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A129444
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Numbers n such that centered triangular number A005448(n) = 3n(n-1)/2 + 1 is a perfect square.
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6
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0, 1, 2, 7, 16, 65, 154, 639, 1520, 6321, 15042, 62567, 148896, 619345, 1473914, 6130879, 14590240, 60689441, 144428482, 600763527, 1429694576, 5946945825, 14152517274, 58868694719, 140095478160, 582740001361, 1386802264322
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Corresponding numbers k>0 such that k^2 is a centered triangular number are listed in A129445(n) = {1, 2, 8, 19, 79, 188, 782, 1861, 7741, 18422, 76628, 182359, ...}.
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FORMULA
| a(n) = 1/2 + Sqrt[ 1/4 + 2/3*(A129445(n)^2 - 1) ].
a(1)=0; a(2)=1; a(3)=2; a(4)=7; a(5)=16; a(6)=65; a(n)=11(a(n-2)-a(n-4))+a(n-6). - Zak Seidov (zakseidov(AT)gmail.com), Apr 17 2007
a(-n) = 1 - a(n+3). - Michael Somos Apr 05 2008
G.f.: (x + x^2 - 5*x^3 - x^4) / ((1 - x) * (1 - 10*x^2 + x^4)). - Michael Somos Apr 05 2008
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MATHEMATICA
| Do[ f = 3n(n-1)/2 + 1; If[ IntegerQ[ Sqrt[f] ], Print[ n ] ], {n, 1, 150000} ]
a[1]=0; a[2]=1; a[3]=2; a[4]=7; a[5]=16; a[6]=65; a[n_]:=a[n]=11(a[n-2]-a[n-4])+a[n-6]; Table[a[n], {n, 100}] - Zak Seidov (zakseidov(AT)gmail.com), Apr 17 2007
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PROG
| (PARI) {a(n) = local(m); m = if( n<1, 2-n, n-1); (n<1) + (-1)^(n<1) * polcoeff( (x + x^2 - 5*x^3 - x^4) / ((1 - x) * (1 - 10*x^2 + x^4)) + x*O(x^m), m)} /* Michael Somos Apr 05 2008 */
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CROSSREFS
| Cf. A005448 = Centered triangular numbers: 3n(n-1)/2 + 1. Cf. A129445 = numbers k>0 such that k^2 is a centered triangular number.
Sequence in context: A084079 A042689 A073998 * A079815 A006883 A023269
Adjacent sequences: A129441 A129442 A129443 * A129445 A129446 A129447
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KEYWORD
| nonn
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AUTHOR
| Alexander Adamchuk (alex(AT)kolmogorov.com), Apr 15 2007
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EXTENSIONS
| More terms from Zak Seidov (zakseidov(AT)gmail.com), Apr 17 2007
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