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%I #9 Feb 10 2018 12:50:52
%S 1,4,9,10,40,90,100,148,400,838,841,900,1000,1111,1440,4000,4058,6430,
%T 7388,8504,8668,8837,9000,10000,10111,10200,11011,11101,11110,11215,
%U 12321,13231,13333,13955,14083,14403,14464,17780,18480,18770,20112,21012,21102,21412,22322,22592,25652,29522,30205,30343,30441
%N Numbers n such that d(1)^1 + d(2)^2 + ... + d(p)^p and d(1)^p + d(2)^p-1 +... + d(p)^1 are squares, where d(i), i=1..p, are the digits of n.
%H Harvey P. Dale, <a href="/A178360/b178360.txt">Table of n, a(n) for n = 1..250</a>
%e 841 is in the sequence because :
%e 8 + 4^2 + 1^3 = 25 = 5^2 ;
%e 8^3 + 4^2 + 1 = 529 = 23^2.
%p with(numtheory):for n from 1 to 50000 do:l:=length(n):n0:=n:s1:=0:s2:=0:for
%p m from 1 to l do:q:=n0:u:=irem(q,10):v:=iquo(q,10):n0:=v :s1:=s1+u^(l-m+1):s2:=s2+u^m:od:s10:= sqrt(s1):s20:=sqrt(s2): if s10=floor(s10) and s20=floor(s20) then printf(`%d, `,n):else fi:od:
%t ndnQ[n_]:=Module[{idn=IntegerDigits[n],len=Range[IntegerLength[n]]},AllTrue[{Sqrt[Total[idn^len]],Sqrt[Total[Reverse[idn]^len]]},IntegerQ]]; Select[Range[31000],ndnQ] (* The program uses the AllTrue function from Mathematica version 10 *) (* _Harvey P. Dale_, Feb 10 2018 *)
%K nonn,base
%O 1,2
%A _Michel Lagneau_, Dec 21 2010
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