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A054214 Numbers n such that n concatenated with n-1 is a square.

%I A054214 #31 Oct 17 2019 20:00:28
%S A054214 82,8242,9802,538277,998002,77837026,99980002,7922547265,8643251345,
%T A054214 9223797610,9999800002,106710893290,453378226757,491023832065,
%U A054214 945958034530,999998000002,11916002265170,15790977390245,24917378001937,25082758752026,36315251812570
%N A054214 Numbers n such that n concatenated with n-1 is a square.
%C A054214 Also, numbers k such that k concatenated with k-2 gives the product of two numbers which differ by 2.
%C A054214 Also, numbers n such that n concatenated with n-5 gives the product of two numbers which differ by 4.
%C A054214 Every term contains an even number of digits. - _Max Alekseyev_, May 14 2007
%C A054214 If n=(10^m-1)^2+1 where m is a positive integer then n is in the sequence. Because then n has 2m digits and n concatenated with n-1 is n*10^(2m)+(n-1) = (10^(2m)-10^m+1)^2. for example, taking m=1 we get 82, the first term of the sequence. - _Farideh Firoozbakht_, Aug 22 2013
%C A054214 As pointed out by _Georg Fischer_, it is very plausible that all of A054214, A116123, and A116142 contain exactly the same terms. If that is indeed true, then A054214 should be edited to mention the alternative constructions, and the other two sequences declared "dead". However, this needs careful analysis to deal with the possibilities that n, n-2, and n-5 may not all have the same number of digits. - _N. J. A. Sloane_, Oct 30 2018. Nov 05 2018: Thanks to Giovanni Resta_ (see below), this has now been done. - _N. J. A. Sloane_, Nov 05 2018
%C A054214 From _Giovanni Resta_, Nov 05 2018: (Start)
%C A054214 For n and n-5 to have a different digit length, we must have n = 10^k+h with 0<=h<=4.
%C A054214 We want to prove that in this case the concatenation of n and n-5 cannot be of the form m(m+4). The numbers m(m+4) modulo 9 can only be equal to 0, 3, 5, or 6, but it is easy to see that the concatenation of 10^k+h and 10^k+h-5 can be equal to one of these values modulo 9 only if h=0.
%C A054214 Now, the concatenation of 10^k and 10^k-5 is equal to 3 modulo 4 for every k>1, but m(m+4) modulo 4 can only be equal to 0 or 1, so A116123 is indeed equal to this sequence.
%C A054214 Using an identical argument (with mods 9 and 4) we can prove that the concatenation of n and n-2, when n and n-2 have a different number of digits, cannot be equal to m(m+2) and so A116142 is equal to this sequence. (End)
%D A054214 Luca, Florian, and Pantelimon Stănică. "Perfect Squares as Concatenation of Consecutive Integers." The American Mathematical Monthly 126.8 (2019): 728-734.
%H A054214 Giovanni Resta, <a href="/A054214/b054214.txt">Table of n, a(n) for n = 1..4751</a>
%e A054214 E.g. '8242' + '8242-1' gives 82428241 which is 9079^2.
%Y A054214 Cf. A054215, A054216, A030465, A030466, A030467, A020339, A020340.
%K A054214 nonn,base
%O A054214 1,1
%A A054214 _Patrick De Geest_, Feb 15 2000
%E A054214 More terms from _Max Alekseyev_, May 14 2007
%E A054214 a(20)-a(21) from _Giovanni Resta_, Nov 05 2018
%E A054214 Edited by _N. J. A. Sloane_, Nov 05 2018