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A068536
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Numbers m such that m^2 + (reversal of m)^2 is a square. (Leading 0's are ignored.)
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5
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88209, 90288, 125928, 196020, 368280, 829521, 1978020, 2328480, 5513508, 8053155, 19798020, 86531940, 197998020, 554344560, 556326540, 1960396020, 1979998020, 5543944560, 5925169800, 8820988209, 9028890288, 12592925928, 14011538112, 19602196020, 19799998020
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OFFSET
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1,1
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COMMENTS
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The sequence is infinite, even if it is restricted to terms that end with a nonzero digit, as any term generates an infinite number of other terms by the following scheme: If m is a term of the sequence and d(m) denotes the number of digits of m, then set m' = m*10^d'+m with d' >= d(m). For d' >= d(m) we have reverse(m') = reverse(m)*10^d' + reverse(m) and thus (m')^2 + reverse(m')^2 = (m*10^d'+m)^2 + (reverse(m)*10^d'+reverse(m))^2 = (m^2+reverse(m)^2)*(10^d'+1)^2. As m^2+reverse(m)^2 is a perfect square by assumption, the product on the right-hand side of the equation is also a perfect square and m' is part of the sequence. The calculation works also with m' = m*(10^(k*d')+...+10^d'+1). As an example take a(1)=88209. All numbers 8820988209, 882098820988209, 88209882098820988209, ... and 88209088209, 882090088209, 8820900088209, ... are also terms of the sequence. - Matthias Baur, May 01 2020
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LINKS
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Sheng Jiang and Rui-Chen Chen, Digits reversed Pythagorean triples, International Journal of Mathematical Education in Science and Technology, volume 29, number 5, 1998, pages 689-696.
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EXAMPLE
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88209^2 + 90288^2 = 126225^2, so 88209 belongs to the sequence.
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MATHEMATICA
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Do[If[IntegerQ[Sqrt[n^2 + FromDigits[Reverse[IntegerDigits[n]]]^2]], Print[n]], {n, 1, 10^6}]
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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