cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A236043 Number of triangular numbers <= 10^n.

Original entry on oeis.org

2, 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: 0

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Author

Derek Orr, Jan 18 2014

Keywords

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

Examples

			There are 4472 triangular numbers less than or equal to 10^7 so a(7) = 4472.
		

Crossrefs

Cf. A000217, A002024, A003056. Essentially the same as A068092.

Programs

  • Magma
    [Floor(Sqrt(2*10^n+1/4) + 1/2): n in [1..30]]; // Vincenzo Librandi, Feb 08 2014
    
  • Maple
    seq(floor(sqrt(2*10^n+1/4)+1/2),n=1..30); # Robert Israel, Dec 22 2024
  • 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 *)
  • PARI
    a236043(n)=floor(sqrt(2*10^n+1/4)+1/2) \\ Chris Boyd, Jan 19 2014
    
  • Python
    from math import isqrt
    def A236043(n): return isqrt(10**n+1<<3)+1>>1 # Chai Wah Wu, Jun 14 2025

Formula

a(n) = floor( sqrt(2*10^n + 1/4) + 1/2 ), adapted from Tanton (see Links section). - Chris Boyd, Jan 19 2014
a(n) = A068092(n + 1) for n >= 2. - R. J. Mathar, Jan 20 2014
a(n) = A003056(10^n) + 1 = A002024(10^n + 1). - Andrew Howroyd, Dec 21 2024

Extensions

More terms from Jon E. Schoenfield, Feb 07 2014
a(0) prepended by Andrew Howroyd, Dec 21 2024