A352756 Positive numbers k such that the centered cube number k^3 + (k+1)^3 is equal to the difference of two positive cubes and to A352755(n).

3, 46, 197, 528, 1111, 2018, 3321, 5092, 7403, 10326, 13933, 18296, 23487, 29578, 36641, 44748, 53971, 64382, 76053, 89056, 103463, 119346, 136777, 155828, 176571, 199078, 223421, 249672, 277903, 308186, 340593, 375196, 412067, 451278, 492901, 537008, 583671, 632962, 684953, 739716, 797323, 857846, 921357

Offset: 1

Author

Vladimir Pletser, Apr 02 2022

Keywords

• nonn
• easy

Comments

Numbers B > 0 such that the centered cube number B^3 + (B+1)^3 is equal to the difference of two positive cubes, i.e., A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 2n - 1, 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 = a(n) (this sequence), C = A352757(n) and D = A352758(n).

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

Subsequence of A352134 and of A352221.

Examples

			a(1) = 3 is a term because 3^3 + 4^3 = 6^3 - 5^3 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 46 is a term because 46^3 + 47^3 = 151^3 - 148^3 and 151 - 148 = 3 = 2*2 - 1.
a(3) = ((2*3 - 1)*(3*(2*3 - 1)^2 + 4) - 1)/2 = 197.
a(4) = 3*197 - 3*46 + 3 + 72 = 528.
		

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 (4,-6,4,-1).

Crossrefs

Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352220, A352222, A352223, A352224, A352225, A352755, A352757, A352758, A352759, A352760, A352761, A352762.

Programs

Maple

restart; for n to 20 do (1/2)* ((2*n - 1)*(3*(2*n - 1)^2 + 4) - 1);  end do;

Formula

a(n)^3 + (a(n)+1)^3 = A352757(n)^3 - A352758(n)^3 and A352757(n) - A352758(n) = 2*n - 1.

a(n) = ((2*n - 1)*(3*(2*n - 1)^2 + 4) - 1)/2.

For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 72, with a(1) = 3, a(2) = 46 and a(3) = 197.

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

G.f.: x*(3 + 34*x + 31*x^2 + 4*x^3)/(1 - x)^4. - Stefano Spezia, Apr 08 2022