A099808 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.

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

Offset: 0

Author

James R. Buddenhagen, Oct 26 2004

Keywords

• nonn
• changed

Comments

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.

Examples

			From 11^3 + 37^3 = 228^2 we get sqrt((a+b)/48) = (11+37)/48 = 1, so 1 is in the sequence. [corrected by _Harvey P. Dale_, Apr 12 2011]
		

Links

• James Buddenhagen, Two Primes Cubed which Sum to a Square.

Crossrefs

Cf. A099806, A099807, A098970, A099809.