A057048 a(n) = A017911(n+1) = round(sqrt(2)^(n+1)).

1, 2, 3, 4, 6, 8, 11, 16, 23, 32, 45, 64, 91, 128, 181, 256, 362, 512, 724, 1024, 1448, 2048, 2896, 4096, 5793, 8192, 11585, 16384, 23170, 32768, 46341, 65536, 92682, 131072, 185364, 262144, 370728, 524288, 741455, 1048576

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

Sequence in context: A317669 A208887 A017911 * A281094 A054782 A261082

Adjacent sequences: A057045 A057046 A057047 * A057049 A057050 A057051

Keyword

nonn,easy

Author

M. F. Hasler, Feb 20 2012

Status

approved