A318779 Smallest n-th power that is pandigital in base n.
4, 64, 625, 248832, 11390625, 170859375, 1406408618241, 3299763591802133, 3656158440062976, 550329031716248441, 766217865410400390625, 15791096563156692195651, 6193386212891813387462761, 243008175525757569678159896851, 3433683820292512484657849089281
Offset: 2
Examples
a(2)=4 because 1^2 = 1 = 1_2 (not pandigital in base 2, since it contains no 0 digit), but 2^2 = 4 = 100_2. a(3)=64 because 1^3 = 1 = 1_3, 2^3 = 8 = 22_3, and 3^3 = 27 = 1000_3 are all nonpandigital in base 3, but 4^3 = 64 = 2101_3. a(16) = 81^16 = 3433683820292512484657849089281 = 2b56d4af8f7932278c797ebd01_16.
Links
- Jon E. Schoenfield, Table of n, a(n) for n = 2..165
Crossrefs
Cf. A049363 (smallest pandigital number in base n), A185122 (smallest pandigital prime in base n), A260182 (smallest square that is pandigital in base n), A260117 (smallest triangular number that is pandigital in base n), A318725 (smallest k such that k! is pandigital in base n), A318780 (smallest k such that k^n is pandigital in base n).
Programs
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Python
from itertools import count from sympy import integer_nthroot from sympy.ntheory import digits def A318779(n): return next(k for k in (k**n for k in count(integer_nthroot((n**n-n)//(n-1)**2+n**(n-2)*(n-1)-1,n)[0])) if len(set(digits(k,n)[1:]))==n) # Chai Wah Wu, Mar 13 2024
Formula
a(n) = A318780(n)^n.
Comments