A066443 -id:A066443 - cp's OEIS frontend

A083234 a(n) = (3*10^n + 2^n)/4.

1, 8, 76, 752, 7504, 75008, 750016, 7500032, 75000064, 750000128, 7500000256, 75000000512, 750000001024, 7500000002048, 75000000004096, 750000000008192, 7500000000016384, 75000000000032768, 750000000000065536

Offset: 0

Paul Barry, Apr 23 2003

Binomial transform of A066443.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (12,-20).

Cf. A066443, A082724.

[(3*10^n+2^n)/4: n in [0..25]]; // Vincenzo Librandi, Jun 29 2011

Table[(3*10^n + 2^n)/4, {n, 0, 20}] (* Wesley Ivan Hurt, Apr 24 2021 *)

a(n)=(3*10^n+2^n)/4 \\ Charles R Greathouse IV, Jun 29 2011

a(n) = (3*10^n + 2^n)/4.
G.f.: (1-4*x)/((1-10*x)*(1-2*x)).
E.g.f.: (3*exp(10*x) + exp(2*x))/4.
a(n) = 12*a(n-1)-20*a(n-2). - Wesley Ivan Hurt, Apr 24 2021

A092896 Related to random walks on the 4-cube.

Original entry on oeis.org

1, 1, 5, 17, 65, 257, 1025, 4097, 16385, 65537, 262145, 1048577, 4194305, 16777217, 67108865, 268435457, 1073741825, 4294967297, 17179869185, 68719476737, 274877906945, 1099511627777, 4398046511105, 17592186044417, 70368744177665, 281474976710657

Offset: 0

Paul Barry, Mar 12 2004

Gives the denominators in the probability that a random walk on the 4-cube returns to its starting corner on the 2n-th step. Partial sums of A092898. Binomial transform of A092897.

Palindromic numbers in base 2 with an odd number of bits that can be written as 2^(2n) + 1, n >= 1. Palindromic numbers in base 2 with an even number of bits that can be written as 2^(2n+1) + 1 are A087289. - Brad Clardy, Feb 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
M. Kac, Random walk and the theory of Brownian motion, Amer. Math. Monthly, 54:369-391, 1947.
Richard M. Low and Ardak Kapbasov, Non-Attacking Bishop and King Positions on Regular and Cylindrical Chessboards, Journal of Integer Sequences, Vol. 20 (2017), Article 17.6.1, Table 13.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A002450, A052539, A066443, A087289.

Cf. A074515, A091775.

[n lt 2 select 1 else 4^(n-1) +1: n in [0..30]]; // G. C. Greubel, Feb 21 2021

A092896:= n -> `if`(n<2, 1, 4^(n-1) +1); seq(A092896(n), n = 0..30); # G. C. Greubel, Feb 21 2021

CoefficientList[Series[(1 -4x +4x^2 -4x^3)/((1-x)(1-4x)), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 20 2014 *)
LinearRecurrence[{5,-4}, {1,1,5,17}, 30] (* Harvey P. Dale, Mar 19 2016 *)

Vec((1-4*x+4*x^2-4*x^3)/((1-x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Nov 25 2016

[1 if n<2 else 4^(n-1) +1 for n in [0..30]]; # G. C. Greubel, Feb 21 2021

G.f.: (1 - 4*x + 4*x^2 - 4*x^3)/((1-x)*(1-4*x)).
a(n) = 1 + 4^n/4 - 0^n/4 + Sum_{k=0..n} binomial(n, k)*k*(-1)^k.
a(n+1) = 4^n + 1 - 0^n = A002450(n+1) - 4*A002450(n-1). - Paul Barry, Mar 13 2008
a(n) = A052539(n-1), n > 1. - R. J. Mathar, Sep 08 2008
Dropping a(0) and interleaving the terms with zeros gives a sequence with e.g.f. (sin(5ix/2)/sin(ix/2) - 3)/2 = cos(2ix) + cos(ix) - 1. Similar expressions apply to A091775 and A074515, which are also power sums representable by the Bernoulli polynomials. - Tom Copeland, Oct 22 2008
a(n) = 4^(n-1) + 1 for n > 1. - Colin Barker, Nov 25 2016
E.g.f.: (exp(4*x) + 4*exp(x) - 1 - 4*x)/4. - G. C. Greubel, Feb 21 2021

A199560 a(n) = (3*9^n + 1)/2.

Original entry on oeis.org

2, 14, 122, 1094, 9842, 88574, 797162, 7174454, 64570082, 581130734, 5230176602, 47071589414, 423644304722, 3812798742494, 34315188682442, 308836698141974, 2779530283277762, 25015772549499854, 225141952945498682, 2026277576509488134, 18236498188585393202

Offset: 0

Vincenzo Librandi, Nov 08 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A066443, A199561.

