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%I #17 Jan 02 2023 12:30:51
%S 0,24,75,99,100,124,175,199,200,224,275,299,300,324,375,399,400,424,
%T 475,499,500,524,575,599,600,624,675,699,700,724,775,799,800,824,875,
%U 899,900,924,975,999,1000,1024,1075,1099,1100,1124,1175,1199,1200,1224
%N a(n) = ((-1)^n - 1)/2 + 25*floor(3*n/2) - 50*floor(n/4).
%C Also: solutions to (2a+1)^2 = 1 mod 400. Occurs in the context of a problem concerning integer-valued percentages (see link): a(n) percent of a(n)+1 is an integer.
%H R. Israel, in reply to E. Angelini, <a href="http://list.seqfan.eu/oldermail/seqfan/2015-December/015798.html">Percentages</a>, SeqFan list, Dec 7, 2015.
%F a(n) = (A265423(n) - 1)/2.
%F G.f.: x*(24 + 51*x + 24*x^2 + x^3)/(1 - x - x^4 + x^5). - _Robert Israel_, Dec 08 2015
%t Table[((-1)^n - 1)/2 + 25 Floor[3 n/2] - 50 Floor[n/4], {n, 0, 50}] (* _Vincenzo Librandi_, Dec 09 2015 *)
%o (PARI) A265424(n)=((-1)^n-1)/2+n*3\2*25-n\4*50
%o (PARI) is_A265424(n)=Mod(n*2+1,400)^2==1
%o (Magma) [((-1)^n-1)/2+25*Floor(3*n/2)-50*Floor(n/4): n in [0..50]]; // _Vincenzo Librandi_, Dec 09 2015
%K nonn,easy
%O 0,2
%A _M. F. Hasler_, Dec 08 2015
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