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.

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

Author

Marian Kraus, Dec 04 2015

Keywords

• nonn
• easy

Examples

			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
		

Links

• Marian Kraus, Illustration for a(4)

Crossrefs

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

Cf. A029744, A063759, A164090.

Programs

R

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

Formula

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)