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A103165
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Pentagonal 13-gonal numbers.
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0
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1, 70, 6469855, 314976376, 28913745325725, 1407627555984530, 129215354378188750651, 6290679101929235185300, 577462643423970343783639225, 28113007162445732840160102526, 2580677088685733707192642808088855, 125636860330920714709176528014392400
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OFFSET
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1,2
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LINKS
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FORMULA
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G.f.: h(z)=(z*(1+70*z+2000860*z^2+2146726*z^3+148995*z^4+19352060*z^5))/((1-z^2)*(1-4468994*z^2+z^4)).
The subsequences of the even-indexed and odd-indexed terms have the same recurrence relation : a(n+2)=4468994*a(n+1)-a(n)+2149856. The g.f. for the first is f(z)=(z*(1+2000860*z+148995*z^2))/((1-z)*(1-4468994*z+z^2)) and for the second g(z)=(z*(70+2146726*z+19352060*z^2))/((1-z)*(1-4468994*z+z^2)).
Therefore h(z)=(1/z)*f(z^2)+g(z^2).
The problem of pentagonal 13-gonal numbers is connected with the Diophantine equation 3*X^2=11*Y^2+232 where X=22*p-9 and Y=6*q-1 ; the parametrization of the conic gives the other equation 33*X^2=Z^2+2552 and the fact that the solutions fall into two sets.
The first uses the values 1, 1085 and the recurrence a(n+2)=2114*a(n+1)-a(n)-864 for p; the second uses 4,7568 and the recurrence a(n+2)=2114*a(n+1)-a(n)-352 for q.
X and Y satisfy the same recurrence a(n+2)=2114*a(n+1)-a(n).
a(n) = 4468995*a(n-2)-4468995*a(n-4)+a(n-6) for n>6. - Colin Barker, Oct 20 2014
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PROG
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(PARI) Vec((x*(1+70*x+2000860*x^2+2146726*x^3+148995*x^4+19352060*x^5))/((1-x^2)*(1-4468994*x^2+x^4)) + O(x^20)) \\ Colin Barker, Oct 20 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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