|
%I #43 Sep 09 2018 12:07:01
%S 1,5,6,963,9867,65766,69714,6317056,90899553,169605719,4270981082,
%T 96528287587,465454256742,692153612536,182921919071841,
%U 655785969669834,650700037578750084,125631041500927357539,673774165549097456624,16719041449406813636569
%N Numbers n such that n^2 ends with the digits of n reversed (A004086(n)).
%C a(29)>10^32 (if it exists)
%H Robert Gerbicz, <a href="/A269588/b269588.txt">Table of n, a(n) for n = 1..28</a>
%H IBM, <a href="https://www.research.ibm.com/haifa/ponderthis/challenges/March2016.html">"Ponder This" puzzle for March 2016</a>
%e 6317056^2 = 39905196507136 which ends with 6507136, so 6317056 is a term.
%t Select[Range[10^7], Function[k, Take[IntegerDigits[#^2], -Length@ k] == Reverse@ k]@ IntegerDigits@ # &] (* _Michael De Vlieger_, Mar 04 2016 *)
%o (PARI) isA269588(n)=dn = digits(n); rn = subst(Polrev(dn), x, 10); nbd = #dn; (n^2 - rn) % 10^nbd == 0; \\ _Michel Marcus_, Mar 01 2016
%o (PARI) \\ printA269588len(d) prints all terms of the sequence with d digits
%o rev(n) = eval(concat(Vecrev(Str(n))));
%o { printA269588len(d) = my(l, u, n); l=ceil(d/2); u=floor(d/2); for(y=0, 10^l-1, n=rev(y^2 % 10^u)*10^l+y; if(#Str(n)==d && Mod(n, 10^d)^2==rev(n), print(n)); ); }
%o \\ _Max Alekseyev_, Mar 07 2016
%Y Subsequence of A115761.
%Y Cf. A003226, A004086.
%K nonn,base
%O 1,2
%A _José Eduardo Gaboardi de Carvalho_, Mar 01 2016
%E a(18)-a(20) from _Max Alekseyev_, Mar 07 2016
%E a(21)-a(27) from _Robert Gerbicz_, Apr 03 2016
%E a(28) from Dieter Beckerle, Jun 09 2016
| 