Magma
```
[(3*9^n+1)/2: n in [0..30]];
```

LinearRecurrence[{10,-9},{2,14},30] (* or *) NestList[9#-4&,2,30] (* Harvey P. Dale, May 30 2012 *)

a(n) = 2*A066443(n).
a(n) = 9*a(n-1) - 4.
a(n) = 10*a(n-1) - 9*a(n-2).
G.f.: 2*(1-3*x)/((1-x)*(1-9*x)).
From Elmo R. Oliveira, Sep 13 2024: (Start)
E.g.f.: exp(x)*(3*exp(8*x) + 1)/2.
a(n) = A199561(n)/2. (End)

A120437 Differences of A037314 (sum of base-3 digits of n = sum of base-9 digits of n).

Original entry on oeis.org

1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 547, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 547, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1

Offset: 1

John W. Layman, Jul 17 2006

It appears that sign(a(n+1) - a(n)) gives A102283. - Filip Zaludek, Oct 29 2016

This is clear: a(n) = 1 for n == 1 or 2 (mod 3), and a(n) >= 7 for n == 0 (mod 3): see comment by Franklin T. Adams-Watters on A037314. - Robert Israel, Nov 06 2016

Michel Marcus, Table of n, a(n) for n = 1..2000

Cf. A000695, A007949, A037314, A066443.

A037314[0]:= 0;
for n from 0 to 33 do for k from 0 to 2 do
  A037314[3*n+k]:= 9*A037314[n]+k
od od:
seq(A037314[i]-A037314[i-1],i=1..100); # Robert Israel, Nov 06 2016

Differences@ Table[FromDigits[RealDigits[n, 3], 9], {n, 1, 100}] (* Michael De Vlieger, Nov 10 2016, after Clark Kimberling at A037314 *)

a037314(n) = {my(d = digits(n, 3)); subst(Pol(d), x, 9); }
a(n) = a037314(n) - a037314(n-1); \\ Michel Marcus, Oct 30 2016

It appears that the sequence is given by the following recursion: a(n)=1 if n=1, a(n)=9a(3^(k-1))-2 if n=3^k for some k>0, a(n)=a(n-3^(k-1)) if 3^(k-1)0. This recursion formula has been verified for n<=2000.
a(n) = A066443(A007949(n)). (This is equivalent to the conjectured recursion above; that recursion is correct.) - Franklin T. Adams-Watters, Jul 24 2006
G.f. g(x) satisfies g(x) = 9 g(x^3) + x*(1+2*x)/(1+x+x^2). - Robert Israel, Nov 06 2016

A158303 Triangle read by rows, A007318 * (A158300 * 0^(n-k)).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 6, 8, 1, 8, 12, 32, 8, 1, 10, 20, 80, 40, 32, 1, 12, 30, 160, 120, 192, 32, 1, 14, 42, 280, 280, 672, 224, 128, 1, 16, 56, 448, 560, 1792, 896, 1024, 128, 1, 18, 72, 672, 1008, 4032, 2688, 4608, 1152, 512

Offset: 0

Gary W. Adamson, Mar 15 2009

			First few rows of the triangle =
1;
1, 2;
1, 4, 2;
1, 6, 6, 8;
1, 8, 12, 32, 8;
1, 10, 20, 80, 40, 32;
1, 12, 30, 160, 120, 192, 32;
1, 14, 42, 280, 280, 672, 224, 128;
1, 16, 56, 448, 560, 1792, 896, 1024, 128;
1, 18, 72, 672, 1008, 4032, 2688, 4608, 1152, 512;
1, 20, 90, 960, 1680, 8064, 6720, 15360, 5760, 5120, 512;
...

Cf. A158300, A122983 (row sums), A054879, A066443

Triangle read by rows, A007318 * (A158300 * 0^(n-k)). Equals binomial transform of an infinite lower triangular matrix with A158300: (1, 2, 2, 8, 8, 32, 32,...) as the main diagonal and the rest zeros.

A362783 Square array A(n,k) = (n^(2*k + 1) + 1)/(n + 1), n >= 0, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 11, 7, 1, 1, 1, 43, 61, 13, 1, 1, 1, 171, 547, 205, 21, 1, 1, 1, 683, 4921, 3277, 521, 31, 1, 1, 1, 2731, 44287, 52429, 13021, 1111, 43, 1, 1, 1, 10923, 398581, 838861, 325521, 39991, 2101, 57, 1, 1, 1, 43691, 3587227, 13421773, 8138021, 1439671

Offset: 0

Juri-Stepan Gerasimov, May 03 2023

			Array begins:
=====================================================================
n/k |  0    1      2       3         4          5            6   ...
----+----------------------------------------------------------------
0   |  1    1      1       1         1          1            1   ...
1   |  1    1      1       1         1          1            1   ...
2   |  1    3     11      43       171        683         2731   ...
3   |  1    7     61     547      4921      44287       398581   ...
4   |  1   13    205    3277     52429     838861     13421773   ...
5   |  1   21    521   13021    325521    8138021    203450521   ...
6   |  1   31   1111   39991   1439671   51828151   1865813431   ...
   ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Columns k=0..3 are A000012, A002061, A060884, A060888.
Rows n=2..4 are A007583, A066443, A299960.
Main diagonal is A179897.

