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 A100682 Floor of 4th root of pentatope numbers. 1
 0, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 30, 31, 31 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Conjecture: a(n) = floor((n - 3/2)/24^(1/4)) for n not in {0, 1, 6, 17, 2403, 5318}. - Charles R Greathouse IV, May 01 2012 REFERENCES J. H. Conway and R. K. Guy, The Book of Numbers, pp. 55-57, Copernicus Press, NY, 1996. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..10000 Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75. Jonathan Vos Post, Table of Polytope Numbers, Sorted, Through 1,000,000. Eric Weisstein's World of Mathematics, Pentatope Number FORMULA a(n) = floor((A000332(n+3))^(1/4)) = floor(Ptop(n)^(1/4)) = floor(C(n+3, 4)^1/4)) = floor(((n * (n+1) * (n+2) * (n+3)/4!)^(1/4)) a(n) = 0.4518... * n + O(1). - Charles R Greathouse IV, Dec 14 2015 EXAMPLE a(3) = 1 because floor((3*4*5*6/24)^(1/4)) = floor(15^(1/4)) = floor(1.96798967)) = 1. MAPLE a:= n-> floor(binomial(n+3, 4)^(1/4)): seq(a(n), n=0..70); # Alois P. Heinz, Dec 14 2015 PROG (PARI) a(n)=binomial(n+3, 4)^(1/4)\1 \\ Charles R Greathouse IV, May 01 2012 (PARI) a(n)=sqrtnint(binomial(n+3, 4), 4) \\ Charles R Greathouse IV, Dec 14 2015 (Magma) [Floor(Binomial(n+3, 4)^(1/4)): n in [3..70]]; // Vincenzo Librandi, Dec 14 2015 CROSSREFS Cf. A000332, A100009, A007501, A099179. Sequence in context: A362131 A095773 A062108 * A075355 A287864 A347578 Adjacent sequences: A100679 A100680 A100681 * A100683 A100684 A100685 KEYWORD easy,nonn AUTHOR Jonathan Vos Post, Dec 06 2004 STATUS approved

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Last modified July 24 05:18 EDT 2024. Contains 374575 sequences. (Running on oeis4.)