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A154140
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Indices k such that 6 plus the k-th triangular number is a perfect square.
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4
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2, 4, 19, 29, 114, 172, 667, 1005, 3890, 5860, 22675, 34157, 132162, 199084, 770299, 1160349, 4489634, 6763012, 26167507, 39417725, 152515410, 229743340, 888924955, 1339042317
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| a(1..4)=(2,4,19,29); a(n>4)=6*a(n-2)-a(n-4)+2. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Nov 10 2009]
a(2n-1)=A001652(n)-A001653(n-1); a(2n)=A001652(n-2)+A001653(n+1).
In general, indices k such that A001109(2j) plus the k-th triangular
number is a perfect square may be found as follows:
b(2n-1)=A001652(n+j-1)-A001653(n-j);
b(2n)=A001652(n-j-1)+A001653(n+j);
indices k such that A001109(2j-1) plus the k-th triangular number is
a perfect square may be found as follows:
b(2n-1)=A001652(n+j-1)-A001653(n-j+1);
b(2n)=A001652(n-j)+A001653(n+j). - Charlie Marion, Mar 10 2011
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LINKS
| F. T. Adams-Watters, SeqFan Discussion, Oct 2009
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FORMULA
| {k: 6+k*(k+1)/2 in A000290}
Conjecture: a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5).
Conjecture: G.f.: x*(-2-2*x-3*x^2+2*x^3+3*x^4)/((x-1)* (x^2-2*x-1)* (x^2+2*x-1)) = =(6+(-1-4*x)/(x^2+2*x-1)+(6+13*x)/(x^2-2*x-1)+1/(x-1))/2.
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EXAMPLE
| 2*(2+1)/2+6 = 3^2. 4*(4+1)/2+6 = 4^2. 19*(19+1)/2+6 = 14^2. 29*(29+1)/2+6 = 21^2.
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CROSSREFS
| Cf. A000217, A000290, A006451.
Sequence in context: A171735 A201379 A056727 * A131578 A018273 A047092
Adjacent sequences: A154137 A154138 A154139 * A154141 A154142 A154143
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KEYWORD
| nonn,more
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AUTHOR
| R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 18 2009
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EXTENSIONS
| a(17)-a(24) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Nov 01 2010
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