A115693 -id:A115693 - cp's OEIS frontend

A115694 Cubes whose digit reversal is a powerful(1) number (A001694).

1, 8, 27, 343, 1000, 1331, 8000, 27000, 343000, 1000000, 1030301, 1331000, 1367631, 8000000, 27000000, 343000000, 1000000000, 1003003001, 1030301000, 1033364331, 1331000000, 1334633301, 1367631000, 8000000000, 10662526601, 27000000000, 343000000000, 1000000000000

Offset: 1

Giovanni Resta, Jan 31 2006

			27 = 3^3 and 72 = 2^3*3^2 is powerful.

Amiram Eldar, Table of n, a(n) for n = 1..265
Index entries for sequences related to powerful numbers.

Cf. A001694, A115693.

Select[Range[2300]^3,Min[FactorInteger[IntegerReverse[#]][[All,2]]]>1 || IntegerReverse[#]==1&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 08 2021 *)

lista(kmax) = {my(c); for(k = 1, kmax, c = k^3; if(ispowerful(fromdigits( Vecrev(digits(c)))), print1(c, ", ")));} \\ Amiram Eldar, Feb 24 2024

a(26)-a(28) from Amiram Eldar, Feb 24 2024

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).

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A115694 Cubes whose digit reversal is a powerful(1) number (A001694).

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

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