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a(n) = (a(n-1) * a(n-6) + 2 * a(n-3) * a(n-4)) / a(n-7). a(1) = ... = a(7) = 1. Somos-7 variation.
1

%I #35 Feb 19 2024 10:33:53

%S 1,1,1,1,1,1,1,3,5,7,13,43,113,521,1241,3681,23657,177721,679505,

%T 4674203,27273277,275517767,3496390229,37043734803,226196947873,

%U 4391322667601,81041508965617,1433151398896001,25397505914206225,472652420405241521,9156799134584424289,499597377081528480243

%N a(n) = (a(n-1) * a(n-6) + 2 * a(n-3) * a(n-4)) / a(n-7). a(1) = ... = a(7) = 1. Somos-7 variation.

%C In general, suppose a(n)*a(n-7) = c1*a(n-1)*a(n-6) + c2*a(n-3)*a(n-4) for all n and constants c1,c2. Define u(n) = a(n)*a(n+5)/(a(n+2)*a(n+3)) which satisfies the generalized Lyness recursion u(n) = (c1*u(n-1) + c2)/u(n-2) for all n. For this sequence c1=1, c2=2, u(n) is (1, 1, 3, 5, 7/3, 13/15, 43/35, ...) and satisfies u(n) = (u(n-1) + 2)/u(n-2). See A076839 for Lyness references. - _Michael Somos_, Sep 26 2022

%H Harry J. Smith, <a href="/A064268/b064268.txt">Table of n, a(n) for n = 1..100</a>

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%F a(8-n) = a(n).

%t RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==a[5]==a[6]==a[7]==1,a[n] == (a[n-1]a[n-6]+2a[n-3]a[n-4])/a[n-7]},a,{n,30}] (* _Harvey P. Dale_, Nov 26 2015 *)

%o (PARI) {a(n) = if( n<1, a(8-n), if( n<8, 1, (a(n-1) * a(n-6) + 2 * a(n-3) * a(n-4)) / a(n-7)))};

%o (PARI) { a7=a6=a5=a4=a3=a2=a1=a=1; for (n=1, 100, if (n>7, a=(a1*a6 + 2*a3*a4)/a7; a7=a6; a6=a5; a5=a4; a4=a3; a3=a2; a2=a1; a1=a); write("b064268.txt", n, " ", a) ) } \\ _Harry J. Smith_, Sep 10 2009

%o (Magma) I:=[1,1,1,1,1,1,1]; [n le 7 select I[n] else (Self(n-1)*Self(n-6) + 2*Self(n-3)*Self(n-4))/Self(n-7): n in [1..30]]; // _G. C. Greubel_, Feb 21 2018

%Y Cf. A006723, A076839.

%K nonn,easy

%O 1,8

%A _Michael Somos_, Sep 24 2001