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A135780
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Squares such that another square can be obtained by a cyclic permutation of the digits, excluding leading zeros.
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2
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144, 196, 256, 441, 625, 961, 11664, 14884, 16384, 16641, 25600, 36864, 38416, 46656, 48841, 60025, 61009, 66564, 86436, 96100, 166464, 214369, 216225, 236196, 272484, 364816, 436921, 481636, 622521, 646416, 842724, 870489, 898704, 962361
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OFFSET
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1,1
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COMMENTS
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It seems that there are never more than two squares having the same digits up to a cyclic permutation (checked up to (10^8)^2).
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LINKS
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EXAMPLE
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a(1) = 144 = 12^2 is the least square such that a cyclic permutation of its decimal digits is again a square, namely 441 = 21.
a(2) = 196 = 14^2 is the second least square having this property, with 961 = 19^2.
A034289(2)=169 does not figure here, since the permutations of its digits yielding squares are 196 and 961, which are not cyclic permutations of 169.
The number 25600 is here since 60025 is also a square.
The fact that 00256 also is a square is irrelevant: permutations with leading zeros are not considered.
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MATHEMATICA
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scpQ[n_]:=Module[{idn=IntegerDigits[n], r}, r=DeleteCases[Rest[ NestList[ RotateRight[ #]&, IntegerDigits[ n], IntegerLength[n]-1]], _?(First[#] == 0&)]; AnyTrue[FromDigits/@r, IntegerQ[Sqrt[#]]&]]; Select[Range[ 1000]^2, scpQ] (* This program uses the AnyTrue function from Mathematica version 10 *) (* Harvey P. Dale, Aug 21 2014 *)
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PROG
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(PARI) for(n=1, 10^8, (t=n^2)/* %10 || next <= this would exclude terms with trailing '0's */; found=0; for(j=1, k=#Str(t)-1, t=divrem(t, 10); t[2] || (t=t[1]) && next /* <= this excludes leading '0's */; issquare(t=t[1]+10^k*t[2]) || next; /* t%10 || next; <= would exclude permutations with trailing '0's */ print1( if(found, "<<<"/* mark multiple permutations: this never happens */, found=1; n^2)", ")))
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CROSSREFS
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KEYWORD
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base,easy,nonn,nice
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AUTHOR
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STATUS
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approved
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