%N Numbers n such that n and n+1 are a pair of consecutive powerful numbers.
%C "Erdős conjectured in 1975 that there do not exist three consecutive powerful integers." - Guy
%C See Guy for Erdős' conjecture and statement that this sequence is infinite. - _Jud McCranie_, Oct 13 2002
%C It is easy to see that this sequence is infinite: if n is in the sequence, so is 4*n*(n+1). - _Franklin T. Adams-Watters_, Sep 16 2009
%C The first of a run of three consecutive powerful numbers (conjectured to be empty) are just those in this sequence and A076445. - _Charles R Greathouse IV_, Nov 16 2012
%C Jaroslaw Wroblewski (see prime puzzles link) shows that there are infinitely many terms in this sequence such that neither a(n) nor a(n+1) is a square. - _Charles R Greathouse IV_, Nov 19 2012
%C Paul Erdős wrote of meeting Kurt Mahler: "I almost immediately posed him the following problem: ... are there infinitely many consecutive powerful numbers? Mahler immediately answered: Trivially, yes! x^2 - 8y^2 = 1 has infinitely many solutions. I was a bit crestfallen since I felt that I should have thought of this myself." - _Jonathan Sondow_, Feb 08 2015
%D J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 288, pp 74, Ellipses, Paris 2008.
%D R. K. Guy, Unsolved Problems in Number Theory, B16
%H Donovan Johnson, <a href="/A060355/b060355.txt">Table of n, a(n) for n = 1..39</a> (terms < 10^22)
%H C. K. Caldwell, <a href="http://primes.utm.edu/glossary/page.php?sort=PowerfulNumber">Powerful Numbers</a>
%H P. Erdős, <a href="http://www.renyi.hu/~p_erdos/1989-34.pdf">Some personal and mathematical reminiscences of Kurt Mahler</a>, Austral. Math. Soc. Gaz., 16 (1) (1989), 1-2.
%H J. J. O'Connor and E. F. Robertson, <a href="http://www-history.mcs.st-and.ac.uk/Biographies/Mahler.html">Biography of Kurt Mahler</a>
%H C. Rivera, <a href="http://www.primepuzzles.net/problems/prob_053.htm">Problem 53. Powerful numbers revisited</a>, Prime Puzzles
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PowerfulNumber.html">Powerful numbers</a>
%e 1825200 belongs to the sequence because 1825200 = 220.127.116.11.18.104.22.168.5.13.13, 1825201 = 22.214.171.124=1351^2 and both are powerful numbers. - _Labos Elemer_, May 03 2001
%t f[n_]:=First[Union[Last/@FactorInteger[n]]];Select[Range,f[#]>1&&f[#+1]>1&] (* _Vladimir Joseph Stephan Orlovsky_, Jan 29 2012 *)
%o (PARI) is(n)=ispowerful(n)&&ispowerful(n+1) \\ _Charles R Greathouse IV_, Nov 16 2012
%o a060355 n = a060355_list !! (n-1)
%o a060355_list = map a001694 $ filter ((== 1) . a076446) [1..]
%o -- _Reinhard Zumkeller_, Jun 03 2015, Nov 30 2012
%o def A060355(n):
%o a = sloane.A001694
%o return a.is_powerful(n) and a.is_powerful(n+1)
%o [n for n in (1..333333) if A060355(n)] # _Peter Luschny_, Feb 08 2015
%Y Primitive elements are in A199801.
%Y Cf. A001694, A060859.
%Y Cf. A076446 (first differences of A001694).
%A Jason Earls (zevi_35711(AT)yahoo.com), Apr 01 2001
%E Corrected and extended by _Jud McCranie_, Jul 08 2001
%E More terms from _Jud McCranie_, Oct 13 2002
%E a(22)-a(23) from _Donovan Johnson_, Jul 29 2011