OFFSET
1,1
COMMENTS
Numbers A > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 3*(2*n - 1) == 3 (mod 6), with (for n > 1) C > D > B > 0, and A = as 27*t^3*(27*t^6 + 1)/4 with t = 2*n-1, and where A = a(n) (this sequence), B = A355751(n), C = A355752(n) and D = A355753(n).
There are infinitely many such numbers a(n) = A in this sequence.
LINKS
Vladimir Pletser, Table of n, a(n) for n = 1..10000
A. Grinstein, Ramanujan and 1729, University of Melbourne Dept. of Math and Statistics Newsletter: Issue 3, 1998.
Vladimir Pletser, Euler's and the Taxi-Cab relations and other numbers that can be written twice as sums of two cubed integers, submitted. Preprint available on ResearchGate, 2022.
Eric Weisstein's World of Mathematics, Centered Cube Number
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
FORMULA
EXAMPLE
a(1) = 189 belongs to the sequence because 189 = 4^3 + 5^3 = 6^3 - 3^3 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 3587409 belongs to the sequence because 3587409 = 121^3 + 122^3 = 369^3 - 360^3 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 27*(2*3 - 1)^3*(27*(2*3 - 1)^6 + 1)/4 = 355957875.
MAPLE
restart; for n from 1 to 20 do 27*(2*n-1)^3*(27*(2*n-1)^6+1)*(1/4); end do;
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Apr 02 2022
STATUS
approved