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%I #30 Dec 20 2020 13:48:40
%S 3,4,11,12,13,14,15,16,17,18,19,21,47,49,51,53,57,59,61,63,67,69,71,
%T 73,77,79,81,83,87,89,91,93,97,99,217,219,221,223,227,229,231,233,237,
%U 239,241,243,247,249,251,253,257,259,261,263,267,269,271,273,277,279,281,283,287,289
%N 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.
%C This is a kind of inverse to A217297.
%C From _David A. Corneth_, Dec 20 2020: (Start)
%C 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.
%C 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)
%H David A. Corneth, <a href="/A339578/b339578.txt">Table of n, a(n) for n = 1..10000</a>
%e 3 is a term because 3^3 = 27 can be obtained by deleting the central digit of 3*3*23 = 207.
%e 4 is a term because 4^3 = 64 can be obtained by deleting the central pair of digits of 2*31*97 = 6014.
%e 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.
%t 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 *)
%o (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
%Y Cf. A014612, A217297.
%K nonn,base
%O 1,1
%A _Harvey P. Dale_ and _N. J. A. Sloane_, Dec 20 2020
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