cp's OEIS Frontend

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.

%I A292687 #18 May 19 2023 04:54:45
%S A292687 1,5,325,85197125,1534774961612150361293125,
%T A292687 8972304477322525702813810177861539421333393918862058319149818714344653125
%N 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.
%C A292687 The next term, a(6), has 202 digits and does not fit on one line.
%C A292687 This is the decimal representation of the terms of A292686 considered as binary numbers.
%C A292687 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.
%F A292687 a(n+1) = (4^(3^n)+1)*a(n).
%F A292687 a(n) = Product_{k=0..n-1} (4^(3^k)+1).
%e A292687 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.
%e A292687 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.
%e A292687 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.
%t A292687 A292687[nmax_]:=FoldList[Times,1,4^(3^Range[0,nmax-1])+1];A292687[6] (* _Paolo Xausa_, May 13 2023 *)
%o A292687 (PARI) A292687(n)=prod(k=0,n-1,4^3^k+1)
%Y A292687 Cf. A292686 for the binary representation of a(n), and for more links, references and motivation.
%K A292687 nonn
%O A292687 0,2
%A A292687 _M. F. Hasler_, Oct 20 2017