cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A099147 Iterated hexagonal numbers, starting at 1.

Original entry on oeis.org

1, 6, 66, 8646, 149497986, 44699295486614406, 3996054033999333969062944766851266, 31936895685284700329548847429175178142518023225832967407199564754246
Offset: 1

Views

Author

Jonathan Vos Post, Nov 14 2004

Keywords

Comments

Agrees with A097419 for n>1.
The next term (a(9)) has 136 digits. - Harvey P. Dale, Nov 24 2024

Examples

			a(4) = b(a(3)) = b(b(a(2))) = b(b(b(2))) = b(b(6)) = b(66) = 8646, where b(n) = A000384(n).

Links

Crossrefs

Cf. A000384, A097419.

Programs

  • Mathematica
    Join[{1},NestList[PolygonalNumber[6,#]&,6,6]] (* Harvey P. Dale, Nov 24 2024 *)
  • PARI
    {hexagonal(n) = n*(2*n-1)}
{a(n) = if(n<=2,hexagonal(n),hexagonal(a(n-1)))} \\ Klaus Brockhaus, Jan 10 2007

Formula

a(n) = b(n) for n<=2, a(n) = b(a(n-1)) for n>2, where b(n) = A000384(n) = n*(2*n-1), the hexagonal numbers.
a(1) = 1, a(2) = 6, a(n) = 2*a(n-1)^2 - a(n-1) for n>2.
Let H(n) = n*(2*n-1) = the n-th hexagonal number. Define A(n, k) recursively by A(1, k) = H(k), A(n, k) = A(1, A(n-1, k)) for n>1. Then a(1) = A(1, 1), a(n) = A(n-1, 2) for n>1.

Extensions

Edited by Klaus Brockhaus, Jan 10 2007

A296374 a(0) = 3; a(n) = a(n-1)*(a(n-1)^2 - 3*a(n-1) + 4)/2.

Original entry on oeis.org

3, 6, 66, 137346, 1295413937737986, 1086915296274625337063297033180803022465442306
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 11 2017

Keywords

Comments

The next term is too large to include.

Examples

			a(0) = 3;
a(1) = 6 and 6 is the 3rd triangular number;
a(2) = 66 and 66 is the 6th hexagonal number;
a(3) = 137346 and 137346 is the 66th 66-gonal number, etc.

Links

Crossrefs

Cf. A001146, A007501, A057145, A060354, A097419, A099137, A099147, A099153.

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 3, a[n] == a[n - 1] (a[n - 1]^2 - 3 a[n - 1] + 4)/2}, a[n], {n, 5}]

Formula

a(0) = 3; a(n) = [x^a(n-1)] x*(1 - 2*x + 4*x^2)/(1 - x)^4.
a(0) = 3; a(n) = a(n-1)! * [x^a(n-1)] exp(x)*x*(1 + x^2/2).
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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