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A340270 a(n) = (n')^A054054(n), where n' is the number resulting from removing the rightmost occurrence of the smallest digit of n, A054054(n).

%I A340270 #37 Jun 30 2025 15:29:00
%S A340270 0,0,0,0,0,0,0,0,0,1,1,2,3,4,5,6,7,8,9,1,2,4,9,16,25,36,49,64,81,1,3,
%T A340270 9,27,64,125,216,343,512,729,1,4,16,64,256,625,1296,2401,4096,6561,1,
%U A340270 5,25,125,625,3125,7776,16807,32768,59049,1
%N A340270 a(n) = (n')^A054054(n), where n' is the number resulting from removing the rightmost occurrence of the smallest digit of n, A054054(n).
%C A340270 Suggested by _Eric Angelini_'s posting to Math-Fun mailing list, Dec 29 2020.
%e A340270 a(201) = 21^0 = 1, a(21) = 2^1 = 2, a(232) = 23^2 = 529.
%p A340270 leastdigit:=proc(n)
%p A340270 min(convert(n,base,10));
%p A340270 end:
%p A340270 locationdigit:=proc(n,d)
%p A340270 local L,i;
%p A340270 L:=convert(n,base,10);
%p A340270 for i from 1 to nops(L) do
%p A340270 if d = L[i] then return (nops(L)+1-i); fi;
%p A340270 od:
%p A340270 end:
%p A340270 cutout:=proc(X,i) [seq(X[j],j=1..i-1),seq(X[j],j=i+1..nops(X))]; end:
%p A340270 ToNum:=proc(X)
%p A340270 add(X[i]*10^(nops(X)-i),i=1..nops(X));
%p A340270 end:
%p A340270 removeleastdigit:=proc(n)
%p A340270 local i,X;
%p A340270 i:=locationdigit(n,leastdigit(n));
%p A340270 X:=ListTools:-Reverse(convert(n,base,10));
%p A340270 ToNum(cutout(X,i));
%p A340270 end proc:
%p A340270 a:=proc(n)
%p A340270 removeleastdigit(n)^leastdigit(n);
%p A340270 end:
%o A340270 (Python)
%o A340270 def A340270(n): return A340184(n)**int(min(str(n)))
%o A340270 print([A340270(n) for n in range(1, 61)]) # _Michael S. Branicky_, Jan 03 2021
%o A340270 (PARI) apply( {A340270(n,m=vecmin(n=digits(n)))=#n>1&& forstep( i=#n,1,-1, n[i]==m && return(fromdigits(n[^i])^m))}, [1..99]) \\ _M. F. Hasler_, Jan 03 2021
%K A340270 base,easy,nonn
%O A340270 1,12
%A A340270 _W. Edwin Clark_, Jan 02 2021
%E A340270 Changed definition as suggested by _M. F. Hasler_