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%I #30 Sep 08 2022 08:46:06
%S 5,14,45,141,447,1414,4472,14142,44721,141421,447214,1414214,4472136,
%T 14142136,44721360,141421356,447213595,1414213562,4472135955,
%U 14142135624,44721359550,141421356237,447213595500,1414213562373,4472135955000,14142135623731
%N Number of triangular numbers <= 10^n.
%C Except for 5, all numbers begin with either a 4 or a 1. If strictly less than, the 5 would become a 4, satisfying this conjecture.
%C This is not a conjecture, it is a fact and it is the result from the square root of 2 and 20 times powers of ten. - _Robert G. Wilson v_, Jan 11 2015
%C Tanton (2012) discusses the equivalent sequence based on excluding zero from the triangular numbers, and presents the relevant formula, which, being asymptotic to floor[sqrt(2*10^n)], explains the observation in the first comment. - _Chris Boyd_, Jan 19 2014
%C Variant of A068092. - _R. J. Mathar_, Jan 20 2014
%H Vincenzo Librandi, <a href="/A236043/b236043.txt">Table of n, a(n) for n = 1..200</a>
%H J. Tanton, <a href="http://www.jamestanton.com/wp-content/uploads/2012/03/Cool-Math-Newsletter_November2012.pdf">Cool Math Newsletter (November 2012)</a>
%F a(n) = floor( sqrt(2*10^n + 1/4) + 1/2 ), adapted from Tanton (see Links section). - _Chris Boyd_, Jan 19 2014
%e There are 4472 triangular numbers less than or equal to 10^7 so a(7) = 4472.
%t Table[ Floor[ Sqrt[2*10^n + 1] + 1/2], {n, 25}] (* _Vincenzo Librandi_, Feb 08 2014; modified by _Robert G. Wilson v_, Jan 11 2015 *)
%o (Python)
%o def Tri(x):
%o ..count = 0
%o ..for n in range(10**40):
%o ....if n*(n+1)/2 <= 10**x:
%o ......count += 1
%o ....else:
%o ......return count
%o x = 1
%o while x < 50:
%o ..print(Tri(x))
%o ..x += 1
%o (PARI) a236043(n)=floor(sqrt(2*10^n+1/4)+1/2) \\ _Chris Boyd_, Jan 19 2014
%o (Magma) [Floor(Sqrt(2*10^n+1/4) + 1/2): n in [1..30]]; // _Vincenzo Librandi_, Feb 08 2014
%Y Cf. A000217.
%K nonn,easy
%O 1,1
%A _Derek Orr_, Jan 18 2014
%E More terms from _Jon E. Schoenfield_, Feb 07 2014
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