|
Search: id:A010036
|
|
|
| A010036 |
|
Sum of 2^n, ..., 2^(n+1) - 1. |
|
+0
10
|
|
| 1, 5, 22, 92, 376, 1520, 6112, 24512, 98176, 392960, 1572352, 6290432, 25163776, 100659200, 402644992, 1610596352, 6442418176, 25769738240, 103079084032, 412316598272, 1649266917376, 6597068718080
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
a(n) = sum of next 2^n natural numbers. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Apr 17 2003
Sum of all proper binary numbers with n digits (i.e. those not beginning with 0). Cf. A101291 Sum of all numbers with n digits [base 10]. - Jonathan Vos Post (jvospost3(AT)gmail.com), Sep 07 2006
a(n)/2^n gives the average eccentricity of the graphs of the Chinese rings puzzle with n+1 rings (also known as baguenaudier). - Daniele Parisse (daniele.parisse(AT)eads.com), Jun 02 2008
|
|
FORMULA
|
a(n+1) = 4*a(n) + 2^n with a(0) = 1 (with a(0)=0, see A006516) . a(n) = 2^(n-1)*A055010(n) . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Feb 20 2004
a(n) = 3*2^(2*n-1) - 2^(n-1) for all n>=0. - Daniele Parisse (daniele.parisse(AT)t-online.de), Jun 10 2007
|
|
MAPLE
|
f := n->3*2^(2*n-3)-2^(n-2);
|
|
CROSSREFS
|
Cf. A010036.
Partial sums are in A006516, A006095.
Sequence in context: A085812 A053297 A071715 this_sequence A127617 A095932 A000346
Adjacent sequences: A010033 A010034 A010035 this_sequence A010037 A010038 A010039
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Steve King (ITTTUCSON(AT)aol.com)
|
|
|
Search completed in 0.002 seconds
|
