login
A236043
Number of triangular numbers <= 10^n.
1
5, 14, 45, 141, 447, 1414, 4472, 14142, 44721, 141421, 447214, 1414214, 4472136, 14142136, 44721360, 141421356, 447213595, 1414213562, 4472135955, 14142135624, 44721359550, 141421356237, 447213595500, 1414213562373, 4472135955000, 14142135623731
OFFSET
1,1
COMMENTS
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.
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
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
Variant of A068092. - R. J. Mathar, Jan 20 2014
FORMULA
a(n) = floor( sqrt(2*10^n + 1/4) + 1/2 ), adapted from Tanton (see Links section). - Chris Boyd, Jan 19 2014
EXAMPLE
There are 4472 triangular numbers less than or equal to 10^7 so a(7) = 4472.
MATHEMATICA
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 *)
PROG
(Python)
def Tri(x):
..count = 0
..for n in range(10**40):
....if n*(n+1)/2 <= 10**x:
......count += 1
....else:
......return count
x = 1
while x < 50:
..print(Tri(x))
..x += 1
(PARI) a236043(n)=floor(sqrt(2*10^n+1/4)+1/2) \\ Chris Boyd, Jan 19 2014
(Magma) [Floor(Sqrt(2*10^n+1/4) + 1/2): n in [1..30]]; // Vincenzo Librandi, Feb 08 2014
CROSSREFS
Cf. A000217.
Sequence in context: A197212 A100059 A270062 * A270661 A222908 A270911
KEYWORD
nonn,easy
AUTHOR
Derek Orr, Jan 18 2014
EXTENSIONS
More terms from Jon E. Schoenfield, Feb 07 2014
STATUS
approved