OFFSET
1,4
COMMENTS
Given a prime p = 6k+1, then there exists a unique pair of integers (x, y) such that 4p = x^2 + 27y^2, x == 1 (mod 3), and y>0. - Michael Somos, Jul 10 2022
REFERENCES
A. J. C. Cunningham, Quadratic Partitions. Hodgson, London, 1904, p. 1.
B. Engquist and Wilfried Schmid, Mathematics Unlimited - 2001 and Beyond, Chapter on Error-correcting codes and curves over finite fields, see pp. 1118-1119. [From Neven Juric, Oct 16 2008.]
D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 55.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Ruperto Corso, Table of n, a(n) for n = 1..1000
A. J. C. Cunningham, Quadratic Partitions, Hodgson, London, 1904 [Annotated scans of selected pages]
Steven R. Finch, Powers of Euler's q-Series, arXiv:math/0701251 [math.NT], 2007.
EXAMPLE
The 7th prime of the form 6k+1 (A002476) is 61 and 4*61 = 244 = 1^2 + 27*3^2 gives a(7) = 3. The 8th prime of the form 6k+1 is 67 and 4*67 = 268 = (-5)^2 + 27*3^2 gives a(8) = 3. - Michael Somos, Jul 10 2022
MATHEMATICA
Reap[For[p = 2, p<2000, p = NextPrime[p], For[x = 1, x <= Floor[2*Sqrt[p]], x++, px = 4*p - x^2; If[Mod[px, 27] == 0, If[IntegerQ[y = Sqrt[px/27]], Sow[y]]]]]][[2, 1]] (* Jean-François Alcover, Sep 06 2018, after Ruperto Corso *)
PROG
(PARI) forprime(p=2, 10000, for(x=1, floor(2*sqrt(p)), px=4*p-x^2; if(px%27==0, if(issquare(px/27, &y), print1(y", "))))) /* Ruperto Corso, Dec 14 2011 */
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Corrected and extended by Ruperto Corso, Dec 14 2011
Name clarified by Michael Somos, Jul 10 2022
STATUS
approved