A099179 -id:A099179 - cp's OEIS frontend

A100010 a(0) = 2 and a(n) = f(a(n-1)) where f(n) = n^2(3n^2-4*n+2).

2, 24, 941184, 2354066797535483525627904

Offset: 0

Jonathan Vos Post, Nov 16 2004

Previous name was: Iterated hyperdiamond numbers, starting with 24-cell(2) = 24. Hyperdiamond numbers, figurate numbers based on the 4-dimensional 24-cell, have the formula 24-cell(n) = n^2*(3*n^2-4*n+2). This sequence is the hyperdiamond number of the hyperdiamond number of ... of 2.
The next term has 98 digits.
This need not start at 24-cell(2) = 24. For example, starting at a(0) = 3, which is not a hyperdiamond number, we have a(1) = 24-cell(3) = 3^2*((3*3^2)-(4*3)+2) = 153; and a(2) = 24-cell(24-cell(3)) = 24-cell(153) = 153^2*((3*153^2)-(4*153)+2) = 1629664353; and a(3) = 24-cell(24-cell(24-cell(3))) = 24-cell(1629664353) = 21159914972910583843562449776792301953.

			a(0) = 2 is the seed for this instance of the more general recurrence;
a(1) = 24-cell(2) = 2^2*(3*2^2-4*2+2) = 24;
a(2) = 24-cell(24-cell(2)) = 24-cell(24) = 24^2*(3*24^2-4*24+2) = 941184.

H. S. M. Coxeter, Regular Polytopes, 3rd ed. New York: Dover, 1973.

Michel Marcus, Table of n, a(n) for n = 0..5
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.

Cf. A092181, A007501, A099179, A000332.

f(n) = n^2*(3*n^2-4*n+2); \\ A092181
a(n) = if (n==0, 2, f(a(n-1))); \\ Michel Marcus, Dec 14 2015

a(0) = 2; hyperdiamond numbers, figurate numbers based on the 4-dimensional 24-cell, have the formula 24-cell(n) = n^2*(3*n^2-4*n+2). a(1) = 24-cell(2) = 24. a(2) = 24-cell(24-cell(2)) = 941184. For k>1, a(k+1) = 24-cell(a(k)).

New name from Joerg Arndt, Feb 23 2022

A100682 Floor of 4th root of pentatope numbers.

Original entry on oeis.org

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

Offset: 0

Jonathan Vos Post, Dec 06 2004

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

			a(3) = 1 because floor((3*4*5*6/24)^(1/4)) = floor(15^(1/4)) = floor(1.96798967) = 1.

J. H. Conway and R. K. Guy, The Book of Numbers, pp. 55-57, Copernicus Press, NY, 1996.

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

Cf. A000332, A100009, A007501, A099179.

[Floor(Binomial(n+3, 4)^(1/4)): n in [3..70]]; // Vincenzo Librandi, Dec 14 2015

a:= n-> floor(binomial(n+3, 4)^(1/4)):
seq(a(n), n=0..70);  # Alois P. Heinz, Dec 14 2015

a(n)=binomial(n+3,4)^(1/4)\1 \\ Charles R Greathouse IV, May 01 2012

a(n)=sqrtnint(binomial(n+3,4),4) \\ Charles R Greathouse IV, Dec 14 2015

from math import comb
from sympy import integer_nthroot
def A100682(n): return integer_nthroot(comb(n+3,4),4)[0] # Chai Wah Wu, Oct 02 2024

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

A100011 a(n+1) = a(n)((261a(n)^3)-(504a(n)^2)+(283a(n))-38)/2 for n > 0, a(0) = 2.

Original entry on oeis.org

2, 600, 16858418928600, 10540904502382779165253193559218987587063928137089696600

Offset: 0

Jonathan Vos Post, Nov 16 2004

The next term (a(4)) has 223 digits and a(5) has 891 digits. - Harvey P. Dale, Dec 22 2024

Cf. A092183, A007501, A099179, A000332.

NestList[# (261 #^3 - 504 #^2 + 283 # - 38)/2 &, 2, 5] (* Harvey P. Dale, Dec 22 2024 *)

A100012 Let h(k) = a(k)((145a(k)^3)-(280a(k)^2)+(179a(k))-38)/6, then a(n) = h(a(n-1)) for n >= 1 and a(0) =2.

Original entry on oeis.org

2, 120, 4930988840, 14287387711051307292599794275187472361080

Offset: 0

Jonathan Vos Post, Nov 17 2004

The next term has 163 digits.

Cf. A092182, A007501, A099179, A000332, A006564.

NestList[#/6*(145#^3-280#^2+179#-38)&,2,3] (* Harvey P. Dale, Apr 09 2015 *)

A100679 Floor of cube root of tetrahedral numbers.

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 33, 33, 34

Offset: 0

Jonathan Vos Post, Dec 06 2004

Tetrahedral numbers Tet(n) = A000292(n) = C(n+2, 3) = n(n+1)(n+2)/6 are obviously of order n^3, varying approximately with the cube of n. Taking the cube root and rounding down, we get the new sequence.

			a(18) = 10 because floor((18*19*20/6)^(1/3)) = floor(1140^(1/3))  = 10.

J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 83.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 4.

R. Jovanovic, First 2500 Tetrahedral numbers.
Eric Weisstein's World of Mathematics, Tetrahedral Number

Cf. A000292, A099179, A099179.

Table[Floor[Binomial[n + 2, 3]^(1/3)], {n, 0, 61}] (* Giovanni Resta, Jun 17 2016 *)

a(n) = floor((A000292(n))^(1/3)) = floor(Tet(n)^(1/3)) = floor(C(n+2, 3)^(1/3)) = floor((n(n+1)(n+2)/6)^(1/3)).

Edited by Giovanni Resta, Jun 17 2016

cp's OEIS Frontend

A100010 a(0) = 2 and a(n) = f(a(n-1)) where f(n) = n^2*(3*n^2-4*n+2).

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A100682 Floor of 4th root of pentatope numbers.

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A100011 a(n+1) = a(n)*((261*a(n)^3)-(504*a(n)^2)+(283*a(n))-38)/2 for n > 0, a(0) = 2.

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A100012 Let h(k) = a(k)*((145*a(k)^3)-(280*a(k)^2)+(179*a(k))-38)/6, then a(n) = h(a(n-1)) for n >= 1 and a(0) =2.

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A100679 Floor of cube root of tetrahedral numbers.

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Extensions

A100010 a(0) = 2 and a(n) = f(a(n-1)) where f(n) = n^2(3n^2-4*n+2).

A100011 a(n+1) = a(n)((261a(n)^3)-(504a(n)^2)+(283a(n))-38)/2 for n > 0, a(0) = 2.

A100012 Let h(k) = a(k)((145a(k)^3)-(280a(k)^2)+(179a(k))-38)/6, then a(n) = h(a(n-1)) for n >= 1 and a(0) =2.