Jon E. Schoenfield: I already did. :-)

Burak Muslu: Thanks :)

Jon E. Schoenfield: Given that, for all primes p > 5, p^2 exceeds a multiple of 120 by either 1 or 49, it's not too hard to show that all terms here other than a(n) == 29 (mod 60) for all terms other than a(1).

Jon E. Schoenfield: Um ... let me try that again.  :-/

Jon E. Schoenfield: Given that, for all primes p > 5, p^2 exceeds a multiple of 120 by either 1 or 49, it's not too hard to show that a(n) == 29 (mod 60) for all terms other than a(1).

Burak Muslu: i missed your help Mr. Schoenfield. Shall I add the last part you wrote to the continuation of the comment section that you made the correction?

Jon E. Schoenfield: I thought maybe you'd like to prove that a(n) = 29 (mod 60), and then -- if you wanted to -- add another line to the Comments section stating and proving that result. :-)  But if you'd rather not, that's fine.  I just noticed that all the listed terms other than 17 had 9 as their last digit, and wondered why that was, and then found the result about a(n) mod 60.

Burak Muslu: Thanks Mr. Schoenfield. I understand. In the following sequence((p^2+25)/2 ), the last digits are 3 and 7, maybe I'll add a general sentence about all of them later. I didn't add all the sequences at the same time, because it might not be suitable for Oeis. However, in parallel with your explanation, I will consider a general statement.

Jon E. Schoenfield: I meant like this. :-)

Burak Muslu: Hi Mr. Schoenfield. Thank you also for your help in the comment section. It's ok for me. If you want, I can press the ready button again.

Jon E. Schoenfield: Does this look okay?

Burak Muslu: Let the comments section be as you, the editors, see fit. If I remember correctly, Fermat had a theorem that the sum of two squares equals 1(mod4).

David A. Corneth: Odd squares are 1 (mod 4) as they are squares of odd numbers. Odd numbers are of the form 2k + 1 and so they are squared (2k + 1)^2 = 4*k^2 + 4k + 1 = 4(k^2 + k) + 1 == 1 (mod 4). For (p^2 + 3)/2 to be an integer, p needs to be odd. As p^2 == 1 (mod 4), p^2 + 9 == 2 (mod 4) and so (p^2 + 9)/2 == 1 (mod 4).

Burak Muslu: Thanks for your helps. If I delete the other two sentences in the comment section and add the proof you wrote to the continuation of the second sentence, is it appropriate for you?

David A. Corneth: I'd only state terms are 1 (mod 4) with proof, if any. The rest is clear from the name or numerology I think.

David A. Corneth: Do we need 3 versions of the name with just small rewordings as comments?

David A. Corneth: this seems quite tedious to do manually, not to speak of the risk of errors.

Burak Muslu: I may have used unnecessary and excessive sentences in the comment section. Honestly, I need your help on that part Mr. Corneth. I can use wolfram. either I couldn't do it or the program didn't find it. Then i tried manually. Sorry for mistakes.