A121263 Descending dungeons: see Comments lines for definition.
10, 11, 13, 16, 20, 25, 31, 38, 46, 55, 65, 87, 135, 239, 463, 943, 1967, 4143, 8751, 18479, 38959, 103471, 306223, 942127, 2932783, 9153583, 28562479, 89028655, 277145647, 861652015, 2675637295, 10173443119, 41132125231, 168836688943, 695134284847
Offset: 10
Examples
For example, 10 ..11 ....12 ......13 ........14 ..........15 ............16 ..............17 ................18 ..................19 ....................20 ......................21 ........................22 ..........................23 is equal to 239.
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..109
- 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, and 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
Programs
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Maple
M:=100; a:=list(10..M): a[10]:=10: lprint(10,a[10]); for n from 11 to M do b:=n; for i from n-1 by -1 to 11 do t1:=convert(i,base,10); b:=add(t1[j]*b^(j-1),j=1..nops(t1)): od: a[n]:=b; lprint(n,a[n]); od: # N. J. A. Sloane asubb := proc(a,b) local t1; t1:=convert(a,base,10); add(t1[j]*b^(j-1),j=1..nops(t1)): end; # asubb(a,b) evaluates a as if it were written in base b # N. J. A. Sloane
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Python
def a(n): a_of_n = [((10 + int(i))) for i in range(n)] while len(a_of_n) != 1: exponent = 0 a_of_n [-2] = list(str(a_of_n [-2])) for i in range(len(a_of_n [-2])): a_of_n [-2] [-(i+1)] = int(a_of_n [-2] [-(i+1)]) a_of_n [-2] [-(i+1)] *= ((a_of_n [-1]) ** exponent) exponent += 1 a_of_n [-2] = sum(a_of_n [-2]) a_of_n = a_of_n [:((len(a_of_n))-1)] return (a_of_n [0]) # Noah J. Crandall, Dec 07 2020
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_{i=1..n} log(9+i)), or equivalently, a_n = 10^10^(n loglog n + O(n)). - David Applegate and N. J. A. Sloane, Aug 25 2006
Comments