A121265 Descending dungeons: a(10)=10; for n>10, a(n) = a(n-1) read as if it were written in base n.

10, 11, 13, 16, 20, 30, 48, 76, 132, 420, 1640, 11991, 249459, 14103793, 5358891675, 19563802363305, 3359230167951561129, 181335944930584275675841374, 54416647690014492928933662292768871352, 6605721238793689879501639879905020611382966457124120828, 360539645288616164606228883801608423987740093330992456820074646988075733781927268

Offset: 10

Author

N. J. A. Sloane, Aug 23 2006

Keywords

• nonn
• base
• nice

Comments

Using N_b to denote "N read in base b", the sequence is given by

......10....10.....10.....10.......etc.

..............11.....11.....11.........

.......................12.....12.......

................................13.....

where the subscripts are evaluated from the top downwards.

More precisely, "N_b" means "Take decimal expansion of N and evaluate it as if it were a base-b expansion".

A "dungeon" of numbers.

Examples

			From _Jianing Song_, May 22 2021: (Start)
a(10) = 10;
a(11) = 10_11 = 11;
a(12) = 11_12 = 13;
a(13) = 13_13 = 16;
a(14) = 16_14 = 20;
a(15) = 20_15 = 30;
a(16) = 30_16 = 48;
... (End)
		

References

• David Applegate, Marc LeBrun and N. J. A. Sloane, Descending Dungeons and Iterated Base-Changing, in "The Mathematics of Preference, Choice and Order: Essays in Honor of Peter Fishburn", edited by Steven Brams, William V. Gehrlein and Fred S. Roberts, Springer, 2009, pp. 393-402.

Links

• N. J. A. Sloane, Table of n, a(n) for n = 10..35
• David Applegate, Marc LeBrun and N. J. A. Sloane, Descending Dungeons and Iterated Base-Changing, arXiv:math/0611293 [math.NT], 2006-2007.
• David Applegate, Marc LeBrun, N. J. A. Sloane, Descending Dungeons, Problem 11286, Amer. Math. Monthly, 116 (2009) 466-467.
• Brady Haran and N. J. A. Sloane, Dungeon Numbers, Numberphile video (2020). (extra)

Crossrefs

Cf. A121263, A121295, A121296, A127744, A122734.

Programs

Maple

M:=35; a:=list(10..M): a[10]:=10: lprint(10,a[10]); for n from 11 to M do t1:=convert(a[n-1],base,10); a[n]:=add(t1[i]*n^(i-1),i=1..nops(t1)); lprint(n,a[n]); od:

Mathematica

nxt[{n_,a_}]:={n+1,FromDigits[IntegerDigits[a],n+1]}; Transpose[ NestList[ nxt,{10,10},20]][[2]] (* Harvey P. Dale, Jul 13 2014 *)

PARI

a(n) = {my(x=10); for (b=11, n, x = fromdigits(digits(x, 10), b);); x;} \\ Michel Marcus, May 26 2019

Formula

If a, b >= 10, then a_b is roughly 10^(log(a)log(b)) (all logs are base 10 and "roughly" means it is an upper bound and using floor(log()) gives a lower bound). Equivalently, there exists c > 0 such that for all a, b >= 10, 10^(c log(a)log(b)) <= a_b <= 10^(log(a)log(b)). Thus a_n is roughly 10^product(log(9+i),i=1..n), or equivalently, a_n = 10^10^(n loglog n + O(n)). - David Applegate and N. J. A. Sloane, Aug 25 2006