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A154138
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Indices k such that 3 plus the k-th triangular number is a perfect square.
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6
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1, 3, 12, 22, 73, 131, 428, 766, 2497, 4467, 14556, 26038, 84841, 151763, 494492, 884542, 2882113, 5155491, 16798188, 30048406, 97907017, 175134947, 570643916, 1020761278, 3325956481, 5949432723, 19385094972, 34675835062, 112984613353
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Also numbers n such that (ceiling(sqrt(n*(n+1)/2)))^2 - n*(n+1)/2 = 3. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Nov 10 2009]
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LINKS
| F. T. Adams-Watters, SeqFan Discussion, Oct 2009
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FORMULA
| {k: 3+k*(k+1)/2 in A000290}
Conjectures: a(n)= +a(n-1) +6*a(n-2) -6*a(n-3) -a(n-4) +a(n-5); also g.f.: x*(-1-2*x-3*x^2+2*x^3+2*x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [Comment from Zak Seidov (zakseidov(AT)yahoo.com), Oct 21 2009: I believe both of these conjectures are correct.]
a(1..4)=(1,3,12,22); a(n>4)=6*a(n-2)-a(n-4)+2. [From Zak Seidov (zakseidov(AT)yahoo.com), Oct 21 2009]
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EXAMPLE
| 1*(1+1)/2+3 = 2^2. 3*(3+1)/2+3 = 3^2. 12*(12+1)/2+3 = 9^2. 22*(22+1)/2+3 = 16^2.
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MATHEMATICA
| a[1]=1; a[2]=3; a[3]=12; a[4]=22; a[n_]:=a[n]=6*a[n-2]-a[n-4]+2; Table[a[n], {n, 35}] [From Zak Seidov (zakseidov(AT)yahoo.com), Oct 21 2009]
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CROSSREFS
| Cf. A000217, A000290, A006451.
Sequence in context: A063598 A063107 A015629 * A012469 A103449 A030571
Adjacent sequences: A154135 A154136 A154137 * A154139 A154140 A154141
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KEYWORD
| nonn
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AUTHOR
| R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 18 2009
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EXTENSIONS
| More terms from Zak Seidov (zakseidov(AT)yahoo.com), Oct 21 2009
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