A093700 Number of 9's immediately following the decimal point in the expansion of (3+sqrt(8))^n.
0, 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 13, 13, 14, 15, 16, 16, 17, 18, 19, 19, 20, 21, 22, 22, 23, 24, 25, 26, 26, 27, 28, 29, 29, 30, 31, 32, 32, 33, 34, 35, 35, 36, 37, 38, 39, 39, 40, 41, 42, 43, 44, 45, 45, 46, 47, 48, 48, 49, 50, 51, 52, 52, 53, 54, 55, 55, 56, 57
Offset: 1
Examples
Let n=10, (3+sqrt(8))^10= 45239073.9999999778... (the fractional part starts with seven 9's), so the 10th element in this sequence is 7. The 132nd element is 100. The 1000th element is 765. The 1307th element is 1000. The arrangement of repeating elements are like A074184 (Index of the smallest power of n >= n!) and A076539 (Numerators a(n) of fractions slowly converging to pi) and A080686 (Number of 19-smooth numbers <= n).
Links
- Henri Cohen, Fernando Rodriguez Villegas, and Don Zagier, Convergence Acceleration of Alternating Series, Experiment. Math. Volume 9, Issue 1 (2000), 3-12, Project Euclid - Cornell Univ (see Proposition 1).
- Robbert Fokkink, The Pell Tower and Ostronometry, arXiv:2309.01644 [math.CO], 2023.
- Math Forum, Triangular Numbers That are Perfect Squares
- Math Pages, On m = sqrt(sqrt(n) + sqrt(kn+1)) [Wrong link]
- Robert Simms, Using counting numbers to generate Pythagorean triples
Programs
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Mathematica
For[n = 1, n < 999, n++, Block[{$MaxExtraPrecision = 50*n}, Print[ -Floor[Log[10, 1 - N[FractionalPart[(3 + 2Sqrt[2])^n], n]]] - 1]]] f[n_] := Block[{}, -MantissaExponent[(3 - Sqrt[8])^n][[2]]]; Table[ f[n], {n, 75}] (* Robert G. Wilson v, Apr 10 2004 *)
Formula
Roughly, floor(3*n/4)
Comments