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A280640
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Numbers k such that k^3 has an odd number of digits and the middle digit is 0.
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12
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0, 30, 40, 42, 100, 101, 115, 116, 123, 126, 135, 163, 164, 171, 199, 200, 201, 214, 468, 479, 487, 498, 500, 502, 513, 520, 525, 543, 557, 562, 564, 575, 576, 577, 578, 579, 585, 596, 600, 615, 623, 642, 656, 661, 666, 690, 695, 697, 700, 705, 709, 717, 721
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internal format)
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OFFSET
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1,2
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COMMENTS
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The sequence of cubes starts: 0, 27000, 64000, 74088, 1000000, 1030301, 1520875, 1560896, ...
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LINKS
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EXAMPLE
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0^3 = (0), 126^3 = 200(0)376, 562^3 = 1775(0)4328.
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MATHEMATICA
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a[n_]:=Part[IntegerDigits[n], (Length[IntegerDigits[n]] + 1)/2];
Select[Range[0, 721], OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==0 &] (* Indranil Ghosh, Mar 06 2017 *)
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PROG
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(PARI)
isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 0);
for(k=0, 721, if(k==0 || isok(k)==1, print1(k, ", "))); \\ Indranil Ghosh, Mar 06 2017
(Python)
i=0
j=1
while i<=721:
n=str(i**3)
l=len(n)
if l%2 and n[(l-1)//2]=="0":
print(str(i), end=", ")
j+=1
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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