login
Numbers n such that the sum of the digits of Lucas(n) is n.
1

%I #36 Sep 07 2015 14:47:29

%S 1,29,43,115,124,198,431,719,979,1496

%N Numbers n such that the sum of the digits of Lucas(n) is n.

%C No more terms < 200000. - _Manfred Scheucher_, Jun 12 2015

%C No more terms < 5400000. - _Dana Jacobsen_, Sep 07 2015

%H Manfred Scheucher, <a href="/A020997/a020997_1.sage.txt">Sage Script</a>

%H David Terr, <a href="http://www.fq.math.ca/Scanned/34-4/terr.pdf">On the Sums of Digits of Fibonacci Numbers,</a> Fibonacci Quarterly 34, Aug. 1996, pp. 349-355.

%t Do[ If[ Mod[ 2Fibonacci[n + 1] - Fibonacci[n], 9] == Mod[n, 9], If[ Plus @@ IntegerDigits[ 2Fibonacci[n + 1] - Fibonacci[n]] == n, Print[n]]], {n, 0, 10^5}] (* _Robert G. Wilson v_ *)

%t Select[Range[0, 10^4], Plus@@IntegerDigits[LucasL[#]]==#&] (* _Vincenzo Librandi_, Feb 18 2015 *)

%o (PARI) isok(n) = sumdigits(fibonacci(n+1)+fibonacci(n-1)) == n; \\ _Michel Marcus_, Feb 18 2015

%o (Perl) use ntheory ":all"; for (1..1e5) { say if $_ == sumdigits(lucasv(1,-1,$_)) } # _Dana Jacobsen_, Sep 06 2015

%K nonn,base,more

%O 1,2

%A _Sven Simon_