A124793 Numbers in a perpendicular plane intersecting a 3D clockwise spiral produced by powers of 2.

1, 5, 35, 157, 1123, 5021, 35939, 160669, 1150051, 5141405, 36801635, 164524957, 1177652323, 5264798621, 37684874339, 168473555869, 1205915978851, 5391153787805, 38589311323235, 172516921209757, 1234857962343523

Offset: 1

Author

Paolo P. Lava and Giorgio Balzarotti, Jun 27 2007

Keywords

• easy
• nonn

Comments

The general formula for powers of k integer is a(n) = k^((1/4)*(10*n - 7 - (-1)^n)) + k^((1/4)*(10*n - 1 + (-1)^n)) - a(n-1), with a(0)=1 and where k is an integer value. If we replace k with "i" or "-i" where i=sqrt(-1), we get a periodic complex sequence (period 8).

Examples

			Write powers of 2 in a sort of 3D clockwise spiral. After the initial 1 (2^0) move right till 2^1=2 (practically only one step); then move down till 2^2=4 (3,4); then left till 2^3=8 (5,6,7,8). When writing number 5 we are in the same column of 1 so 5 is the second number of the sequence. Then move up till 2^4=16. Then move up perpendicularly to the plane till 2^5=32 and again right till 2^6=64. The number 35 is in the sequence because it lies in the same line as 1 and 5. The process continues down, left, up, perpendicular, right and so on.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..1329
• Index entries for linear recurrences with constant coefficients, signature (-1,32,32).

Crossrefs

Cf. A108981, A001107.

Programs

Maple

P:=proc(n) local a,i,x,y; a:=1; print(a); for i from 1 by 1 to n do x:=1/4*(10*i-7-(-1)^i); y:=1/4*(10*i-1+(-1)^i); a:=2^x+2^y-a; print(a); od; end: P(100);

Mathematica

LinearRecurrence[{-1, 32, 32}, {1, 5, 35}, 25] (* Paolo Xausa, Feb 23 2024 *)

Formula

a(n) = 2^((1/4)*(10*n - 7 - (-1)^n)) + 2^((1/4)*(10*n - 1 + (-1)^n)) - a(n-1), with a(0)=1.

From Colin Barker, Jul 07 2012: (Start)

a(n) = -a(n-1) + 32*a(n-2) + 32*a(n-3).

G.f.: x*(1+2*x)*(1+4*x)/((1+x)*(1-32*x^2)). (End)

a(2n) = 3/31 + 19*32^n/124, a(2n+1) = -3/31 + 136*32^n/124. [R. J. Mathar, Jul 10 2012]