A269965 Start with a square; at each stage add a square at each expandable vertex so that the ratio of the side of the squares at stage n+1 and at stage n is the golden ratio phi=0.618...; a(n) is the number of squares in a portion of the n-th stage (see below).
1, 3, 10, 26, 63, 145, 332, 760, 1745, 4007, 9198, 21102, 48403, 111021, 254656, 584132, 1339893, 3073459, 7049906, 16171066, 37093175, 85084313, 195166404, 447672720, 1026871705, 2355438303, 5402904310, 12393181766, 28427480091, 65206953349, 149571708488
Offset: 5
Links
- Colin Barker, Table of n, a(n) for n = 5..1000
- Paolo Franchi, Illustration of initial terms
- Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,4,0,-2).
Programs
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Mathematica
RecurrenceTable[{a[n + 1] == 2 a[n] + a[n - 1] - 2 a[n - 2] + 2 a[n - 3] + 2 a[n - 4] + 5, a[5] == 1, a[6] == 3, a[7] == 10, a[8] == 26, a[9] == 63}, a, {n, 5, 30}] RecurrenceTable[{a[n + 1] == 3 a[n] - a[n - 1] - 3 a[n - 2] + 4 a[n - 3] - 2 a[n - 5], a[5] == 1, a[6] == 3, a[7] == 10, a[8] == 26, a[9] == 63, a[10] == 145}, a, {n, 5, 30}]
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PARI
Vec(x^5*(1+2*x^2+2*x^3)/((1-x)*(1+x)*(1-3*x+2*x^2-2*x^4)) + O(x^50)) \\ Colin Barker, Mar 09 2016
Formula
a(1)=a(2)=a(3)=a(4)=0, for n>= 5, a(n) = A269963(n-4)+a(n-1).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) + 2*a(n-4) + 2*a(n-5) + 5.
a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + 4*a(n-4) - 2*a(n-6).
G.f.: x^5*(1+2*x^2+2*x^3) / ((1-x)*(1+x)*(1-3*x+2*x^2-2*x^4)). - Colin Barker, Mar 09 2016
Comments