A057048 - OEIS

%I #32 Jun 19 2024 01:52:42

%S 1,2,3,4,6,8,11,16,23,32,45,64,91,128,181,256,362,512,724,1024,1448,

%T 2048,2896,4096,5793,8192,11585,16384,23170,32768,46341,65536,92682,

%U 131072,185364,262144,370728,524288,741455,1048576

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

%C 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]

%H Vincenzo Librandi, <a href="/A057048/b057048.txt">Table of n, a(n) for n = 0..1000</a>

%F a(2n-1) = 2^n, n > 0. - _M. F. Hasler_, Feb 20 2012

%e Write the natural numbers A000027 as a triangle:

%e row 1: 1 . . . <- 2^0 in row 1=a(0)

%e row 2: 2 3 . . . <- 2^1 in row 2=a(1)

%e row 3: 4 5 6 . . . <- 2^2 in row 3=a(2)

%e row 4: 7 8 9 10 . . <- 2^3 in row 4=a(3)

%e row 5: 11 12 13 14 15

%e row 6: 16 17 18 19 20 21 <- 2^4 in row 6=a(4).

%t Table[Round[Sqrt[2]^(n+1)], {n, 0, 50}] (* _Vincenzo Librandi_, Mar 24 2013 *)

%o (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

%o (PARI) a(n)=sqrtint(2^(n+1)) \\ _Charles R Greathouse IV_, Aug 19 2016

%o (Magma) [Round(Sqrt(2)^(n+1)): n in [0..50]]; // _Vincenzo Librandi_, Mar 24 2013

%o (Python)

%o from math import isqrt

%o def A057048(n): return -isqrt(m:=1<<n+1)+isqrt(m<<2) # _Chai Wah Wu_, Jun 18 2024

%K nonn,easy

%O 0,2

%A _M. F. Hasler_, Feb 20 2012