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A339578
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Numbers k such that k^3 can be obtained by taking a number m that is the product of exactly three primes (cf. A014612) and deleting the central digit if m has an odd number of digits, or by deleting the central pair of digits if m has an even number of digits.
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2
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3, 4, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99, 217, 219, 221, 223, 227, 229, 231, 233, 237, 239, 241, 243, 247, 249, 251, 253, 257, 259, 261, 263, 267, 269, 271, 273, 277, 279, 281, 283, 287, 289
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OFFSET
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1,1
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COMMENTS
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This is a kind of inverse to A217297.
If a(n) > 100 then gcd(a(n), 10) = 1. Proof: Suppose gcd(a(n), 10)) > 1 then a(n) is divisible by 2 or 5. As the last three digits of m (as defined in name) will have the last three digits such that m is divisible by 2^3 = 8 or 5^3 = 125 we have a number that is a product of strictly more than three primes. Contradiction.
Furthermore, a(n)^3 has an even number of digits as after deletion of the middle digits there always is an even number of digits left. (End)
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LINKS
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EXAMPLE
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3 is a term because 3^3 = 27 can be obtained by deleting the central digit of 3*3*23 = 207.
4 is a term because 4^3 = 64 can be obtained by deleting the central pair of digits of 2*31*97 = 6014.
20 is not a term because 20^3 = 8000, and any candidate for m will end in 00, and therefore will have at least four prime factors.
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MATHEMATICA
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p3oQ[x_]:=AnyTrue[Table[With[{td=TakeDrop[IntegerDigits[x], IntegerLength[ x]/2]}, FromDigits[ Flatten[ Join[ {td[[1]], {v}, td[[2]]}]]]], {v, 0, 9}], PrimeOmega[#] == 3&]; p3eQ[x_]:= AnyTrue[ Table[With[ {td=TakeDrop[ IntegerDigits[x], IntegerLength[ x]/2]}, FromDigits[Flatten[ Join[ {td[[1]], PadLeft[{w}, 2], td[[2]]}]]]], {w, 0, 99}], PrimeOmega[#]==3&]; p3Q[m_]:=p3oQ[m] || p3eQ[m]; Module[{rng={Ceiling[Reduce[k^3>#, k][[-1]]], Floor[Reduce[k^3<10#, k][[-1]]] }&/@ Table[10^(2n-1), {n, 4}]}, Surd[#, 3]&/@Select[Flatten[Table[w^3, {w, #[[1]], #[[2]]}]&/@ rng], p3Q]] // Quiet (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 20 2020 *)
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PROG
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(PARI) is(n) = { my(d = digits(n^3), half); if(#d % 2 == 1, return(0)); half = #d \ 2; left = vector(half, i, d[i]); right = vector(half, i, d[i + half]); for(j = 1, 2, c = fromdigits(left) * 10^(half + j) + fromdigits(right); for(i = 0, 10^j - 1, if(bigomega(c + i*10^half) == 3, print(c + i*10^half); return(1) ) ) ); 0 } \\ David A. Corneth, Dec 20 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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STATUS
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approved
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