|
%I #24 Mar 19 2023 15:59:23
%S 1,1,2,3,3,4,5,6,6,7,8,8,9,10,10,11,12,13,13,14,15,15,16,17,18,18,19,
%T 20,20,21,22,22,23,24,25,25,26,27,27,28,29,30,30,31,32,32,33,34,35,35,
%U 36,37,37,38,39,39,40,41,42,42,43,44,44,45,46,47,47,48,49,49,50,51,51
%N Integer part of square root of n-th triangular number.
%H Winston de Greef, <a href="/A061288/b061288.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = floor(sqrt(n*(n+1)/2)). - _Zak Seidov_, May 25 2015
%F a(n) = a(-1-n) = A186221(n) - n for all n in Z. - _Michael Somos_, Aug 19 2018
%e a(10) = 7, the 10th triangular number is 55 and floor(sqrt(55)) = floor(7.4161) = 7.
%p for n from 1 to 150 do printf("%d,",floor(sqrt(n*(n+1)/2))) od;
%t Table[Floor[Sqrt[n*(n + 1)/2]], {n, 100}] (* _Zak Seidov_, May 25 2015 *)
%t IntegerPart[Sqrt[#]]&/@Accumulate[Range[80]] (* _Harvey P. Dale_, May 13 2018 *)
%o (PARI) {a(n) = sqrtint(n * (n+1) \ 2)}; /* _Michael Somos_, Aug 19 2018 */
%Y Cf. A000217, A186221.
%K nonn,easy
%O 1,3
%A _Amarnath Murthy_, Apr 25 2001
%E Corrected and extended by Larry Reeves (larryr(AT)acm.org), May 07 2001
|
