OFFSET
1,2
COMMENTS
Each number > 90 contains at least two identical digits because the sequence A197125 contains a subset of numbers all of whose digits are distinct and are all the permutations of 1567890. But 1^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 = 1926 is not square. Consequently, it is impossible to find numbers > 90 with distinct digits in this sequence.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
EXAMPLE
4669 is in the sequence because:
4 + 6 + 6 + 9 = 25 = 5^2;
4^2 + 6^2 + 6^2 + 9^2 = 169 = 13^2;
4^3 + 6^3 + 6^3 + 9^3 = 1225 = 35^2.
MAPLE
for n from 1 to 10000 do:l:=evalf(floor(ilog10(n))+1):n0:=n:s1:=0:s2:=0:s3:=0:for m from 1 to l do:q:=n0:u:=irem(q, 10):v:=iquo(q, 10): n0:=v :s1:=s1+u:s2:=s2+u^2:s3:=s3+u^3:od:if sqrt(s1)=floor(sqrt(s1)) and sqrt(s2)=floor(sqrt(s2)) and sqrt(s3)=floor(sqrt(s3))then printf(`%d, `, n): else fi:od:
MATHEMATICA
sdQ[n_]:=Module[{idn=IntegerDigits[n]}, IntegerQ[Sqrt[Total[idn]]] && IntegerQ[Sqrt[Total[idn^2]]]&&IntegerQ[Sqrt[Total[idn^3]]]]; Select[ Range[ 10000], sdQ] (* Harvey P. Dale, Oct 25 2011 *)
PROG
(PARI) is(n)=my(v=eval(Vec(Str(n)))); issquare(sum(i=1, #v, v[i]))&&issquare(sum(i=1, #v, v[i]^2))&&issquare(sum(i=1, #v, v[i]^3)) \\ Charles R Greathouse IV, Oct 10 2011
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Oct 10 2011
STATUS
approved