A355751 -id:A355751 - cp's OEIS frontend

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

Vladimir Pletser, Apr 02 2022

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.

			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.

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).

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

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;

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).

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

Original entry on oeis.org

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

Offset: 1

Vladimir Pletser, Apr 02 2022

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.

			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.

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).

Cf. A005898, A001235, A272885, A352133, A352134, A352135, A352136, A352221, A352222, A352223, A352224, A352225, A352755, A352756, A352757, A352758, A355751, A355752, A355753.

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

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).

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

Original entry on oeis.org

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

Vladimir Pletser, Apr 02 2022

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.

			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.

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

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

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

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

A355752 a(n) = 3(2n - 1)( 3(2*n - 1)^3 + 1) / 2.

Original entry on oeis.org

6, 369, 2820, 10815, 29538, 65901, 128544, 227835, 375870, 586473, 875196, 1259319, 1757850, 2391525, 3182808, 4155891, 5336694, 6752865, 8433780, 10410543, 12715986, 15384669, 18452880, 21958635, 25941678, 30443481, 35507244, 41177895, 47502090, 54528213, 62306376, 70888419, 80327910, 90680145, 102002148, 114352671, 127792194

Offset: 1

Vladimir Pletser, Jul 15 2022

Numbers C > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference (C - D) == 3 (mod 6), C - D = 3 (2n - 1) for 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 = 27*t^3 *(27*t^6+1)/4 with t = 2*n-1, and where A = A352759(n), B = A355751(n), C = a(n) (this sequence), and D = A355753(n).

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

			a(1) = 6 belongs to the sequence as 6^3 - 3^3 = 4^3 + 5^3 = 189 and 6 - 3 = 3 = 3 (2*1 - 1).
a(2) = 369 belongs to the sequence as 369^3 - 360^3 = 121^3 + 122^3 = 3587409 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 3*(2*3 - 1)*( 3*(2*3 - 1)^3 + 1) / 2 = 2820.
a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2820 - 3*369 + 6 + 1728*2  = 10815.

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, A352757, A352758, A352759, A355751, A355753.

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

a(n)=3*(2*n-1)*(3*(2*n-1)^3+1)/2 \\ Charles R Greathouse IV, Oct 21 2022

a(n)^3 - A355753(n)^3 = A355751(n)^3 + (A355751(n) + 1)^3 = A352759(n) and a(n) - A355753(n) = 3*(2*n - 1).
a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1728*(n - 2), with a(1) = 6, a(2) = 369 and a(3) = 2820.
a(n) can be extended for negative n such that a(-n) = a(n+1) - 3*(2*n + 1).
G.f.: -3*x*(2+113*x+345*x^2+115*x^3+x^4) / (x-1)^5 . - R. J. Mathar, Aug 03 2022

A355753 a(n) = 3(2n - 1)( 3(2*n - 1)^3 - 1) / 2 for n > 0.

Original entry on oeis.org

3, 360, 2805, 10794, 29511, 65868, 128505, 227790, 375819, 586416, 875133, 1259250, 1757775, 2391444, 3182721, 4155798, 5336595, 6752760, 8433669, 10410426, 12715863, 15384540, 18452745, 21958494, 25941531, 30443328, 35507085, 41177730, 47501919, 54528036, 62306193, 70888230, 80327715, 90679944, 102001941, 114352458, 127791975

Offset: 1

Vladimir Pletser, Jul 15 2022

Numbers D > 0 such that A = B^3 + (B+1)^3 = C^3 - D^3 such that the difference C - D == 3 (mod 6), C - D = 3*(2*n - 1) for 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 = 27*t^3 *(27*t^6+1)/4 with t = 2*n-1, and where A = A352759(n), B = A355751(n), C = A355752(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.

			a(1) = 3 belongs to the sequence as 6^3 - 3^3 = 4^3 + 5^3 = 189 and 6 - 3 = 3 = 3*(2*1 - 1).
a(2) = 360 belongs to the sequence as 369^3 - 360^3 = 121^3 + 122^3 = 3587409 and 369 - 360 = 9 = 3*(2*2 - 1).
a(3) = 3*(2*3 - 1)*( 3*(2*3 - 1)^3 - 1) / 2 = 2805.
a(4) = 3*a(3) - 3*a(2) + a(1) + 1728*2 = 3*2805 - 3*360 + 3 + 1728*2  = 10794.

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

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

A355752(n)^3 - a(n)^3 = A355751(n)^3 + (A355751(n) + 1)^3 = A352759(n) and A355752(n) - a(n) = 3*(2*n - 1).
a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 - 1) / 2 for n > 0.
For n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 1728*(n - 2), with a(1) = 3, a(2) = 360 and a(3) = 2805.
a(n) can be extended for negative n such that a(-n) = a(n+1) + (2n + 1).
G.f.: -3*x*(1+115*x+345*x^2+113*x^3+2*x^4) / (x-1)^5 . - R. J. Mathar, Aug 03 2022

cp's OEIS Frontend

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.

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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.

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

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A355752 a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 + 1) / 2.

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A355753 a(n) = 3*(2*n - 1)*( 3*(2*n - 1)^3 - 1) / 2 for n > 0.

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

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

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

A355752 a(n) = 3(2n - 1)( 3(2*n - 1)^3 + 1) / 2.

A355753 a(n) = 3(2n - 1)( 3(2*n - 1)^3 - 1) / 2 for n > 0.