A352757 a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1)/2 for n > 0.

6, 151, 1016, 3753, 10090, 22331, 43356, 76621, 126158, 196575, 293056, 421361, 587826, 799363, 1063460, 1388181, 1782166, 2254631, 2815368, 3474745, 4243706, 5133771, 6157036, 7326173, 8654430, 10155631, 11844176, 13735041, 15843778, 18186515, 20779956, 23641381, 26788646, 30240183, 34015000, 38132681

Offset: 1

Author

Vladimir Pletser, Apr 02 2022

Keywords

• nonn
• easy

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.

Examples

			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.
		

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).

Crossrefs

Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352221, A352223, A352224, A352225, A352755, A352756, A352758, A352759, A352760, A352761, A352762.

Programs

Maple

restart; for n to 20 do (1/2)*(3*(2*n - 1)^2*((2*n - 1)^2 + 2) + 2*n + 1); end do;

Python

def A352757(n): return n*(n*(n*(24*n - 48) + 48) - 23) + 5 # Chai Wah Wu, Jul 10 2022

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