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A077291 Second member of Diophantine pair (m,k) that satisfies 6*(m^2 + m) = k^2 + k: a(n) = k. 12
0, 3, 8, 35, 84, 351, 836, 3479, 8280, 34443, 81968, 340955, 811404, 3375111, 8032076, 33410159, 79509360, 330726483, 787061528, 3273854675, 7791105924, 32407820271, 77123997716, 320804348039, 763448871240, 3175635660123, 7557364714688, 31435552253195 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The corresponding m are given in A077288.
LINKS
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
FORMULA
Let b(n) be A077290. Then a(n) = (-1 + sqrt(8*b(n) + 1))/2.
G.f.: x*(x^3+3*x^2-5*x-3) / ((x-1)*(x^4-10*x^2+1)). - Colin Barker, Mar 09 2012
From Vladimir Pletser, Jul 26 2020: (Start)
a(n) = 10*a(n-2) - a(n-4) + 4 with a(-2)=-4, a(-1)=-1, a(0)=0, a(1)=3.
Can be defined for negative n by setting a(-n) = - a(n-1) - 1 for all n in Z.
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). (End)
EXAMPLE
b(3)=630 so a(3) = (-1 + sqrt(8*630 + 1))/2 = (-1 + sqrt(5041))/2 = (71 - 1)/2 = 35.
MAPLE
f := gfun:-rectoproc({a(-2) = -4, a(-1) = -1, a(0) = 0, a(1) = 3, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 26 2020
MATHEMATICA
LinearRecurrence[{1, 10, -10, -1, 1}, {0, 3, 8, 35, 84}, 30] (* Harvey P. Dale, Oct 11 2019 *)
PROG
(PARI) concat(0, Vec(x*(x^3+3*x^2-5*x-3)/((x-1)*(x^4-10*x^2+1)) + O(x^100))) \\ Colin Barker, May 15 2015
CROSSREFS
Sequence in context: A218154 A349968 A204451 * A192212 A148918 A304726
KEYWORD
easy,nonn
AUTHOR
Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002
STATUS
approved

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Last modified March 28 13:35 EDT 2024. Contains 371254 sequences. (Running on oeis4.)