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A278316
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Odd numbers n such that q(n)^2 = q(n^2) != 0, where q(n) is the digit product on base 10.
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0
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1, 3, 661, 983, 2631, 2893, 12385, 12893, 14661, 18615, 27519, 35383, 36213, 38691, 46215, 49231, 83631, 87291, 92843, 113865, 116683, 123415, 129815, 136423, 139261, 152619, 161683, 162435, 166817, 178119, 194725, 244635, 247941, 254663, 274165, 276941
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OFFSET
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1,2
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REFERENCES
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Michael Huke, Solution to exercise psi-15 (German language article), WURZEL 11/2016, November 2016, page 252, http://wurzel.org/
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LINKS
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EXAMPLE
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For n=3, a(3)=661: q(661)^2 = (6*6*1)^2 = 36^2 = 1296 = 4*3*6*9*2*1 = q(436921) = q(661^2).
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MATHEMATICA
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Select[Range[1, 10^6, 2], And[MatchQ @@ #, Times @@ # != 0] &@{(Times @@ IntegerDigits@ #)^2, Times @@ IntegerDigits[#^2]} &] (* Michael De Vlieger, Dec 06 2016 *)
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PROG
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(PARI) pd(n) = my(d=digits(n)); prod(k=1, #d, d[k]);
isok(n) = (n % 2) && (p = pd(n)^2) && (p == pd(n^2)); \\ Michel Marcus, Dec 04 2016
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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