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A261778
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Positive numbers n such that (digitsum(n))^2 equals (product of digits(n))^3.
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1
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1, 11114, 11141, 11411, 14111, 41111, 111122, 111212, 111221, 112112, 112121, 112211, 121112, 121121, 121211, 122111, 211112, 211121, 211211, 212111, 221111, 1111111111111111119, 1111111111111111191, 1111111111111111911, 1111111111111119111, 1111111111111191111, 1111111111111911111
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OFFSET
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1,2
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COMMENTS
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Sequence is infinite because it contains all the numbers made of k fours and 8^k-4k ones. - Giovanni Resta, Sep 01 2015
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LINKS
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EXAMPLE
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11114 appears in the sequence because (1 + 1 + 1 + 1 + 4)^2 = (1*1*1*1*4)^3 = 64.
111122 appears in the sequence because (1 + 1 + 1 + 1 + 2 + 2)^2 = (1*1*1*1*2*2)^3 = 64.
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MATHEMATICA
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Select[Range[20000000], Plus @@ IntegerDigits[#]^2 == Times @@ IntegerDigits[#]^3 &]
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PROG
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(PARI) for(n = 1, 1000000, d = digits(n); if((sumdigits(n))^2 == prod(i = 1, #d, d[i])^3, print1(n, ", ")));
(PARI) proddigits(n)=my(d=digits(n)); prod(i=1, #d, d[i])
is(n)=my(s=sumdigits(n)); if(!ispower(s, 3), return(0)); s^2==proddigits(n)^3 \\ Charles R Greathouse IV, Aug 31 2015
(Magma) [n : n in [1..1000000] | (&+Intseq(n))^2 eq (&*Intseq(n))^3 ];
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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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