OFFSET
1,1
COMMENTS
Numbers C > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference C - D is odd, C - D = 2*n - 1, and the difference between the positive cubes C^3 - D^3 is equal to a centered cube number, C^3 - D^3 = B^3 + (B+1)^3, with C > D > B > 0, and A > 0, A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = A352755(n), B = A352756(n), C = a(n) (this sequence), and D = A352758(n).
There are infinitely many such numbers a(n) = C 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 (5,-10,10,-5,1).
FORMULA
a(n)^3 - A352758(n)^3 = A352756(n)^3 + (A352756(n) + 1)^3 = A352755(n) and a(n) - A352758(n) = 2*n - 1.
a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1)/2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 576*(n - 2), with a(1) = 6, a(2) = 151 and a(3) = 1016.
a(n) can be extended for negative n such that a(-n) = a(n+1) - (2*n + 1).
G.f.: x*(6 + 121*x + 321*x^2 + 123*x^3 + 5*x^4)/(1 - x)^5. - Stefano Spezia, Apr 08 2022
EXAMPLE
a(1) = 6 belongs to the sequence as 6^3 - 5^3 = 3^3 + 4^3 = 91 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 151 belongs to the sequence as 151^3 - 148^3 = 46^3 + 47^3 = 201159 and 151 - 148 = 3 = 2*2 - 1.
a(3) = (3(2*3 - 1)^2*((2*3 - 1)^2 + 2) + 2*3 + 1)/2 = 1016.
a(4) = 3*a(3) - 3*a(2) + a(1) + 576*2 = 3*1016 - 3*151 + 6 + 576*2 = 3753.
MAPLE
restart; for n to 20 do (1/2)*(3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1); end do;
PROG
(Python)
def A352757(n): return n*(n*(n*(24*n - 48) + 48) - 23) + 5 # Chai Wah Wu, Jul 10 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Apr 02 2022
STATUS
approved