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Search: id:A103625
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| A103625 |
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Define a(1)=0, a(2)=0, a(3)=2, a(4)=4, a(5)=34, a(6)=62, a(7)=480, a(8)=870 such that from i=1 to 8 : 48*(a(i)^2)+48*a(i)+1=j(i)^2 with j(1)=1, j(2)=1, j(3)=17, j(4)=31, j(5)=239, j(6)=433, j(7)=3329, j(8)=6031 Then a(n)=a(n-8)+28*sqrt(48*(a(n-4)^2)+48*a(n-4)+1). |
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| 0, 0, 2, 4, 34, 62, 480, 870, 6692, 12124, 93214, 168872, 1298310, 2352090, 18083132, 32760394
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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j(n)=sqrt(48*(a(n)^2)+48*a(n)+1) j(n) is prime for n=3, 4, 5, 6, 7, 25, 28, 32, 35, 48, 65, 66, 88, 96, 113, 119, 151, 155, 182, 220, 231, 316, 488, 531, 599, 722, 1049, 1176
For n>1, first member of the Diophantine pair (m,k) that solves 12(m^2+m)=k^2+k; a(n)=m. - Herbert Kociemba (kociemba(AT)t-online.de), May 12 2008
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FORMULA
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G.f.: 2(x^2+x+1)/(1-x-14x^2+14x^3+x^4-x^5). - Ralf Stephan, May 18 2007
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MATHEMATICA
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a[1]=0; a[2]=0; a[3]=2; a[n_]:=(6+14a[n-2]+2Sqrt[1+48a[n-2]+48a[n-2]^2])/2; Table[a[i], {i, 1, 20}] - Herbert Kociemba (kociemba(AT)t-online.de), May 12 2008
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CROSSREFS
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Cf. A053141, A103200, A103737.
Sequence in context: A114642 A099433 A051225 this_sequence A006989 A132529 A140984
Adjacent sequences: A103622 A103623 A103624 this_sequence A103626 A103627 A103628
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KEYWORD
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uned,nonn
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AUTHOR
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Pierre CAMI (pierrecami(AT)tele2.fr), Mar 29 2005
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