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Length of list created by n substitutions k -> Range(-abs(k+1), abs(k-1)) starting with {1}.
3

%I #18 Nov 27 2022 10:43:12

%S 1,3,11,41,159,633,2575,10657,44735,190017,815231,3527681,15378687,

%T 67478401,297777407,1320753665,5884652543,26326301697,118211192831,

%U 532574203905,2406726828031,10906541371393

%N Length of list created by n substitutions k -> Range(-abs(k+1), abs(k-1)) starting with {1}.

%H Vincenzo Librandi, <a href="/A084077/b084077.txt">Table of n, a(n) for n = 0..200</a>

%F invOGF satisfies n - (1+3*n)*a(n) - 2*n*(1+n)*a(n)^2 - 2*n^2*a(n)^3 = 0. [Is it true?]

%F Recurrence: (n+3)*(7*n-4)*a(n) = 3*(7*n^2+3*n+2)*a(n-1) + 2*(28*n^2+5*n-9)*a(n-2) + 4*(n-1)*(7*n+3)*a(n-3). - _Vaclav Kotesovec_, Oct 14 2012

%F a(n) ~ sqrt(52+34*sqrt(2))*(2+2*sqrt(2))^n/(sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 14 2012

%e {1}, {-2,-1,0}, {-1,0,1,2,3,0,1,2,-1,0,1}

%t Length/@Flatten/@NestList[ # /. k_Integer:>Range[ -Abs[k+1], Abs[k-1]]&, {1}, 8]

%t Flatten[{1,RecurrenceTable[{(n+3)*(7*n-4)*a[n] == 3*(7*n^2+3*n+2)*a[n-1] + 2*(28*n^2+5*n-9)*a[n-2] + 4*(n-1)*(7*n+3)*a[n-3],a[1]==3,a[2]==11,a[3]==41},a,{n,20}]}] (* _Vaclav Kotesovec_, Oct 14 2012 *)

%o (Magma) I:=[1,3,11]; [n le 3 select I[n] else (3*(7*n^2 -11*n +6)*Self(n-1) + 2*(28*n^2 -51*n +14)*Self(n-2) + 4*(n-2)*(7*n-4)*Self(n-3))/((n+2)*(7*n-11)): n in [1..41]]; // _G. C. Greubel_, Nov 23 2022

%o (SageMath)

%o @CachedFunction

%o def a(n): # a = A084077

%o if (n<3): return (1,3,11)[n]

%o else: return (3*(7*n^2 +3*n +2)*a(n-1) + 2*(28*n^2 +5*n -9)*a(n-2) + 4*(n-1)*(7*n+3)*a(n-3))/((n+3)*(7*n-4))

%o [a(n) for n in range(31)] # _G. C. Greubel_, Nov 23 2022

%Y Cf. A084075, A084076, A084078.

%K nonn

%O 0,2

%A _Wouter Meeussen_, May 11 2003