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

0,2

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.

Links

Table of n, a(n) for n=0..52.

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

Example

From 11^3 + 37^3 = 228^2 we get sqrt((a+b)/48)=(11+37)/48=1, so 1 is in the sequence.

Crossrefs

Cf. A099806, A099807, A098970, A099809.

Sequence in context: A223449 A031334 A178958 * A291054 A134621 A106499

Adjacent sequences: A099805 A099806 A099807 * A099809 A099810 A099811

Keyword

nonn

Author

James R. Buddenhagen, Oct 26 2004

Extensions

Example corrected by Harvey P. Dale, Apr 12 2011.

Status

approved