A283874 The Pierce expansion of the number Sum_{k>=1} 1/3^((2^k) - 1).

2, 3, 4, 9, 10, 81, 82, 6561, 6562, 43046721, 43046722, 1853020188851841, 1853020188851842, 3433683820292512484657849089281, 3433683820292512484657849089282, 11790184577738583171520872861412518665678211592275841109096961, 11790184577738583171520872861412518665678211592275841109096962

Offset: 0

Author

Kutlwano Loeto, Mar 24 2017

Keywords

• nonn

Comments

This sequence is the Pierce expansion of the number 3*s(3) - 1 = 0.370827687432918983346475478500709113969827799141493576... where s(u) = Sum_{k>=0} 1/u^(2^k) for u=3 has been considered by N. J. A. Sloane in A004200.

The continued fraction expansion of the number 3*s(3)-1 is essentially A081771.

Examples

			The Pierce expansion of 0.3708276874329189833 starts as 1/2 - 1/(2*3) + 1/(2*3*4) - 1/(2*3*4*9) + 1/(2*3*4*9*10) - 1/(2*3*4*9*10*81) + ...
		

Links

• Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209-217.

Programs

Maple

L:=[2]: for k from 0 to 6 do: L:=[op(L),3^(2^k),3^(2^k)+1]: od: print(L);

PARI

a(n) = if (n==0, 2, if (n%2, 3^(2^((n-1)/2)), 3^(2^((n-2)/2))+1)); \\ Michel Marcus, Mar 31 2017

Formula

a(0) = 2, a(2k+1) = 3^(2^k), a(2k+2) = 3^(2^k) + 1, k >= 0.