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A236043
Number of triangular numbers <= 10^n.
1
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
0,1
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
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
EXAMPLE
There are 4472 triangular numbers less than or equal to 10^7 so a(7) = 4472.
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 *)
PROG
(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
KEYWORD
nonn,easy
AUTHOR
Derek Orr, Jan 18 2014
EXTENSIONS
More terms from Jon E. Schoenfield, Feb 07 2014
a(0) prepended by Andrew Howroyd, Dec 21 2024
STATUS
approved