A138147 -id:A138147 - cp's OEIS frontend

A020522 a(n) = 4^n - 2^n.

0, 2, 12, 56, 240, 992, 4032, 16256, 65280, 261632, 1047552, 4192256, 16773120, 67100672, 268419072, 1073709056, 4294901760, 17179738112, 68719214592, 274877382656, 1099510579200, 4398044413952, 17592181850112, 70368735789056, 281474959933440

Offset: 0

N. J. A. Sloane, Simon Plouffe

Number of walks of length 2*n+2 between any two diametrically opposite vertices of the cycle graph C_8. - Herbert Kociemba, Jul 02 2004
If we consider a(4*k+2), then 2^4 == 3^4 == 3 (mod 13); 2^(4*k+2) + 3^(4*k+2) == 3^k*(4+9) == 3*0 == 0 (mod 13). So a(4*k+2) can never be prime. - Jose Brox, Dec 27 2005
If k is odd, then a(n*k) is divisible by a(n), since: a(n*k) = (2^n)^k + (3^n)^k = (2^n + 3^n)*((2^n)^(k-1) - (2^n)^(k-2) (3^n) + - ... + (3^n)^(k-1)). So the only possible primes in the sequence are a(0) and a(2^n) for n>=1. I've checked that a(2^n) is composite for 3 <= n <= 15. As with Fermat primes, a probabilistic argument suggests that there are only finitely many primes in the sequence. - Dean Hickerson, Dec 27 2005
Let x,y,z be elements from some power set P(n), i.e., the power set of a set of n elements. Define a function f(x,y,z) in the following manner: f(x,y,z) = 1 if x is a subset of y and y is a subset of z and x does not equal z; f(x,y,z) = 0 if x is not a subset of y or y is not a subset of z or x equals z. Now sum f(x,y,z) for all x,y,z of P(n). This gives a(n). - Ross La Haye, Dec 26 2005
Number of monic (irreducible) polynomials of degree 1 over GF(2^n). - Max Alekseyev, Jan 13 2006
Let P(A) be the power set of an n-element set A and B be the Cartesian product of P(A) with itself. Then a(n) = the number of (x,y) of B for which x does not equal y. - Ross La Haye, Jan 02 2008
For n>1: central terms of the triangle in A173787. - Reinhard Zumkeller, Feb 28 2010
Pronic numbers of the form: (2^n - 1)*2^n, which is the n-th Mersenne number times 2^n, see A000225 and A002378. - Fred Daniel Kline, Nov 30 2013
Indices where records of A037870 occur. - Philippe Beaudoin, Sep 03 2014
Half the total edge length for a minimum linear arrangement of a hypercube of dimension n. (See Harper's paper below for proof). - Eitan Frachtenberg, Apr 07 2017
Number of pairs in GF(2)^{n+1} whose dot product is 1. - Christopher Purcell, Dec 11 2021

			n=5: a(5) = 4^5 - 2^5 = 1024 - 32 = 992 -> '1111100000'.

Vincenzo Librandi, Table of n, a(n) for n = 0..170
M. Archibald, A. Blecher, A. Knopfmacher, and M. E. Mays, Inversions and Parity in Compositions of Integers, J. Int. Seq., Vol. 23 (2020), Article 20.4.1.
Tom Copeland, The Kervaire-Milnor formula
John Elias, Illustration of initial terms: Twin 2^n hexagonal numbers
L. H. Harper, Optimal Assignment of Numbers to Vertices, J. SIAM 12(1), p. 131--135, March 1964; alternative link.
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
The Sixtieth William Lowell Putnam Mathematical Competition, Question A6, Amer. Math. Monthly 107 (Oct 2000), 721-732; see p. 725.
Index entries for linear recurrences with constant coefficients, signature (6,-8).

Ratio of successive terms of A028365.
Cf. A000225, A060867, A161168, A006516, A059153, A065442.
Cf. A000079, A265736, A020988, A047849.
Cf. A000079, A000302.

a020522 = (* 2) . a006516  -- Reinhard Zumkeller, Dec 15 2015

[4^n - 2^n: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

A020522:=n->4^n-2^n; seq(A020522(n), n=0..50); # Wesley Ivan Hurt, Nov 29 2013

Table[4^n - 2^n, {n, 40}] (* or *) LinearRecurrence[{6, -8}, {0, 2}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n)=4^n-2^n \\ Charles R Greathouse IV, Jan 30 2012

