OFFSET
0,2
COMMENTS
If the natural numbers A000027 are written as a triangle, then a(n) gives the row of the triangle in which the number 2^n can be found. See A017911 for a more elaborate explanation and relation with A000217. [Original definition by Clark Kimberling, Jul 30 2000, clarified by M. F. Hasler, Feb 20 2012, following an observation from T. D. Noe, Apr 27 2003]
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
FORMULA
a(2n-1) = 2^n, n > 0. - M. F. Hasler, Feb 20 2012
EXAMPLE
Write the natural numbers A000027 as a triangle:
row 1: 1 . . . <- 2^0 in row 1=a(0)
row 2: 2 3 . . . <- 2^1 in row 2=a(1)
row 3: 4 5 6 . . . <- 2^2 in row 3=a(2)
row 4: 7 8 9 10 . . <- 2^3 in row 4=a(3)
row 5: 11 12 13 14 15
row 6: 16 17 18 19 20 21 <- 2^4 in row 6=a(4).
MATHEMATICA
Table[Round[Sqrt[2]^(n+1)], {n, 0, 50}] (* Vincenzo Librandi, Mar 24 2013 *)
PROG
(PARI) A057048(n)=round(sqrt(2^(n+1))) /* for large values, an implementation using integer arithmetic would be preferable */ \\ M. F. Hasler, Feb 20 2012
(PARI) a(n)=sqrtint(2^(n+1)) \\ Charles R Greathouse IV, Aug 19 2016
(Magma) [Round(Sqrt(2)^(n+1)): n in [0..50]]; // Vincenzo Librandi, Mar 24 2013
(Python)
from math import isqrt
def A057048(n): return -isqrt(m:=1<<n+1)+isqrt(m<<2) # Chai Wah Wu, Jun 18 2024
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, Feb 20 2012
STATUS
approved