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Search: id:A099808
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| A099808 |
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If a,b are primes which satisfy the Diophantine equation a^3 + b^3 = c^2, then this sequence consists of the numbers sqrt((a+b)/48), sorted by the magnitude of c. |
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5
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| 1, 15, 28, 35, 44, 44, 55, 56, 91, 90, 88, 119, 161, 165, 200, 184, 273, 319, 285, 357, 377, 400, 380, 434, 550, 517, 592, 615, 638, 667, 682, 666, 740, 697, 784, 688, 825, 682, 846, 770, 893, 814, 868, 925, 775, 899, 885, 1007, 1045, 1040, 1078, 1184, 1015
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OFFSET
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0,2
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COMMENT
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For each n let a=A099806[n], b=A099807[n]. Then sqrt((a+b)/48) is an integer and equals A099808[n]. Note that a^3 + b^3 = c^2 factors as (a+b)*(a^2-a*b+b^2). The first factor (a+b) is 48*d^2, some d. This sequence tabulates the d values. Remember, a and b are prime numbers.
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LINKS
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James Buddenhagen, Two Primes Cubed which Sum to a Square.
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EXAMPLE
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From 11^3 + 37^3 = 228^3 we get sqrt((a+b)/48)=(11+37)/48=1, so 1 is in the sequence.
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CROSSREFS
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Cf. A099806, A099807, A098970, A099809.
Sequence in context: A116070 A073766 A031334 this_sequence A134621 A106499 A043108
Adjacent sequences: A099805 A099806 A099807 this_sequence A099809 A099810 A099811
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KEYWORD
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nonn
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AUTHOR
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James Buddenhagen (jbuddenh(AT)gmail.com), Oct 26 2004
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