A010701 Constant sequence: the all 3's sequence.
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 0
Examples
1/3 = 0.33333333333333333333333333333333333333333333... - _Bruno Berselli_, Mar 21 2014
References
- Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.3.1, p. 281.
Links
- Tanya Khovanova, Recursive Sequences.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1011.
- Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seqs., Vol. 6 (2003), Article 03.1.6.
- Rick Mabry, Proof without words: 1/4+(1/4)^2+(1/4)^3+...=1/3, Math. Mag., Vol. 72, No. 1 (1999), p. 63.
- Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.
- Index entries for linear recurrences with constant coefficients, signature (1).
- Index to divisibility sequences.
Programs
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Haskell
a010701 = const 3 a010701_list = repeat 3 -- Reinhard Zumkeller, May 07 2012
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Maple
evalf(1/3, 100); # Michal Paulovic, Mar 02 2023
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Mathematica
Table[3, {100}] (* Wesley Ivan Hurt, Jul 16 2014 *) -
Maxima
makelist(3, n, 0, 30); /* Martin Ettl, Nov 09 2012 */
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PARI
a(n)=3 \\ Felix Fröhlich, Jul 16 2014
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Python
def A010701(n): return 3 # Chai Wah Wu, Nov 10 2022
Formula
G.f.: 3/(1-x). - Bruno Berselli, Mar 15 2011
E.g.f.: 3*e^x. - Vincenzo Librandi, Jan 24 2012
a(n) = 3*A000012(n). - Michel Marcus, Dec 18 2015
a(n) = floor(1/(n - cot(1/n))). - Clark Kimberling, Mar 10 2020
Equals Sum_{k>=1} (1/4)^k (as a constant). - Michel Marcus, Jun 11 2020
Equals Sum_{k>=2} (k-1)/binomial(2*k,k) (as a constant). - Amiram Eldar, Jun 05 2021
Equals Sum_{k>=1} (-1)^(k+1)/2^k. - Michal Paulovic, Mar 02 2023
Comments