A260819 -id:A260819 - cp's OEIS frontend

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

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

A261116 Pairs of integers (a,b) such a^2 + (a+1)^2 = b^2.

Original entry on oeis.org

0, 1, 3, 5, 20, 29, 119, 169, 696, 985, 4059, 5741, 23660, 33461, 137903, 195025, 803760, 1136689, 4684659, 6625109, 27304196, 38613965, 159140519, 225058681, 927538920, 1311738121, 5406093003, 7645370045, 31509019100, 44560482149, 183648021599, 259717522849

Offset: 1

Marco Ripà, Aug 08 2015

Can also be seen as a table with two columns, read by rows: T[n,1] = a(2n-1) = A001652(n), T[n,2] = a(2n) = A001653(n).

The conjectured recurrence formula and g.f. are proved by the formulas for A001652. - M. F. Hasler, Aug 11 2015

			a(5) = 20 and a(6) = 29, because 20^2 + 21^2 = 29^2.

Colin Barker, Table of n, a(n) for n = 1..1000
V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 3 p. 8.
Index entries for linear recurrences with constant coefficients, signature (0,7,0,-7,0,1).

Cf. A001652, A001653, A260819.

LinearRecurrence[{0, 7, 0, -7, 0, 1}, {0, 1, 3, 5, 20, 29}, 50] (* Paolo Xausa, Jan 31 2024 *)

concat(0, Vec(-x^2*(x^4-x^3-2*x^2+3*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^50))) \\ Colin Barker, Aug 12 2015

From Colin Barker, Aug 09 2015: (Start)
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6) for n>6.
G.f.: -x^2*(x^4-x^3-2*x^2+3*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)).
(End)
a(2n-1)=A001652(n), a(2n)=A001653(n). - M. F. Hasler, Aug 11 2015

Edited by M. F. Hasler, Aug 11 2015

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

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