1, 257, 65793, 16843009, 4311810305, 1103823438081, 282578800148737, 72340172838076673, 18519084246547628289, 4740885567116192841985, 1213666705181745367548161, 310698676526526814092329217, 79538861190790864407636279553, 20361948464842461288354887565569

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 256 (A133752), q-integers for q=256.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. Cites this sequence.
Index entries related to partial sums.
Index entries related to q-numbers.
Index entries for linear recurrences with constant coefficients, signature (257,-256).

Cf. A133752.

Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[n le 2 select n-1 else 257*Self(n-1) - 256*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 07 2012

LinearRecurrence[{257, -256}, {0, 1}, 30] (* Vincenzo Librandi, Nov 07 2012 *)

PARI
```
A218723(n)=256^n\255
```

def A218723(n): return (1<<(n<<3))//255 # Chai Wah Wu, Nov 10 2022

a(n) = floor(256^n/255).
From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 256*x)).
a(n) = 257*a(n-1) - 256*a(n-2). (End)
E.g.f.: exp(x)*(exp(255*x) - 1)/255. - Stefano Spezia, Mar 23 2023

A218752 a(n) = (50^n - 1)/49.

Original entry on oeis.org

0, 1, 51, 2551, 127551, 6377551, 318877551, 15943877551, 797193877551, 39859693877551, 1992984693877551, 99649234693877551, 4982461734693877551, 249123086734693877551, 12456154336734693877551, 622807716836734693877551, 31140385841836734693877551

Offset: 0

M. F. Hasler, Nov 04 2012

Partial sums of powers of 50 (A165800).

Converges in a 10-adic sense to ...734693877551.

Vincenzo Librandi, Table of n, a(n) for n = 0..600
Index entries related to partial sums.
Index entries for linear recurrences with constant coefficients, signature (51,-50).

Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

Cf. A165800.

[n le 2 select n-1 else 51*Self(n-1) - 50*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 08 2012

LinearRecurrence[{51, -50}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)
(50^Range[0,20]-1)/49 (* Harvey P. Dale, Sep 12 2022 *)

makelist(floor(50^n/49), n, 0, 30); /* Martin Ettl, Nov 06 2012 */

PARI
```
a(n)=50^n\49
```

a(n) = floor(50^n/49).
G.f.: x/((1-x)(1-50x)).
a(0)=0, a(n) = 50*a(n-1) + 1. - Vincenzo Librandi, Nov 08 2012
E.g.f.: exp(x)*(exp(49*x) - 1)/49. - Elmo R. Oliveira, Aug 29 2024

A261544 a(n) = Sum_{k=0..n} 1000^k.

Original entry on oeis.org

1, 1001, 1001001, 1001001001, 1001001001001, 1001001001001001, 1001001001001001001, 1001001001001001001001, 1001001001001001001001001, 1001001001001001001001001001, 1001001001001001001001001001001, 1001001001001001001001001001001001

Offset: 0

Ilya Gutkovskiy, Aug 24 2015

A sequence of palindromic numbers.

			From _Bruno Berselli_, Aug 25 2015: (Start)
a(n)   is the binary representation of    A023001
-------------------------------------------------
1  ...........................................  1
1001  ........................................  9
1001001 .....................................  73
1001001001  ................................  585
1001001001001  ............................  4681
1001001001001001  ........................  37449
1001001001001001001  ....................  299593
1001001001001001001001  ................  2396745
1001001001001001001001001  ............  19173961, etc.
(End)

Colin Barker, Table of n, a(n) for n = 0..333 (corrected by Michel Marcus, Jan 19 2019)
John Rafael M. Antalan, A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle, arXiv:1908.06014 [math.HO], 2019.
Eric Weisstein's World of Mathematics, Palindromic Number.
Index entries for linear recurrences with constant coefficients, signature (1001,-1000).

Cf. A000533, A002113, A023001.
Subsequence of A033146.
Sums of 100^k: A094028; sums of 10^k: A000042.
Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[(1000^(n+1)-1)/999: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

Table[(1000^(n + 1) - 1)/999, {n, 0, 15}]
LinearRecurrence[{1001, -1000}, {1, 1001}, 20] (* Vincenzo Librandi, Aug 24 2015 *)

Vec(1 / ((x-1)*(1000*x-1)) + O(x^20)) \\ Colin Barker, Aug 24 2015

a(n) = (1000^(n + 1) - 1)/999.
a(n) = 1001*a(n-1) - 1000*a(n-2). - Colin Barker, Aug 24 2015
G.f.: 1 / ((x-1)*(1000*x-1)). - Colin Barker, Aug 24 2015
E.g.f.: (1/999)*(1000000*exp(1000*x) - exp(x)). - G. C. Greubel, Aug 29 2015

cp's OEIS Frontend

A094028 Expansion of 1/((1-x)*(1-100*x)).

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A133853 a(n) = (64^n - 1)/63.

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A218723 a(n) = (256^n - 1)/255.

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A218752 a(n) = (50^n - 1)/49.

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Maxima

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A261544 a(n) = Sum_{k=0..n} 1000^k.

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Magma

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A094028 Expansion of 1/((1-x)(1-100x)).