A230400 -id:A230400 - cp's OEIS frontend

A369951 Volumes of integer-sided cuboids in which either the surface area divides the volume or vice versa (assuming dimensionless unit of length).

1, 2, 4, 8, 16, 18, 27, 32, 36, 216, 250, 256, 288, 400, 432, 450, 486, 576, 882, 1728, 1800, 1944, 2000, 2048, 2304, 2744, 2916, 3200, 3456, 3528, 3600, 3888, 4608, 6144, 6174, 6750, 6912, 7056, 7200, 7350, 7776, 7986, 8000, 8100, 8232, 9000, 9216, 9600, 9800

Offset: 1

Felix Huber, Feb 12 2024

For n <= 9, the surface area divides the volume. The 9 triples with the edge lengths (u,v,w) are (1,1,1), (2,1,1), (2,2,1), (2,2,2), (4,4,1), (6,3,1), (3,3,3), (4,4,2), (6,3,2).
For 10 <= n <= 19 the surfaces and volumes are equal. This is sequence A230400.
For n >= 20 the volume divides the surface area.

			a(9) = 36, because V = 6*3*2 = 36 and S = 2*(6*3+3*2+6*2) = 72 and S/V = 2.
a(12) = 256, because V = 8*8*4 = 256 and S = 2*(8*8+8*4+8*4) = 256 and S=V.
a(20) = 1728, because V = 12*12*12 = 1728 and S = 6*12*12 = 864 and V/S = 2.

Cf. A230400 (subsequence), A066955.

A369951 := proc(V) local a, b, c, k; for a from ceil(V^(1/3)) to V do if V/a = floor(V/a) then for b from ceil(sqrt(V/a)) to floor(V/a) do c := V/(a*b); if c = floor(c) then k := 2*(a*b + c*b + a*c)/(a*b*c); if k = floor(k) or 1/k = floor(1/k) then return V; end if; end if; end do; end if; end do; end proc; seq(A369951(V), V = 1 .. 10000);

For 10 <= n <= 19, a(n) = A230400(n - 9).

A229941 Sequence of triples: the 10 solutions of 1/p + 1/q + 1/r = 1/2 with 0 < p <= q <= r, lexicographically sorted.

Original entry on oeis.org

3, 7, 42, 3, 8, 24, 3, 9, 18, 3, 10, 15, 3, 12, 12, 4, 5, 20, 4, 6, 12, 4, 8, 8, 5, 5, 10, 6, 6, 6

Offset: 1

Jean-François Alcover, Oct 04 2013

As noted by John Baez, "each of [the 10 solutions of 1/p + 1/q + 1/r = 1/2] gives a way for three regular polygons to snugly meet at a point".
Among the 14 4-term Egyptian fractions with unit sum, there are 10 of the form 1/2 + 1/p + 1/q + 1/r.
Also integer values of length, width and height of a rectangular prism whose surface area is equal to its volume: pqr = 2(pq+pr+qr). - John Rafael M. Antalan, Jul 05 2015

			a(1) = 3, a(2) = 7, a(3) = 42, since 1/3 + 1/7 + 1/42 = 1/2.
The 10 solutions are:
3,  7, 42;
3,  8, 24;
3,  9, 18;
3, 10, 15;
3, 12, 12;
4,  5, 20;
4,  6, 12;
4,  8,  8;
5,  5, 10;
6,  6,  6

John Baez, The answer is 42.
J. F. T. Rabago and R. P. Tagle, On the Area and Volume of a certain Rectangular Solid and the Diophantine Equation 1/2=1/x+1/y+1/z, Notes on Number Theory and Discrete Mathematics, 19-3 (2013), 28-32.
Wikipedia, Hurwitz's automorphisms theorem.

Cf. A230400, A260819.

{p, q, r} /. {ToRules[Reduce[0 < p <= q <= r && 1/p + 1/q + 1/r == 1/2, {p, q, r}, Integers] ]} // Flatten

cp's OEIS Frontend

A369951 Volumes of integer-sided cuboids in which either the surface area divides the volume or vice versa (assuming dimensionless unit of length).

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