A292687 a(n) = Product_{k=0..n-1} (4^(3^k) + 1) = decimal value of the Sierpinski-type iteration result A292686(n) (replace 0 with 000 and 1 with 101) considered as a binary number.

1, 5, 325, 85197125, 1534774961612150361293125, 8972304477322525702813810177861539421333393918862058319149818714344653125

Offset: 0

Author

M. F. Hasler, Oct 20 2017

Keywords

• nonn

Comments

The next term, a(6), has 202 digits and does not fit on one line.

This is the decimal representation of the terms of A292686 considered as binary numbers.

To get a(n+1) from a(n), write a(n) in binary, replace digits 0 with 000 and 1 with 101, and convert back to decimal. Equivalently, consider the binary expansion of a(n) as base-8 expansion, multiply it by 5, and convert back from octal to decimal.

Examples

			a(0) = 1 is already written in binary; multiplied by 5 it yields 5, read in octal is the same as in decimal, a(1) = 5.
a(1) = 5  = 101[2] in binary; consider 101 in base 8 (or base 10), multiply by 5 to get 505, convert from octal to decimal to get a(2) = 5*8^2 + 5 = 325.
a(2) = 325  = 101000101[2] in binary; consider this in base 8 (or base 10), multiply by 5 to get 505000505, convert from octal to decimal to get a(2) = 325*8^6 + 325 = 85197125.
		

Crossrefs

Cf. A292686 for the binary representation of a(n), and for more links, references and motivation.

Programs

Mathematica

A292687[nmax_]:=FoldList[Times,1,4^(3^Range[0,nmax-1])+1];A292687[6] (* Paolo Xausa, May 13 2023 *)

PARI

A292687(n)=prod(k=0,n-1,4^3^k+1)

Formula

a(n+1) = (4^(3^n)+1)*a(n).

a(n) = Product_{k=0..n-1} (4^(3^k)+1).