A115676 -id:A115676 - cp's OEIS frontend

A115675 Brilliant numbers (A078972) whose digit reversal is a powerful(1) number (A001694).

4, 9, 10, 121, 169, 869, 961, 1273, 1843, 10201, 10609, 12769, 16171, 44521, 48361, 48613, 70597, 76121, 94249, 96721, 106009, 108853, 121879, 129307, 211591, 255953, 276491, 278699, 291149, 445051, 526567, 613927, 616571, 1026169

Offset: 1

Giovanni Resta, Jan 31 2006

			869 = 11*79 is brilliant and 968 = 2^3*11^2 is powerful.

Chai Wah Wu, Table of n, a(n) for n = 1..853

Cf. A001694, A078972, A115676.

A173518 Solutions z of the Diophantine equation x^3 + y^3 = 6z^3.

Original entry on oeis.org

21, 960540, 16418498901144294337512360, 436066841882071117095002459324085167366543342937477344818646196279385305441506861017701946929489111120

Offset: 1

Michel Lagneau, Feb 20 2010

nonn

A. Nitaj proved Erdős's conjecture (1975) and claimed that there exist infinitely many triples of 3-powerful numbers a,b,c with (a,b) = 1, such that a+b=c, because the equation x^3 + y^3 = 6z^3 admits an infinite number of solutions, and given by the recurrence equations (see formula). It is proved that a=x(k)^3, b=y(k)^3, and c=6c^3, and are 3-powerful numbers for each k >= 1.

			37^3 + 17^3 = 6*21^3.

J. M. De Koninck, Ces nombres qui nous fascinent, Ellipses, 2008, p. 348.
Mordell, L. J. (1969). Diophantine equations. Academic Press. ISBN 0-12-506250-8

P. Erdős, C. Pomerance, and A. Sárközy, On locally repeated values of certain arithmetic functions, II, Acta Math. Hungarica 49 (1987), pp. 251-259. [alternate link]
A. Nitaj, On a conjecture of Erdős on 3-powerful numbers, Bull. London Math. Soc. 27 (1995), no. 4, 317-318.
Wikipedia, Diophantine equation

Cf. A050240, A050241, A057521, A060859, A113839, A115645, A115651, A115676, A115686, A115687, A115689, A115691, A115693, A115695, A115697, A116064, A140172.

x0:=37:y0:=17:z0:=21: for p from 1 to 5 do: x1:=x0*(x0^3+ 2*y0^3):y1:=-y0*(2*x0^3+ y0^3):z1:=z0*(x0^3- y0^3): print(z1) : x0 :=x1 :y0 :=y1 :z0 :=z1 :od :

We generate the solutions (x(k),y(k),z(k)) from the initial solution x(0) = 37, y(0)=17, z(0)=21 x(k+1) = x(k)*(x(k)^3 + 2*y(k)^3) y(k+1) = -y(k)*(2*x(k)^3 + y(k)^3) z(k+1) = z(k)*(x(k)^3 - y(k)^3).

cp's OEIS Frontend

A115675 Brilliant numbers (A078972) whose digit reversal is a powerful(1) number (A001694).

Views

Author

Keywords

Examples

Links

Crossrefs

A173518 Solutions z of the Diophantine equation x^3 + y^3 = 6z^3.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Formula