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A328780
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Nonnegative integers k such that k and k^2 have the same number of nonzero digits.
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4
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0, 1, 2, 3, 10, 20, 30, 100, 200, 245, 247, 249, 251, 253, 283, 300, 448, 548, 949, 1000, 1249, 1253, 1416, 1747, 1749, 1751, 1753, 1755, 2000, 2245, 2247, 2249, 2251, 2253, 2429, 2450, 2451, 2470, 2490, 2498, 2510, 2530, 2647, 2830, 3000, 3747, 3751, 4480, 4899
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OFFSET
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1,3
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COMMENTS
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The idea of this sequence comes from the 1st problem of the 28th British Mathematical Olympiad in 1992 (see the link).
This sequence is infinite because the family of integers {10^k, k >= 0} (A011557) belongs to this sequence.
The numbers m, m + 1, m + 2 where m = 49*10^k - 3, or m = 99*10^k - 3, k >= 3 are terms with all nonzero digits. - Marius A. Burtea, Dec 21 2020
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REFERENCES
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A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 1 pp. 57 and 109 (1992)
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LINKS
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EXAMPLE
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247^2 = 61009, hence 247 and 61009 both have 3 nonzero digits, 247 is a term.
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MAPLE
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q:= n->(f->f(n)=f(n^2))(t->nops(subs(0=[][], convert(t, base, 10)))):
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MATHEMATICA
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Select[Range[0, 5000], Equal @@ Total /@ Sign@ IntegerDigits[{#, #^2}] &] (* Giovanni Resta, Feb 27 2020 *)
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PROG
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(Magma) nz:=func<n|#Intseq(n)-Multiplicity(Intseq(n), 0)>; [k:k in [0..5000] | nz(k) eq nz(k^2)]; // Marius A. Burtea, Dec 21 2020
(PARI) isok(k) = hammingweight(digits(k)) == hammingweight(digits(k^2)); \\ Michel Marcus, Dec 22 2020
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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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