A006895 Parenthesized one way gives the powers of 2: (1), (2), (1+3), ..., another way the powers of 3: (1), (2+1), (3+6), ....

1, 2, 1, 3, 6, 2, 16, 9, 23, 58, 6, 128, 109, 147, 512, 70, 954, 1233, 815, 4096, 1650, 6542, 13141, 3243, 32768, 23038, 42498, 131072, 3577, 258567, 272874, 251414, 1048576, 294333, 1802819, 2980150, 1214154, 8388608, 4746145, 12031071, 31015650, 2538782

Offset: 0

Author

N. J. A. Sloane, K. S. Brown [ kevin2003(AT)delphi.com ]

Keywords

• nonn
• nice

Comments

Powers of 2 need 1 term or 2 terms parenthesized, whereas powers of 3 need 2 or 3 terms parenthesized, when 3 then the middle term is a power of 2. See A227928. - Reinhard Zumkeller, Oct 09 2013

Examples

			.  a(0)                                              =  _^0
.  a(1)                                              =  2^1
.  a(1) + a(2)            =    2 +   1               =  3^1
.  a(2) + a(3)            =    1 +   3        =    4 =  2^2
.  a(3) + a(4)            =    3 +   6        =    9 =  3^2
.  a(4) + a(5)            =    6 +   2        =    8 =  2^3
.  a(6)                                       =   16 =  2^4
.  a(5) + a(6) + a(7)     =    2 +  16 +   9  =   27 =  3^3
.  a(7) + a(8)            =    9 +  23        =   32 =  2^5
.  a(8) + a(9)            =   23 +  58        =   81 =  3^4
.  a(9) + a(10)           =   58 +   6        =   64 =  2^6
.  a(11)                                      =  128 =  2^7
.  a(10) + a(11) + a(12)  =    6 + 128 + 109  =  243 =  3^5
.  a(12) + a(13)          =  109 + 147        =  256 =  2^8
.  a(14)                                      =  512 =  2^9
.  a(13) + a(14) + a(15)  =  147 + 512 +  70  =  3^6 =  729 .
		

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000079, A000244, A006899, A108906, A085239.

Programs

Haskell

a006895 n = a006895_list !! n
a006895_list = 1 : f 0 0 (tail a000079_list) (tail a000244_list) where
   f x y us'@(u:us) vs'@(v:vs)
     | x > 0     = (u - x) : f 0 (u - x + y) us vs'
     | y > v - u = (v - y) : f (v + x - y) 0 us' vs
     | otherwise =       u : f 0 (u + y) us vs'
-- Reinhard Zumkeller, Oct 09 2013