def A020522(n): return (1<Chai Wah Wu, Mar 10 2025

[4^n - 2^n for n in range(0,23)] # Zerinvary Lajos, Jun 05 2009

From Herbert Kociemba, Jul 02 2004: (Start)
G.f.: 2*x/((-1 + 2*x)*(-1 + 4*x)).
a(n) = 6*a(n-1) - 8*a(n-2). (End)
E.g.f.: exp(4*x) - exp(2*x). - Mohammad K. Azarian, Jan 14 2009
From Reinhard Zumkeller, Feb 07 2006, Jaroslav Krizek, Aug 02 2009: (Start)
a(n) = A099393(n)-A000225(n+1) = A083420(n)-A099393(n).
In binary representation, n>0: n 1's followed by n 0's (A138147(n)).
A000120(a(n)) = n.
A023416(a(n)) = n.
A070939(a(n)) = 2*n.
2*a(n)+1 = A030101(A099393(n)). (End)
a(n) = A085812(n) - A001700(n). - John Molokach, Sep 28 2013
a(n) = 2*A006516(n) = A000079(n)*A000225(n) = A265736(A000225(n)). - Reinhard Zumkeller, Dec 15 2015
a(n) = (4^(n/2) - 4^(n/4))*(4^(n/2) + 4^(n/4)). - Bruno Berselli, Apr 09 2018
Sum_{n>0} 1/a(n) = E - 1, where E is the Erdős-Borwein constant (A065442). - Peter McNair, Dec 19 2022
a(n) = A000302(n) - A000079(n). - John Reimer Morales, Aug 04 2025

A135577 Numbers that have only the digit "1" as first, central and final digit. For numbers with 5 or more digits the rest of digits are "0".

Original entry on oeis.org

1, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001

Offset: 1

Omar E. Pol, Feb 24 2008

Also, equal to A135576(n), written in base 2.
Essentially the same as A066138. - R. J. Mathar Apr 29 2008
a(n) has 2n-1 digits.

			----------------------------
n ............ a(n)
----------------------------
1 ............. 1
2 ............ 111
3 ........... 10101
4 .......... 1001001
5 ......... 100010001
6 ........ 10000100001
7 ....... 1000001000001
8 ...... 100000010000001
9 ..... 10000000100000001
10 ... 1000000001000000001

G. C. Greubel, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A001576, A066138, A135576, A138118, A138119, A138120, A138147, A138826.

Join[{1}, LinearRecurrence[{111, -1110, 1000}, {111, 10101, 1001001}, 25]] (* G. C. Greubel, Oct 19 2016 *)
Join[{1},Table[FromDigits[Join[{1},PadRight[{},n,0],{1},PadRight[{},n,0],{1}]],{n,0,10}]] (* Harvey P. Dale, Aug 15 2022 *)

Vec(-x*(2000*x^3-1110*x^2+1)/((x-1)*(10*x-1)*(100*x-1))  + O(x^100)) \\ Colin Barker, Sep 16 2013

a(n) = A135576(n), written in base 2.
Also, a(1)=1, for n>1; a(n)=(concatenation of 1, n-2 digits 0, 1, n-2 digits 0 and 1).
From Colin Barker, Sep 16 2013: (Start)
a(n) = 1 + 10^(n-1) + 100^(n-1) for n>1.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>4.
G.f.: x*(2000*x^3 - 1110*x^2 + 1)/((1-x)*(10*x-1)*(100*x-1)). (End)
E.g.f.: (-111 - 200*x + 100*exp(x) + 10*exp(10*x) + exp(100*x))/100. - Elmo R. Oliveira, Jun 13 2025

A138118 Concatenation of 2n-1 digits 1 and n digits 0.

Original entry on oeis.org

10, 11100, 11111000, 11111110000, 11111111100000, 11111111111000000, 11111111111110000000, 11111111111111100000000, 11111111111111111000000000, 11111111111111111110000000000

Offset: 1

Omar E. Pol, Mar 29 2008

a(n) is also A147537(n) written in base 2. [From Omar E. Pol, Nov 08 2008]

			n .......... a(n)
1 ........... 10
2 ......... 11100
3 ....... 11111000
4 ..... 11111110000
5 ... 11111111100000

Index entries for linear recurrences with constant coefficients, signature (1010,-10000)

Cf. A138147, A138721.

Cf. A147537.

FromDigits/@Table[Join[PadRight[{},2n-1,1],PadRight[{},n,0]],{n,15}] (* Harvey P. Dale, Dec 09 2011 *)

O.g.f.: 10*(1+100x)/[(-1+1000x)*(-1+10x)]. a(n)=A100706(n)*10^n = 10*a(n-1)+11*1000^n. - R. J. Mathar, Apr 03 2008

A109241 Expansion of 1/((1-10x)(1-100*x)).

Original entry on oeis.org

1, 110, 11100, 1111000, 111110000, 11111100000, 1111111000000, 111111110000000, 11111111100000000, 1111111111000000000, 111111111110000000000, 11111111111100000000000, 1111111111111000000000000, 111111111111110000000000000, 11111111111111100000000000000

Offset: 0

Paul Barry, Jun 23 2005

a(n) has n+1 1's and n 0's. Partial sums are A109242.
a(n) = A171476(n) converted from decimal to binary. - Robert Price, Jan 19 2016
Also the binary representation of the n-th iteration of the elementary cellular automaton starting with a single ON (black) cell for Rules 206 and 238. - Robert Price, Feb 21 2016

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. 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 (110,-1000)

Cf. A006516, A138147.

