A097547 -id:A097547 - cp's OEIS frontend

A099129 Let T(n) be the n-th triangular number n*(n+1)/2; then a(n) = n-th iteration T(T(T(...(n)))).

0, 1, 6, 231, 1186570, 347357071281165, 2076895351339769460477611370186681, 143892868802856286225154411591351342616163027795335641150249224655238508171

Offset: 0

Jonathan Vos Post, Nov 14 2004

The next term, a(8), has 162 digits. - Harvey P. Dale, May 29 2013

			a(3) = 231 because we can write the 3-time iterated expression on T(3), the triangular number sequence n*(n+1)/2, namely: T(T(T(3))) = 231.

Alois P. Heinz, Table of n, a(n) for n = 0..10

Cf. A000217, A007501, A058009 (analog with primes), A097547.

a:= n-> (t-> (t@@n)(n))(j-> j*(j+1)/2):
seq(a(n), n=0..7);  # Alois P. Heinz, Sep 29 2023

Table[Nest[(#(#+1))/2&,n,n],{n,8}] (* Harvey P. Dale, May 29 2013 *)

a(n) = my(k = n); for (j=1, n, k = k*(k+1)/2;); k; \\ Michel Marcus, Jan 01 2017

a(n) = A000217^n(n).

The sequence grows like O(n^2^n*1/2^n). This can be derived from the growth O(n^2*1/2) of the triangle sum by iteration. - Hieronymus Fischer, Jan 21 2006

Offset changed to 1 by Georg Fischer, Jun 20 2022

a(0)=0 prepended by Alois P. Heinz, Sep 29 2023

A020955 a(n) = n^(2^n - n - 1).

Original entry on oeis.org

1, 1, 2, 81, 4194304, 1490116119384765625, 226267027688376192080197927193400943822503936, 258086210989349276047917817413172383631691140276099547911280598425927853437317437263620645695945672001

Offset: 0

Yasutoshi Kohmoto

nonn

Number of finite models of natural number on the free class with n members.

			For n=2, a(2)=2, universe class={0,1}.
1 : { }ma 0, {0}ma 1, {0,1}ma 1, {1}ma 0,
2 : { }ma 0, {0}ma 1, {0,1}ma 1, {1}ma 1.

Yasutoshi Kohmoto, The Other Side of Mathematics [broken link?]
Yasutoshi Kohmoto, The Other Side of Mathematics [via Internet Archive Wayback-machine]

Cf. A097547.

a(n) = A097547(n)/A007778(n), n > 0. - R. J. Mathar, Jan 12 2017

a(0)=1 and a(7) corrected by Vincenzo Librandi, Apr 25 2011

A216992 Decimal expansion of Sum_{n = 1, ..., infinity } 1/n^(2^n).

Original entry on oeis.org

1, 0, 6, 2, 6, 5, 2, 4, 1, 6, 0, 2, 3, 1, 0, 6, 5, 1, 6, 2, 3, 4, 3, 1, 1, 9, 0, 7, 9, 4, 9, 7, 3, 2, 7, 8, 6, 1, 6, 0, 6, 4, 6, 2, 4, 2, 9, 5, 0, 7, 8, 5, 4, 8, 7, 4, 8, 1, 2, 5, 0, 5, 8, 3, 2, 4, 0, 8, 9, 3, 8, 4, 6, 2, 0, 9, 3, 6, 6, 0, 5, 1, 9, 3, 9, 6, 8, 7, 1, 9, 6, 6, 4, 4, 4, 2, 4, 9, 8, 0, 4, 5, 8, 9, 3

Offset: 1

Nicolas M. Perrault, Sep 21 2012

The sum converges very quickly and therefore just a few summands are quite enough to get the value accurate to hundreds of decimal places. For example, 1/10^(2^10) = 10^(-1024), meaning that the impact of n = 10 on the sum can't be seen among the first thousand decimal digits. - Alonso del Arte, Sep 21 2012

			1.0626524160231065162343119079497327861...

Cf. A097547.

evalf(sum(1/n^(2^n), n=1..infinity), 140);  # Alois P. Heinz, Sep 29 2023

RealDigits[Sum[1/n^(2^n), {n, 10}], 10, 105][[1]] (* T. D. Noe, Sep 21 2012 *)

suminf(n=1,1/n^2^n) \\ Charles R Greathouse IV, Apr 21 2016

cp's OEIS Frontend

A099129 Let T(n) be the n-th triangular number n*(n+1)/2; then a(n) = n-th iteration T(T(T(...(n)))).

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A020955 a(n) = n^(2^n - n - 1).

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A216992 Decimal expansion of Sum_{n = 1, ..., infinity } 1/n^(2^n).

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Maple

Mathematica

PARI