A115695 -id:A115695 - cp's OEIS frontend

A115696 Pentagonal numbers (A000326) whose digit reversal is a powerful(1) number (A001694).

1, 925, 1426, 6112, 9801, 140301, 251945, 578151, 2112860, 78883630, 466005001, 52244922262, 394240923000, 527042773467, 706743866647, 797657661680, 869119510700, 1445531162626, 3905876132501, 5205252778462, 63253297234426, 135399021815552

Offset: 1

Giovanni Resta, Jan 31 2006

			925 is the 25th pentagonal number and 529=23^2 is powerful.

Cf. A000326, A001694, A115695.

a(19)-a(22) from Donovan Johnson, Jun 29 2011

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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A115696 Pentagonal numbers (A000326) 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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