cp's OEIS Frontend

Also, numbers k such that k concatenated with k-2 gives the product of two numbers which differ by 2.

Also, numbers n such that n concatenated with n-5 gives the product of two numbers which differ by 4.

Every term contains an even number of digits. - Max Alekseyev, May 14 2007

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

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

From Giovanni Resta, Nov 05 2018: (Start)

For n and n-5 to have a different digit length, we must have n = 10^k+h with 0<=h<=4.

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.

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.

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)

Also called Sastry numbers. - Lekraj Beedassy, Jul 18 2008

All numbers of the form f(n)=9(n).5.0(2n).5.0(n-1).1 where n>0 are in the sequence because if k(n)=9(n).0(n).25.0(n-1).9(n).6 then f(n)^2=k(n).(k(n)+5). For example f(2)=9950000501; k(2)=9900250996 and f(2)^2=9950000501^2=9900250996.9900251001 =k(2).(k(2)+5). - Farideh Firoozbakht, Nov 26 2006

m^2 = (k)|(k+5) = (k)|(k) + 5 = (10^q + 1)*k + 5 where | denotes concatenation and q is the number of digits of k gives a nonlinear equation that can be solved using the solver below. - David A. Corneth, Jan 02 2021

From Farideh Firoozbakht, Nov 26 2006: (Start)

1. All numbers of the form f(n)=3(n).4.6(n).5.3(n+1) are in the sequence because if k(n)=1(n).2.0(n+1).8(n).5 then f(n)^2= k(n).(k(n)+4). For example f(3)=333466653333; k(3)=111200008885 and f(3)^2=333466653333^2=k(3).(k(3)+4)=111200008885.111200008889.

2. All numbers of the form g(n)=6(n).5.3(n).4.6(n).8 are in the sequence because g(0)=548 is in the sequence(548^2=300.304) and for n>0 if h(n)=4(n).2.6(n-1).70.2(n).0 then g(n)^2=h(n).(h(n)+4). For example g(5)=666665333334666668; h(5)=444442666670222220 and g(5)^2=h(5).(h(5)+4)=444442666670222220.444442666670222224. (End)