A267414 -id:A267414 - cp's OEIS frontend

A226955 Number of representations of n! as a sum of 3 positive cubes.

0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 4, 0, 2, 2, 6, 2, 11, 2, 13, 20, 24, 9

Offset: 1

Zak Seidov, Jun 26 2013

Conjecture: For any t >= 0, there are only finitely many values of n such that a(n) = t. - Altug Alkan, Aug 16 2020

			n = 12, n! = 479001600 = x^3 + y^3 + z^3 with {x,y,z} = {35,309,766}, {47,214,777}, {60,486,714}, {240,504,696}; 4 solutions, hence a(12) = 4;
n = 16, n! = x^3 + y^3 + z^3 with {x,y,z} = {7644,21192,22212}, {8240,8400,27040}, {10980,15288,25212}, {11648,18016,23808}, {12096,19968,22368}, {13030,18330,23240}; 6 solutions, hence a(16) = 6.
From _Chai Wah Wu_, May 21 2017: (Start)
n = 22, n! = x^3 + y^3 + z^3 with (x,y,z) = (286272, 8168832, 8334144), (443100, 4806340, 10042760), (663040, 7882560, 8590400), (720720, 5343408, 9902592), (757890, 8108100, 8389710), (854812, 2888886, 10320506), (861120, 3584160, 10251360), (1025640, 2784600, 10326960), (1266408, 4510296, 10099728), (1443806, 7114569, 9129295), (1792350, 6013602, 9657648), (1814400, 3689280, 10221120), (1871415, 4292190, 10126305), (1926720, 5685120, 9771840), (2419200, 7506240, 8823360), (2517424, 7223832, 9008552), (2779200, 3144960, 10232640), (2870532, 6957468, 9140040), (3021408, 4549080, 10007592), (3244410, 7888800, 8429190), (3776535, 6384105, 9321480), (5083936, 5242592, 9467136), (5681592, 7233408, 8253000), (6391665, 6719895, 8239770)
n = 23, n! = x^3 + y^3 + z^3 with (x,y,z) = (136080, 8250480, 29352960), (5369910, 6098890, 29422400), (5766592, 18082176, 27029696), (6151320, 19606860, 26247060), (7572485, 23185155, 23485930), (8255520, 10856160, 28848960), (8678304, 19104696, 26316360), (11959740, 19850400, 25365060), (13799880, 22091640, 23172240)
(End)

Lars Blomberg, Table of n, a(n) for n = 1..21, with solutions

Cf. A003072 (numbers that are the sum of 3 positive cubes), A267414.

a(n) = A025456(n!). - Charles R Greathouse IV, Oct 27 2013

a(17)-a(18) from Giovanni Resta, Jun 26 2013
a(4) corrected and a(19)-a(21) from Lars Blomberg, Sep 07 2013
a(22)-a(23) from Chai Wah Wu, May 21 2017

A337046 Integers n such that n! = x^2 + y^3 + z^6 where x, y and z are nonnegative integers, is soluble.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 10, 14, 16, 17, 19, 20, 21, 22, 23, 24, 25

Offset: 1

Altug Alkan, Aug 12 2020

Conjecture I: Natural density of this sequence is 1.
Conjecture II: Any sufficiently large n is in the sequence.
Conjecture III: There is a fixed value of t such that all integers >= t are terms.
If k is of the form x^2 + y^3 + z^6 then so is k*m^6 = (x*m^3)^2 + (y*m^2)^3 + (z*m)^6. - David A. Corneth, Aug 13 2020

			6 is a term since 6! = 12^2 + 8^3 + 2^6.

David A. Corneth, PARI program

Cf. A267414, A273553 (subsequence).

\\ See Corneth link. David A. Corneth, Aug 13 2020

a(12)-a(18) from David A. Corneth, Aug 12 2020

A343491 Number of representations of n! as a sum of 3 tetrahedral numbers (A000292).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 1, 3, 5, 2, 3, 6, 5, 8, 8, 7, 2, 7, 8, 3, 11, 2, 2

Offset: 1

Altug Alkan, Apr 17 2021

Conjecture I: There are infinitely many n such that a(n) >= 1.
Conjecture II: Natural density of numbers n such that a(n) >= 1 is 1.
Conjecture III: Numbers n such that a(n) = 0 is a finite sequence.
Conjecture IV: a(n) >= 1 for all n.
See Links section for some solutions.

			a(4) = 2 because 4! = 0 + 4 + 20 = 4 + 10 + 10.
a(24) = 2 because 24! = f(11393630) + f(118661018) + f(127041924) = f(81298034) + f(61098204) + f(143537134) where f = A000292.

Altug Alkan, A note on sequence

Cf. A000142, A000292, A102796, A104246, A102799, A267414, A308852, A341794.

Table[Length[Solve[{i*(i + 1)*(i + 2) + j*(j + 1)*(j + 2) + k*(k + 1)*(k + 2) == 6*n!, i >= 0, j >= 0, k >= 0, i <= j, j <= k, k < (6*n!)^(1/3)}, Integers]], {n, 1, 10}] (* Vaclav Kotesovec, Apr 19 2021 *)

A333841 Integers n such that n! = x^2 + y^3 + z^4 where x, y and z are nonnegative integers, is soluble.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24

Offset: 1

Altug Alkan, Aug 14 2020

			6! = 11^2+7^3+4^4; 8! = 192^2+15^3+3^4;  9! = 443^2+55^3+4^4; 10! = 1888^2+40^3+4^4; 11! = 5896^2+172^3+16^4, so 6, 8, 9, 10 and 11 are in the sequence. - _R. J. Mathar_, Dec 15 2020

Cf. A123053, A267414, A337046.

{k: k! in A123053}. - R. J. Mathar, Dec 15 2020

A337366 Number of representations of A036691(n) as a sum of 3 nonnegative cubes.

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 1, 2, 1, 4, 6, 3, 8, 8, 14, 7

Offset: 0

Altug Alkan, Aug 25 2020

Conjecture I: a(n) = 0 only for n = 1. That is, any product of first n > 1 composite numbers is a sum of at most 3 positive cubes. For example,
A036691(100) = 2563573191821442299652988946477367093137353211904000000000^3 + 21431289850849406740917647451954098598503667204096000000000^3 + 26409890400237152457638095665189553529771293409280000000000^3.
Conjecture II: For any term t >= 1, there are only finitely many values of n such that a(n) = t.

			a(4) = 2 because A036691(4) = 1728 = 12^3 = 6^3 + 8^3 + 10^3.

Cf. A025447, A036691, A226955, A267414.

A036691 = Join[{1}, FoldList[Times, Select[Range[20], CompositeQ]]];
Table[Length@ PowersRepresentations[A036691[[n]], 3, 3], {n, 10}] (* Robert Price, Sep 08 2020 *)

a(n) = A025447(A036691(n)).

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A226955 Number of representations of n! as a sum of 3 positive cubes.

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A337046 Integers n such that n! = x^2 + y^3 + z^6 where x, y and z are nonnegative integers, is soluble.

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A343491 Number of representations of n! as a sum of 3 tetrahedral numbers (A000292).

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A333841 Integers n such that n! = x^2 + y^3 + z^4 where x, y and z are nonnegative integers, is soluble.

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A337366 Number of representations of A036691(n) as a sum of 3 nonnegative cubes.

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