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Search: id:A152829
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| A152829 |
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Numbers k whose squares can be written in exactly one way as a sum of three squares: k^2 = a^2 + b^2 + c^2 with 1 <= a <= b <= c. |
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+0
1
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| 3, 6, 7, 12, 13, 14, 24, 26, 28, 48, 52, 56, 96, 104, 112, 192, 208, 224, 384, 416, 448, 768, 832, 896
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Guessed continuation (see ogf below):
1536,1664,1792,3072,3328,3584,6144,6656,7168,12288,13312,14336,24576,26624,28672,
49152,53248,57344,98304,106496,114688,196608,212992,229376,393216,425984,458752
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LINKS
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inspired by a posting to sci.math.symbolic by M. A. Fajjal
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FORMULA
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Guessed ogf: x*(x^4 + 6*x^3 + 7*x^2 + 6*x + 3)/(1 - 2*x^3)
{k: A025427(k^2)=1}. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 15 2008]
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EXAMPLE
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9 is not in this sequence because 9^2 = 1^2+4^2+8^2 3^2+6^2+6^2 = 4^2+4^2+7^2
7 is in this sequence because 7^2 = 2^2+3^2+6^2 is the only way to write 7^2 as a sum of three squares.
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PROGRAM
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(C-program 3sq.c) #include <stdio.h> int main (int argc, char *argv[]) { long n, n2, a, a2, b, b2, c, c2; int s = 0; n=atol(argv[1]); n2=n*n;
for (a=1; a<n; a++) { a2 = a*a; for (b=a; (b2=b*b)<n2-a2; b++) for (c=b; (c2=c*c)<=n2-a2-b2; c++) if (a2+b2+c2==n2) s++; }
return (s==1?0:1); }
Then the bash-line
for (( i=1; i<1000; i++ )); do ./3sq $i && echo -n $i, ; done >3sq.txt
gives the elements less than 1000 of this sequence
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CROSSREFS
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Sequence in context: A004755 A004760 A093906 this_sequence A104463 A072757 A032849
Adjacent sequences: A152826 A152827 A152828 this_sequence A152830 A152831 A152832
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KEYWORD
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nonn
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AUTHOR
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Peter Pein (petsie(AT)dordos.net), Dec 13 2008
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