/* as array */ [[&+[(-n)^j: j in [0..2*k]]: k in [0..6]]: n in [0..6]]; // Juri-Stepan Gerasimov, May 06 2023

A(n,k) = (n^(2*k + 1) + 1)/(n + 1) \\ Andrew Howroyd, May 03 2023

A(n,k) = Sum_{j=0..2*k} (-n)^j.

a(49) corrected by Andrew Howroyd, Jan 20 2024

A075878 Sum of coefficients of (x1)^(2i(1))(x2)^(2i(2))(x3)^(2i(3))*(x4)^(2i(4)) for {(i1),(i2),(i3),(i4)}=0,1,2,... : sum(i)=2n in the expansion of (x1+x2+x3+x4)^(2n) where n >= 1.

Original entry on oeis.org

4, 40, 544, 8320, 131584, 2099200, 33562624, 536903680, 8590065664, 137439477760, 2199025352704, 35184380477440, 562949986975744, 9007199388958720, 144115188612726784, 2305843011361177600, 36893488156009037824, 590295810393065390080, 9444732965876729380864, 151115727452378402652160, 2417851639231457372667904

Offset: 1

Jan Hagberg (jan.hagberg(AT)stat.su.se), Oct 16 2002

For k=3, the sequence divided by 3 is equal to A066443.

Cf. A066443.

Essentially the same as A092812. - Kang Seonghoon (lifthrasiir(AT)gmail.com), Oct 09 2008

a(n, k=4) = 2^(1-k)*sum(r=0,floor((k-1)/2), binomial(k, r)*(k-2*r)^(2*n));
vector(33,n,a(n)) \\ Joerg Arndt, Apr 21 2025

a(n, 4) = 2^(1-4)*Sum_{r=0..floor((4-1)/2)} binomial(4, r)*(4-2*r)^(2*n).

a(n, k) = 2^(1-k)*Sum_{r=0..floor((k-1)/2)} binomial(k, r)*(k-2*r)^(2*n) for k>=1.

Corrected by T. D. Noe, Nov 07 2006

A102592 a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*5^(n-k).

Original entry on oeis.org

1, 8, 80, 832, 8704, 91136, 954368, 9994240, 104660992, 1096024064, 11477712896, 120196169728, 1258710630400, 13181388849152, 138037296103424, 1445545331654656, 15137947242201088, 158526641599938560

Offset: 0

Paul Barry, Jan 22 2005

In general, Sum_{k=0..n} binomial(2n+1,2k)*r^(n-k) has g.f. (1-(r-1)x)/(1-2(r+1)+(r-1)^2x^2) and a(n) = ((sqrt(r)-1)^(2n+1) + (sqrt(r)+1)^(2n+1))/(2*sqrt(r)).

Index entries for linear recurrences with constant coefficients, signature (12,-16).

Cf. A066443, A102591.

G.f.:(1-4x)/(1-12x+16x^2);
a(n) = 12*a(n-1) - 16*a(n-2);
a(n) = sqrt(5)*(sqrt(5)-1)^(2n+1)/10 + sqrt(5)*(sqrt(5)+1)^(2n+1)/10.
a(n) = Sum_{k=0..n} binomial(2n+1, k+1)*5^k. - Paul Barry, May 27 2005
a(n) = 4^(n+1)*A001519(n+1). - N. J. A. Sloane, Apr 13 2011
a(n) = 5^n* 2F1(-n-1/2, -n ; 1/2 ; 1/5). - R. J. Mathar, Aug 23 2024

cp's OEIS Frontend

A083234 a(n) = (3*10^n + 2^n)/4.

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A092896 Related to random walks on the 4-cube.

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A199560 a(n) = (3*9^n + 1)/2.

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A120437 Differences of A037314 (sum of base-3 digits of n = sum of base-9 digits of n).

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Maple

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A158303 Triangle read by rows, A007318 * (A158300 * 0^(n-k)).

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A362783 Square array A(n,k) = (n^(2*k + 1) + 1)/(n + 1), n >= 0, k >= 0, read by antidiagonals.

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A075878 Sum of coefficients of (x1)^(2i(1))*(x2)^(2i(2))*(x3)^(2i(3))*(x4)^(2i(4)) for {(i1),(i2),(i3),(i4)}=0,1,2,... : sum(i)=2n in the expansion of (x1+x2+x3+x4)^(2n) where n >= 1.

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Comments

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Programs

PARI

Formula

Extensions

A102592 a(n) = Sum_{k=0..n} binomial(2n+1, 2k)*5^(n-k).

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A075878 Sum of coefficients of (x1)^(2i(1))(x2)^(2i(2))(x3)^(2i(3))*(x4)^(2i(4)) for {(i1),(i2),(i3),(i4)}=0,1,2,... : sum(i)=2n in the expansion of (x1+x2+x3+x4)^(2n) where n >= 1.