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A121296
Descending dungeons: like A121295 but read subscripts from top downwards.
12
10, 11, 13, 16, 20, 28, 45, 73, 133, 348, 4943, 22779, 537226, 11662285, 46524257772, 1092759075796059, 159271598072111595659, 3317896028408943302861454961, 594387514787460257685718548861374076357, 91930654519343922607883279072515432244874866615525276
OFFSET
10,1
COMMENTS
A "dungeon" of numbers.
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
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)
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)).
EXAMPLE
a(13) = ((13_12)_11)_10 = (15_11)_10 = 16_10 = 16.
MAPLE
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
s2:=[10]; for n from 11 to 35 do t1:=n; for i from 1 to n-10 do t1:=asubb(t1, n-i); od: s2:=[op(s2), t1]; od;
CROSSREFS
KEYWORD
nonn,base
AUTHOR
STATUS
approved