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Search: id:A077398
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| A077398 |
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First member of the Diophantine pair (m,k) that solves 7(m^2+m)=k^2+k; a(n)=m. |
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+0
6
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| 0, 2, 5, 39, 87, 629, 1394, 10032, 22224, 159890, 354197, 2548215, 5644935, 40611557, 89964770, 647236704, 1433791392, 10315175714, 22850697509, 164395574727, 364177368759, 2620014019925, 5803987202642, 41755828744080
(list; graph; listen)
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OFFSET
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0,2
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FORMULA
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G.f.: x(2+3x+2x^2)/((1-x)(1-16x^2+x^4)).
a(n)=16*a(n-2)+a(n-4)+7, n>=3.
Let b(n) be A077397 then a(n+2)=2*a(n+1)-a(n)+b(n) with a(0)=0 a(1)=2
a(0)=0, a(1)=2, a(n+2)=(7+16a(n)+3Sqrt[1+28a(n)+28a(n)^2])/2 - Herbert Kociemba (kociemba(AT)t-online.de), May 12 2008
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PROGRAM
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(PARI) a(n)=if(n<0, 0, polcoeff(x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4))+x*O(x^n), n))
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CROSSREFS
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Cf. A077397, A077399, A077400. The k values are in A077401.
Cf. A053141.
Sequence in context: A036780 A051501 A135378 this_sequence A067083 A109003 A120031
Adjacent sequences: A077395 A077396 A077397 this_sequence A077399 A077400 A077401
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KEYWORD
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easy,nonn
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AUTHOR
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Bruce Corrigan (scentman(AT)myfamily.com), Nov 05 2002
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