A352759 Centered cube numbers that are the difference of two positive cubes; a(n) = 27*t^3*(27*t^6 + 1)/4 with t = 2*n-1.

189, 3587409, 355957875, 7354447191, 70607389041, 429735975669, 1932670025559, 7006302268875, 21612640524741, 58809832966521, 144757538551899, 328260072633759, 695228576765625, 1389765141771741, 2643927354266751, 4818621138983379, 8458493032498509, 14364150148238625, 23685527077994691, 38040743821584231

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 = 3*(2*n - 1) == 3 (mod 6), with (for n > 1) C > D > B > 0, and A = as 27*t^3*(27*t^6 + 1)/4 with t = 2*n-1, and where A = a(n) (this sequence), B = A355751(n), C = A355752(n) and D = A355753(n).

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

Subsequence of A005898, and of A352133.

Examples

			a(1) = 189 belongs to the sequence because 189 = 4^3 + 5^3 = 6^3 - 3^3 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 3587409 belongs to the sequence because 3587409 = 121^3 + 122^3 = 369^3 - 360^3 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 27*(2*3 - 1)^3*(27*(2*3 - 1)^6 + 1)/4 = 355957875.
		

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, A352755, A352756, A352757, A352758, A355751, A355752, A355753.

Programs

Maple

restart; for n from 1 to 20 do 27*(2*n-1)^3*(27*(2*n-1)^6+1)*(1/4); end do;

Formula

a(n) = A355751(n)^3 + (A355751(n) + 1)^3 = A355752(n)^3 - A355753(n)^3 and A355752(n) - A355753(n) = 3*(2*n - 1).

a(n) = 27*(2*n - 1)^3*(27*(2*n - 1)^6 + 1)/4.

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