A352755 Positive centered cube numbers that can be written as the difference of two positive cubes: a(n) = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1 and n > 0.

91, 201159, 15407765, 295233841, 2746367559, 16448122691, 73287987409, 264133278045, 811598515091, 2202365761759, 5410166901741, 12249942682409, 25914353312575, 51755729480091, 98389720844009, 179211321358741, 314429627203659, 533744613620855, 879807401606341, 1412624924155809

Offset: 1

Author

Vladimir Pletser, Apr 02 2022

Keywords

• nonn
• easy

Comments

Numbers A > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 and such that C - D = 2n - 1, with C > D > B > 0, and A = t*(3*t^2 + 4)*(t^2*(3*t^2 + 4)^2 + 3)/4 with t = 2*n-1, and where A = a(n) (this sequence), B = A352756(n), C = A352757(n) and D = A352758(n).

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

Subsequence of A005898, of A352133 and of A352220.

Examples

			a(1) = 91 belongs to the sequence because 91 = 3^3 + 4^3 = 6^3 - 5^3 and 6 - 5 = 1 = 2*1 - 1.
a(2) = 201159 belongs to the sequence because 201159 = 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)*((2*3 - 1)^2*(3*(2*3 - 1)^2 + 4)^2 + 3)/4 = 15407765.
		

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).

Crossrefs

Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352221, A352222, A352223, A352224, A352225, A352756, A352757, A352758, A352759, A355751, A355752, A355753.

Programs

Maple

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

Formula

a(n) = A352756(n)^3 + (A352756(n) + 1)^3 = A352757(n)^3 - A352758(n)^3 and A352757(n) - A352758(n) = 2n - 1.

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

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