cp's OEIS Frontend

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.

%I A099808 #11 Aug 29 2025 10:13:41
%S A099808 1,15,28,35,44,44,55,56,91,90,88,119,161,165,200,184,273,319,285,357,
%T A099808 377,400,380,434,550,517,592,615,638,667,682,666,740,697,784,688,825,
%U A099808 682,846,770,893,814,868,925,775,899,885,1007,1045,1040,1078,1184,1015
%N 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.
%C A099808 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.
%H A099808 James Buddenhagen, <a href="https://web.archive.org/web/20170330231120/http://www.buddenbooks.com/jb/num_theory/sum_of_2_cubes_a_square.htm">Two Primes Cubed which Sum to a Square</a>.
%e A099808 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]
%Y A099808 Cf. A099806, A099807, A098970, A099809.
%K A099808 nonn,changed
%O A099808 0,2
%A A099808 _James R. Buddenhagen_, Oct 26 2004