A352761 -id:A352761 - cp's OEIS frontend

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

Vladimir Pletser, Apr 02 2022

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.

			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.

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).

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

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

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

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

Original entry on oeis.org

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

Vladimir Pletser, Apr 02 2022

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.

			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.

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).

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

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

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

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

cp's OEIS Frontend

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).

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

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cp's OEIS Frontend

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Python

Formula

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