OFFSET
19730,1
COMMENTS
2^0 = 1, 2^1 = 2, 2^2 = 4, 2^2^2 = 2^^3 = (2^2)^2 = 16,
2^2^2^2 = 2^^4 = (((2^2)^2)^2)^2 = 65536
so that 2^2^2^2^2 = 2^^5 = 2^(2^(2^(2^2))) = 2^65536 = 20035299304068464649790723515602557504478254755697514192650169737108940595563...
LINKS
Chai Wah Wu, Table of n, a(n) for n = 19730..39458
Googology Wiki, Tetration.
Pointless Large numbers stuff by Cookiefonster, 2.02 Knuth's Up-Arrows and the Hyper-Operators.
Robert P. Munafo, Sequence A094358, 2^^N = 1 mod N.
FORMULA
Equals 2^2^2^2^2 = 2^^5 = (((((((((((((((2^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2)^2.
EXAMPLE
2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348
(...19529 digits omitted...)
5775699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156736.
The above example line shows the first one hundred decimal digits and the last one hundred digits with the number of unrepresented digits in parentheses.
MATHEMATICA
nbrdgt = 100; f[base_, exp_] := RealDigits[ 10^FractionalPart[ N[ exp*Log10[ base], nbrdgt + Floor[ Log10[ exp]] + 2]], 10, nbrdgt][[1]]; f[2, 2^2^2^2]
IntegerDigits[2^65536][[;; 100]] (* Paolo Xausa, Jan 31 2024 *)
PROG
(Python)
def A361100(n): return (1<<(1<<(1<<(1<<(1<<1)))))//10**(39458-n)%10 # Chai Wah Wu, Apr 03 2023
CROSSREFS
KEYWORD
AUTHOR
Marco Ripà, Mar 03 2023
STATUS
approved