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Powers of sqrt(5) rounded up.
3

%I #30 Jun 19 2024 10:36:18

%S 1,3,5,12,25,56,125,280,625,1398,3125,6988,15625,34939,78125,174693,

%T 390625,873465,1953125,4367321,9765625,21836602,48828125,109183007,

%U 244140625,545915034,1220703125,2729575168

%N Powers of sqrt(5) rounded up.

%H Vincenzo Librandi, <a href="/A017921/b017921.txt">Table of n, a(n) for n = 0..1000</a>

%e sqrt(5)^3 = 11.18033988749895... so a(3) = 12.

%e sqrt(5)^4 = 25, so a(4) = 25.

%e sqrt(5)^5 = 55.90169943749474241... so a(5) = 56.

%t Ceiling[Sqrt[5]^Range[0, 40]] (* _Vincenzo Librandi_, Nov 20 2011 *)

%o (Magma) [Ceiling(Sqrt(5^n)): n in [0..40]]; // _Vincenzo Librandi_, Nov 20 2011

%o (PARI) a(n) = ceil(sqrt(5^n)); \\ _Michel Marcus_, Dec 14 2017

%o (Python)

%o from math import isqrt

%o def A017921(n): return isqrt(5**n)+(n&1) # _Chai Wah Wu_, Jun 19 2024

%Y Cf. A017919, A017920.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_