A352758 a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) - 2*n + 3)/2 for n > 0.

5, 148, 1011, 3746, 10081, 22320, 43343, 76606, 126141, 196556, 293035, 421338, 587801, 799336, 1063431, 1388150, 1782133, 2254596, 2815331, 3474706, 4243665, 5133728, 6156991, 7326126, 8654381, 10155580, 11844123, 13734986, 15843721, 18186456, 20779895, 23641318, 26788581, 30240116, 34014931, 38132610

Offset: 1

Author

Vladimir Pletser, Apr 02 2022

Keywords

• nonn
• easy

Comments

Numbers D > 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 of the positive cubes C^3 - D^3 is equal to centered cube numbers, 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 = A352757(n), and D = a(n) (this sequence).

There are infinitely many such numbers a(n) = D in this sequence.

Subsequence of A352136 and of A352223.

Examples

			a(1) = 5 belongs to the sequence as 6^3 - 5^3 = 3^3 + 4^3 = 91 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 148 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 + 3)/2 = 1011.
a(4) = 3*a(3) - 3*a(2) + a(1) + 576*2 = 3*1011 - 3*148 + 5 + 576*2  = 3746.
		

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, A352222, A352224, A352225, A352755, A352756, A352757, A352759, A355751, A355752, A355753.

Programs

Maple

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

Python

def A352758(n): return n*(n*(n*(24*n - 48) + 48) - 25) + 6 # Chai Wah Wu, Jul 11 2022

Formula

A352757(n)^3 - a(n)^3 = A352756(n)^3 + (A352756(n) + 1)^3 = A352755(n) and A352757(n) - a(n) = 2*n - 1.

a(n) = (3*(2*n - 1)^2*((2*n - 1)^2 + 2) - 2*n + 3)/2.

For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 576*(n - 2), with a(1) = 5, a(2) = 148 and a(3) = 1011.

a(n) can be extended for negative n such that a(-n) = a(n+1) + (2n + 1).

G.f.: x*(5 + 123*x + 321*x^2 + 121*x^3 + 6*x^4)/(1 - x)^5. - Stefano Spezia, Apr 08 2022