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A133216
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Integers that are simultaneously triangular (A000217) and decagonal (A001107).
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2
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0, 1, 10, 1540, 11935, 1777555, 13773376, 2051297326, 15894464365, 2367195337045, 18342198104230, 2731741367653000, 21166880717817451, 3152427171076225351, 24426562006163234620, 3637898223680596402450, 28188231388231654934425, 4198131397700237172202345
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OFFSET
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1,3
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COMMENTS
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Positive terms are of the form (m^2-9)/16 where m runs over the elements of A077443 that are congruent to 5 modulo 8. Correspondingly, for n>1, sqrt(16*a(n)+9) form a subsequence of A077443, while sqrt(8*a(n)+1) form a subsequence of A077442 with indices congruent to 2,3 modulo 4. [Max Alekseyev]
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LINKS
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FORMULA
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For n>5, a(n) = 1154*a(n-2) - a(n-4) + 396.
For n>6, a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/64 * ( (9 + 4* sqrt(2)*(-1)^n)*(1+sqrt(2))^(4*n-6) + (9 - 4* sqrt(2)*(-1)^n)*(1-sqrt(2))^(4*n-6) - 22).
a(n) = floor ( 1/64 * (9 + 4*sqrt(2)*(-1)^n) * (1+sqrt(2))^(4*n-6) ).
G.f.: (x^5 + 9*x^4 + 376*x^3 + 9*x^2 + x)/((1 - x)*(x^2 - 34*x + 1)*(x^2 + 34*x + 1)). [corrected by Peter Luschny, Apr 04 2019]
Lim (n -> Infinity, a(2n+1)/a(2n)) = (1/49)*(3649+2580*sqrt(2)).
Lim (n -> Infinity, a(2n)/a(2n-1)) = (1/49)*(193+132*sqrt(2)).
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EXAMPLE
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The initial terms of the sequences of triangular (A000217) and decagonal (A001107) numbers are 0, 1, 3, 6, 10, 15, ... and 0, 1, 10, 27, ... respectively. As the third number which is common to both sequences is 10, we have a(3) = 10.
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MATHEMATICA
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LinearRecurrence[{1, 1154, -1154, -1, 1} , {0, 1, 10, 1540, 11935, 1777555}, 17] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Term 0 prepended and entry revised accordingly by Max Alekseyev, Nov 06 2011
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STATUS
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approved
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