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%I #19 Jul 04 2019 11:37:06
%S 1,1,1,1,16,361,143641,20741472361,26888415586959536281,
%T 2002733778095476250641191709976062096,
%U 27923382501685315585533445603599269911720565853675615809277429923281
%N a(1) = a(2) = a(3) = a(4) = 1; for n>4, a(n) = ( a(n-1)+a(n-2)+a(n-3)+1 )^2 / a(n-4).
%C All terms are perfect squares.
%C The next term (a(12)) has 125 digits. - _Harvey P. Dale_, Jul 04 2019
%H Seiichi Manyama, <a href="/A276257/b276257.txt">Table of n, a(n) for n = 1..15</a>
%F a(n) = A276259(n)^2
%F a(n) = 25*a(n-1)*a(n-2)*a(n-3) - 2*a(n-1) - 2*a(n-2) - 2*a(n-3) - 2 - a(n-4).
%F a(n)*a(n-1)*a(n-2)*a(n-3) = ((a(n) + a(n-1) + a(n-2) + a(n-3) + 1)/5)^2.
%t RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1,a[n]==(a[n-1]+a[n-2]+ a[n-3]+ 1)^2/a[n-4]},a,{n,11}] (* _Harvey P. Dale_, Jul 04 2019 *)
%Y Cf. A276097, A276256, A276259.
%K nonn
%O 1,5
%A _Seiichi Manyama_, Aug 25 2016
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