[10^(2*n+1)/9-10^n/9: n in [0..40]]; // Vincenzo Librandi, Feb 22 2016

A109241 := proc(n)(10^(2*n+1)-10^n)/9 ; end proc:
seq(A109241(n),n=0..20) ; # R. J. Mathar, Mar 21 2011

Table[(10^(2*n+1)-10^n)/9, {n, 0, 100}] (* Robert Price, Feb 21 2016 *)
CoefficientList[Series[1/((1 - 100 x) (1 - 10 x)), {x, 0, 33}], x] (* Vincenzo Librandi, Feb 22 2016 *)

a(n)=10^(2*n+1)/9-10^n/9 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (10^(2n+1) - 10^n)/9.
a(n) = A006516(n+1) written in base 2. - Omar E. Pol, Feb 24 2008
a(n) = A138147(n+1)/10. - Omar E. Pol, Nov 08 2008
a(n) = 110*a(n-1) -1000*a(n-2), n>=2. - Vincenzo Librandi, Mar 18 2011
a(n) = A002275(n+1)*10^n. - Wesley Ivan Hurt, Jun 22 2013
E.g.f.: (1/9)*(10*exp(100*x) - exp(10*x)). - G. C. Greubel, Aug 01 2017

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)

A276352 a(n) = 100^n - 10^n.

Original entry on oeis.org

0, 90, 9900, 999000, 99990000, 9999900000, 999999000000, 99999990000000, 9999999900000000, 999999999000000000, 99999999990000000000, 9999999999900000000000, 999999999999000000000000, 99999999999990000000000000, 9999999999999900000000000000, 999999999999999000000000000000

Offset: 0

Carauleanu Marc, Aug 31 2016

Index entries for linear recurrences with constant coefficients, signature (110,-1000).

Cf. A002283, A011557, A138147, A168624.

[100^n-10^n : n in [0..20]]; // Wesley Ivan Hurt, Sep 07 2016

A276352:=n->100^n-10^n: seq(A276352(n), n=0..20); # Wesley Ivan Hurt, Sep 07 2016

Table[100^n - 10^n, {n, 0, 10}] (* Alonso del Arte, Aug 31 2016 *)

a(n)=100^n-10^n \\ Charles R Greathouse IV, Sep 01 2016

a(n) = floor((1000^n)/(10^n + 1)).
a(n) = A002283(n)*A011557(n).
a(n) = 9*A138147(n), for n>0.
a(n) = A168624(n) - 1.
a(n) = Sum_{k=1..n} 9*10^(2n-k).
a(n) = ((10^n)*A002283(2n))/(10^n + 1).
From Chai Wah Wu, Sep 01 2016: (Start)
a(n) = 110*a(n-1) - 1000*a(n-2) for n > 1.
G.f.: 90*x/((10*x - 1)*(100*x - 1)). (End)
E.g.f.: exp(10*x)*(exp(90*x) - 1). - Elmo R. Oliveira, Aug 14 2024

A276758 Numbers n such that A045876(n) = A045876(n+1).

Original entry on oeis.org

10, 1010, 1100, 1119, 1339, 1519, 3139, 5119, 8899, 27799, 46699, 48499, 50559, 55059, 64699, 72799, 84499, 100110, 101010, 101100, 110010, 110100, 111000, 111229, 112129, 117799, 121129, 136699, 147499, 163699, 168199, 171799, 174499, 177199, 186199

Offset: 1

Altug Alkan, Sep 17 2016

A138147 is a subsequence. Therefore, the sequence is infinite. - David A. Corneth, Sep 17 2016
Suppose a term is of the form SDN, where S is a sequence of digits without leading zeros, D is a digit less than 9 and N is a sequence of digits 9 (possibly 0 nines; terms from A002283) and SDN is a concatenation of S, D and N. Let S' be a permutation of digits of S without leading zeros. Then S'DN is also in the sequence. To search terms one may choose S from A179239. - David A. Corneth, Sep 18 2016
Since (n + 8*k) = (n - k + 1)*(n - k) has solutions that are n = k + 3*sqrt(k) and n = k - 3*sqrt(k), for square values of k there are infinitely many terms such that: 1119, 1111119999, 111111111999999999, ...

			1339 is a term because A045876(1339) = A045876(1340).
See 2nd comment. As 27799 is in the sequence, we can see S = 27, D = 7 and N = 99. Now all permutations S' (distinct) of S without leading zeros give terms. They are 72, giving term 72799. - _David A. Corneth_, Sep 18 2016

Cf. A045876, A179239.

A047726(n) = n=digits(n); (#n)!/prod(i=0, 9, sum(j=1, #n, n[j]==i)!);
A007953(n) = sumdigits(n);
lista(nn) = for(n=1, nn, if(A047726(n)*A007953(n) == A047726(n+1)*A007953(n+1), print1(n, ", ")))

cp's OEIS Frontend

A020522 a(n) = 4^n - 2^n.

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A135577 Numbers that have only the digit "1" as first, central and final digit. For numbers with 5 or more digits the rest of digits are "0".

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A138118 Concatenation of 2n-1 digits 1 and n digits 0.

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A109241 Expansion of 1/((1-10*x)*(1-100*x)).

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

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A276352 a(n) = 100^n - 10^n.

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A276758 Numbers n such that A045876(n) = A045876(n+1).

A109241 Expansion of 1/((1-10x)(1-100*x)).