A256229 Powering the decimal digits of n (right-associative) with 0^0 = 1 by convention.

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, 1, 6, 36, 216, 1296

Offset: 1

Author

M. F. Hasler, Mar 19 2015

Keywords

• nonn
• base
• hear
• look

Comments

See A075877 for the left-associative version (which grows much more slowly). Usually the "^" operator is considered right-associative (so this is the "natural" version), i.e., a^b^c = a^(b^c) since (a^b)^c could be written a^(b*c) instead, while there is no such simplification for a^(b^c).

If n's first digit is succeeded by an odd number of consecutive 0's, a(n) is 1. If it is by an even number, a(n) is the first digit of n (A000030). - Alex Costea, Mar 27 2019

Examples

			a(253) = 2^5^3 = 2^(5^3) = 2^125 = 42535295865117307932921825928971026432.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..237
• Wikipedia, Zero to the power of zero

Crossrefs

Cf. A007953, A007954, A075877, A133500, A185969, A248907, A256179.

Programs

Maple

a:= proc(n) local m, r; m, r:= n, 1;
      while m>0 do r:= irem(m, 10, 'm')^r od; r
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 19 2015

Mathematica

Power @@ IntegerDigits@ # & /@ Range@ 64 /. Indeterminate -> 1 (* Michael De Vlieger, Mar 21 2015 *)

PARI

A256229(n,p=1)={until(!n\=10,p=(n%10)^p);p}

Python

def A256229(n):
    y = 1
    for d in reversed(str(n)):
        y = int(d)**y
    return y # Chai Wah Wu, Mar 21 2015

Formula

a(n) = A075877(n) for n < 212.

a(n) = A133500(n) for n < 100.

a(10n+1) = a(n).

Extensions

Incorrect comments deleted by Alex Costea, Mar 24 2019