|
|
A056008
|
|
Difference between (smallest square strictly greater than 2^n) and 2^n.
|
|
3
|
|
|
3, 2, 5, 1, 9, 4, 17, 16, 33, 17, 65, 68, 129, 89, 257, 356, 513, 697, 1025, 1337, 2049, 2449, 4097, 4001, 8193, 4417, 16385, 17668, 32769, 24329, 65537, 4633, 131073, 18532, 262145, 74128, 524289, 296512, 1048577, 1186048, 2097153, 1778369
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
If n is even, a(n) = 2*2^(n/2) + 1, since 2^n = (2^(n/2))^2, and a(n) = (2^(n/2) + 1)^2 - (2^(n/2))^2 = 2*2^(n/2) + 1. - Jean-Marc Rebert, Mar 02 2016
If n is odd, a(n) = 4*a(n-2) or 4*a(n-2) - 4*sqrt(a(n-2) + 2^(n-2)) + 1. - Robert Israel, Mar 02 2016
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
a(5)=6^2-2^5=4; a(6)=9^2-2^6=17
|
|
MAPLE
|
f:= proc(n) local m;
if n::even then m:= 2*2^(n/2)+1
else m:= ceil(sqrt(2)*2^((n-1)/2))
fi;
m^2-2^n
end proc:
|
|
MATHEMATICA
|
ssg[n_]:=Module[{s=2^n}, (1+Floor[Sqrt[s]])^2-s]; Array[ssg, 50, 0] (* Harvey P. Dale, Aug 22 2015 *)
Table[((Floor[2^(n/2)] + 1)^2 - 2^n), {n, 0, 50}] (* Vincenzo Librandi, Mar 03 2016 *)
|
|
PROG
|
(Python)
from math import isqrt
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|