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%I #10 Jul 15 2021 10:18:10
%S 767,1839,1975,2358,2813,3089,3495,3824,4052,5596,5877,6292,6311,6386,
%T 6633,6666,6839,6886,7142,7670,8007,8241,9396,9796,10817,10839,11076,
%U 11644,14675,15069,15198,15923,16611,16615,16894,17179,17269,17751
%N Numbers k such that k^4 contains a pandigital substring.
%e 6666^4 = 7197(4518637029)136.
%o (Python)
%o def haspan(s): return any(len(set(s[i:i+10]))==10 for i in range(len(s)-9))
%o print([m for m in range(20000) if haspan(str(m**4))]) # _Michael S. Branicky_, Feb 28 2021
%Y Cf. A115933, A115935, A115936, A115937, A115938.
%K nonn,base
%O 1,1
%A _Giovanni Resta_, Feb 06 2006