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 A133141 Numbers which are both centered pentagonal and centered hexagonal numbers. 0
 1, 331, 159391, 76825981, 37029963301, 17848365484951, 8602875133782931, 4146567966117887641, 1998637156793688059881, 963338963006591526974851, 464327381532020322313818151, 223804834559470788763733373781 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The problem is to find p and r such that 6*(2*p-1)^2=5*(2*r+1)^2+1 equivalent to 3*p^2-3*p+1=0.5*(5*r^2+5*r+2). The Diophantine equation (6*X)^2=30*Y^2+6 is such that X is given by 1, 21, 461, 10121,... with a(n+2)=22*a(n+1)-a(n) and also a(n+1)=11*a(n)+(120*a(n)^2-20)^0.5 Y is given by 1, 23, 805, 11087,... with a(n+2)=22*a(n+1)-a(n) and also a(n+1)=11*a(n)+(120*a(n)^2+24)^0.5 r is given by 0,11, 252, 5543, 121704,... with a(n+2)=22*a(n+1)-a(n)+10 and also a(n+1)=11*a(n)+5+(120*a(n)^2+120*a(n)+36)^0.5 p is given by 1, 11, 231, 5061,... with a(n+2)=22*a(n+1)-a(n)-10 and also a(n+1)=11*a(n)-5+(120*a(n)^2-120*a(n)+25)^0.5 LINKS Index to sequences with linear recurrences with constant coefficients, signature (483,-483,1). FORMULA a(n+2)=482*a(n+1)-a(n)-150 a(n+1)=241*a(n)-75+11*(480*a(n)^2-300*a(n)+45)^0.5 G.f.: z*(1-152*z+z^2)/((1-z)*(1-482*z+z^2)) a(n)=(5/16)+(11/32)*[241+44*sqrt(30)]^n-(1/16)*sqrt(30)*[241-44*sqrt(30)]^n+(11/32)*[241-44 *sqrt(30)]^n+(1/16)*[241+44*sqrt(30)]^n*sqrt(30), with n>=0 [From Paolo P. Lava, Sep 26 2008] CROSSREFS Sequence in context: A060894 A002228 A199820 * A097401 A114084 A095199 Adjacent sequences:  A133138 A133139 A133140 * A133142 A133143 A133144 KEYWORD nonn AUTHOR Richard Choulet, Sep 21 2007 EXTENSIONS More terms from Paolo P. Lava, Sep 26 2008 STATUS approved

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Last modified May 24 06:29 EDT 2013. Contains 225617 sequences.