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A046194
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Heptagonal triangular numbers.
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3
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1, 55, 121771, 5720653, 12625478965, 593128762435, 1309034909945503, 61496776341083161, 135723357520344181225, 6376108764003055554511, 14072069153115290487843091, 661087708807868029661744485
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OFFSET
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1,2
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COMMENTS
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From Ant King, Oct 18 2011: (Start)
lim(n->Infinity, u(2n+1)/u(2n)) = 1/2(2207+987*sqrt(5)),
lim(n->Infinity, u(2n)/u(2n-1)) = 1/2(47+21*sqrt(5)).
(End)
From Raphie Frank, Nov 30 2012: (Start)
Where L_n is a Lucas number and F_n is Fibonacci number:
lim(n->Infinity, u(2n+1)/u(2n)) = 1/2(L_16+F_16*sqrt(5)),
lim(n->Infinity, u(2n)/u(2n-1)) = 1/2(L_8+F_8*sqrt(5)),
a(n) = L_1*a(n-1)+ L_24*a(n-2) - L_24*a(n-3)- L_1*a(n-4) + L_1*a(n-5).
(End)
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LINKS
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Table of n, a(n) for n=1..12.
Eric Weisstein's World of Mathematics, Heptagonal Triangular Number
Index to sequences with linear recurrences with constant coefficients, signature (1,103682,-103682,-1,1).
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FORMULA
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The two bisections satisfy the same recurrence relation: a(n+2)=103682*a(n+1)-a(n)+18144 or a(n+1)=51841*a(n)+9072+2898*(320*a(n)^2+112*a(n)+9)^0.5. The g.f. satisfies f(z)=(z+55*z^2+18088*z^3+18088*z^4+55*z^5+z^6)/((1-z^2)*(1-103682*z^2+z^4)=1*z+55*z^2+121771*z^3+... - Richard Choulet, Sep 20 2007
From Ant King, Oct 18 2011: (Start)
a(n) = a(n-1)+103682a(n-2)-103682a(n-3)-a(n-4)+a(n-5)
a(n) = 1/80*((3-sqrt(5)*(-1)^n)*(2+sqrt(5))^(4n-2)+(3+sqrt(5)*(-1)^n)*(2-sqrt(5))^(4n-2)-14)
a(n) = floor(1/80*(3-sqrt(5)*(-1)^n)*(2+sqrt(5))^(4n-2))
G.f.: x(1+54*x+18034*x^2+54*x^3+x^4)/((1-x)(1-322*x+x^2)(1+322*x+x^2))
(End)
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MATHEMATICA
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LinearRecurrence[{1, 103682, -103682, -1, 1}, {1, 55, 121771, 5720653, 12625478965}, 12] (* Ant King, Oct 18 2011 *)
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PROG
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(PARI) a(n)=((3-sqrt(5)*(-1)^n)*(2+sqrt(5))^(4*n-2)+(3+sqrt(5)*(-1)^n)*(2-sqrt(5))^(4*n-2)-14)\/80 \\ Charles R Greathouse IV, Oct 18 2011
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CROSSREFS
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Cf. A039835, A046193.
Sequence in context: A221000 A196428 A027580 * A172808 A172856 A093255
Adjacent sequences: A046191 A046192 A046193 * A046195 A046196 A046197
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KEYWORD
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nonn,easy,changed
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AUTHOR
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Eric W. Weisstein
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STATUS
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approved
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