A163978 -id:A163978 - cp's OEIS frontend

A164090 a(n) = 2*a(n-2) for n > 2; a(1) = 2, a(2) = 3.

2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 98304, 131072, 196608, 262144, 393216, 524288, 786432, 1048576, 1572864, 2097152, 3145728

Offset: 1

Klaus Brockhaus, Aug 09 2009

Interleaving of A000079 without initial 1 and A007283.
Agrees from a(2) onward with A145751 for all terms listed there (up to 65536). Apparently equal to 2, 3 followed by A090989. Equals 2 followed by A163978.
Binomial transform is A000129 without first two terms, second binomial transform is A020727, third binomial transform is A164033, fourth binomial transform is A164034, fifth binomial transform is A164035.
Number of achiral necklaces or bracelets with n beads using up to 2 colors. For n=5, the eight achiral necklaces or bracelets are AAAAA, AAAAB, AAABB, AABAB, AABBB, ABABB, ABBBB, and BBBBB. - Robert A. Russell, Sep 22 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A000079 (powers of 2), A007283 (3*2^n), A029744, A145751, A090989, A163978, A000129, A020727, A164033, A164034, A164035, A052955, A027383, A060482, A070875.

Second column of A284855.

[ n le 2 select n+1 else 2*Self(n-2): n in [1..42] ];

a[n_] := If[EvenQ[n], 3*2^(n/2 - 1), 2^((n + 1)/2)]; Array[a, 42] (* Jean-François Alcover, Oct 12 2017 *)
RecurrenceTable[{a[1]==2,a[2]==3,a[n]==2a[n-2]},a,{n,50}] (* or *) LinearRecurrence[{0,2},{2,3},50] (* Harvey P. Dale, Mar 01 2018 *)

a(n) = if(n%2,2,3) * 2^((n-1)\2); \\ Andrew Howroyd, Oct 07 2017

a(n) = A029744(n+1).
a(n) = A052955(n-1) + 1.
a(n) = A027383(n-2) + 2 for n > 1.
a(n) = A060482(n-1) + 3 for n > 3.
a(n) = A070875(n) - A070875(n-1).
a(n) = (7 - (-1)^n)*2^((1/4)*(2*n - 1 + (-1)^n))/4.
G.f.: x*(2+3*x)/(1-2*x^2).
a(n) = A063759(n-1), n>1. - R. J. Mathar, Aug 17 2009
Sum_{n>=1} 1/a(n) = 5/3. - Amiram Eldar, Mar 28 2022

A265207 Draw a square and follow these steps: Take a square and place at its edges isosceles right triangles with the edge as hypotenuse. Draw a square at every new edge of the triangles. Repeat for all the new squares of the same size. New figures are only placed on empty space. The structure is symmetric about the first square. The sequence gives the numbers of squares of equal size in successive rings around the center.

Original entry on oeis.org

1, 8, 20, 36, 60, 92, 140, 204, 300, 428, 620, 876, 1260, 1772, 2540, 3564, 5100, 7148, 10220, 14316, 20460, 28652, 40940, 57324, 81900, 114668, 163820, 229356, 327660, 458732, 655340, 917484, 1310700, 1834988, 2621420, 3669996, 5242860, 7340012, 10485740, 14680044, 20971500, 29360108, 41943020, 58720236

Offset: 1

Marian Kraus, Dec 04 2015

			By recursion:
a(3)=2*a(1)+20=2*8+20=36
a(4)=2*a(2)+20=2*20+20=60
By function:
a(3)=4*sum_{k=1}^{[(3+1)/2]}(2^k)+6*sum_{k=1}^{[3/2]}(2^k)
=4*sum_{k=1}^{[2]}(2^k)+6*sum_{k=1}^{[1.5]}(2^k)
=4*sum_{k=1}^{2}(2^k)+6*sum_{k=1}^{1}(2^k)
=4*(2^1+2^2)+6*(2^1)
=4*(2+4)+6*(2)=24+12=36
a(4)=4*sum_{k=1}^{[(4+1)/2]}(2^k)+6*sum_{k=1}^{[4/2]}(2^k)
=4*sum_{k=1}^{[2.5]}(2^k)+6*sum_{k=1}^{[2]}(2^k)
=4*sum_{k=1}^{2}(2^k)+6*sum_{k=1}^{2}(2^k)
=4*(2^1+2^2)+6*(2^1+2^2)
=4*(2+4)+6*(2+4)=24+36=60

Marian Kraus, Illustration for a(4)

For the differences (a(n)-a(n-1))/4, n>2, see A163978.

Cf. A029744, A063759, A164090.

rm(a)
a <- vector() powerof2 <- vector()
x <- 300
n <- x/2
for (i in 1:x){
   powerof2[i] <- 2^(i-1)}
powerof2 for (i in 1:n){
   a[2*i]   <- 8*(sum(powerof2[1:i]))+12*(sum(powerof2[1:i]))}
for (i in 1:(n+1)){
   a[2*i+1] <- 8*(sum(powerof2[1:(i+1)]))+12*(sum(powerof2[1:i]))}
a[1]<-8
a

Conjectured recurrence:
a(0)=1,
a(1)=8,
a(2)=20, and thereafter
a(n)=2*a(n-2)+20.
Conjectured formula: ("[]" is the floor function)
a(n)=4*sum_{k=1}^{[(n+1)/2]}(2^k)+6*sum_{k=1}^{[n/2]}(2^k).
Conjectures from Colin Barker, Dec 07 2015: (Start)
a(n) = (-20+2^(1/2*(-1+n))*(10-10*(-1)^n+7*sqrt(2)+7*(-1)^n*sqrt(2))) for n>1.
a(n) = 5*2^(n/2+1/2)-5*(-1)^n*2^(n/2+1/2)+7*2^(n/2)+7*(-1)^n*2^(n/2)-20 for n>1.
a(n) = a(n-1)+2*a(n-2)-2*a(n-3) for n>4.
G.f.: x*(1+7*x+10*x^2+2*x^3) / ((1-x)*(1-2*x^2)).
(End)

cp's OEIS Frontend

A164090 a(n) = 2*a(n-2) for n > 2; a(1) = 2, a(2) = 3.

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