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A280642
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Numbers k such that k^3 has an odd number of digits and the middle digit is 2.
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3
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5, 9, 103, 113, 133, 146, 151, 154, 165, 180, 198, 202, 470, 473, 493, 496, 504, 507, 521, 531, 538, 542, 566, 569, 581, 591, 593, 599, 612, 618, 620, 650, 654, 673, 681, 686, 703, 711, 715, 728, 729, 732, 740, 779, 801, 829, 841, 850, 855, 856, 857, 858, 874
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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The sequence of cubes starts: 125, 729, 1092727, 1442897, 2352637, 3112136, 3442951, 3652264, ...
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LINKS
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EXAMPLE
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5^3 = 1(2)5, 180^3 = 583(2)000, 618^3 = 2360(2)9032.
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MATHEMATICA
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a[n_]:=Part[IntegerDigits[n], (Length[IntegerDigits[n]]+1)/2];
Select[Range[0, 874], OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==2 &] (* 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] == 2);
for(k=0, 874, if(isok(k)==1, print1(k, ", "))); \\ Indranil Ghosh, Mar 06 2017
(Python)
i=0
j=1
while i<=874:
n=str(i**3)
l=len(n)
if l%2 and n[(l-1)//2]=="2":
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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