cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A275695 a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1; for n>6, a(n) = ( a(n-1)+a(n-3)+a(n-5) )*( a(n-2)+a(n-4)+a(n-6) ) / a(n-7).

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%I A275695 #54 Sep 08 2022 08:46:17
%S A275695 1,1,1,1,1,1,1,9,33,385,13825,5474305,75873853441,415386585427968001,
%T A275695 3501887406773528570406162401,
%U A275695 44079910680970588907541344275243042224979209,400942556117903539711475671972145122347091674105174721165559627509313
%N A275695 a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1; for n>6, a(n) = ( a(n-1)+a(n-3)+a(n-5) )*( a(n-2)+a(n-4)+a(n-6) ) / a(n-7).
%H A275695 Seiichi Manyama, <a href="/A275695/b275695.txt">Table of n, a(n) for n = 0..21</a>
%F A275695 a(n) = (8-4*(-1)^n)*a(n-1)*a(n-3)*a(n-5) - a(n-2) - a(n-4) - a(n-6).
%t A275695 RecurrenceTable[{a[n] == (a[n - 1] + a[n - 3] + a[n - 5]) (a[n - 2] + a[n - 4] + a[n - 6])/a[n - 7], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 0, 16}] (* _Michael De Vlieger_, Aug 25 2016 *)
%t A275695 nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,((g+e+c)(f+d+b))/a}; NestList[ nxt,{1,1,1,1,1,1,1},20][[All,1]] (* _Harvey P. Dale_, May 04 2019 *)
%o A275695 (PARI) a(n) = if (n <=6, 1, (a(n-1)+a(n-3)+a(n-5))*(a(n-2)+a(n-4)+a(n-6))/a(n-7)); \\ _Michel Marcus_, Aug 25 2016
%o A275695 (Ruby)
%o A275695 def A(m, n)
%o A275695   a = Array.new(2 * m + 1, 1)
%o A275695   ary = [1]
%o A275695   while ary.size < n + 1
%o A275695     i = (1..m).inject(0){|s, i| s + a[2 * i - 1]} * (1..m).inject(0){|s, i| s + a[2 * i]}
%o A275695     break if i % a[0] > 0
%o A275695     a = *a[1..-1], i / a[0]
%o A275695     ary << a[0]
%o A275695   end
%o A275695   ary
%o A275695 end
%o A275695 def A275695(n)
%o A275695   A(3, n)
%o A275695 end # _Seiichi Manyama_, Aug 27 2016
%o A275695 (Magma) I:=[1,1,1,1,1,1,1]; [n le 7 select I[n] else (Self(n-1) + Self(n-3) + Self(n-5))*(Self(n-2) + Self(n-4) + Self(n-6))/Self(n-7): n in [1..17]]; // _G. C. Greubel_, Feb 21 2018
%Y A275695 Cf. A006723, A276130.
%K A275695 nonn
%O A275695 0,8
%A A275695 _Bruno Langlois_, Aug 21 2016