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A077288
First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.
15
0, 1, 3, 14, 34, 143, 341, 1420, 3380, 14061, 33463, 139194, 331254, 1377883, 3279081, 13639640, 32459560, 135018521, 321316523, 1336545574, 3180705674, 13230437223, 31485740221, 130967826660, 311676696540, 1296447829381, 3085281225183, 12833510467154
OFFSET
0,3
COMMENTS
Also nonnegative m such that 24*m^2 + 24*m + 1 is a square. - Gerald McGarvey, Apr 02 2005
FORMULA
Let b(n) be A072256. Then a(2*n+2) = 2*a(2*n+1) - a(2*n) + b(n+1), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + b(n+2), with a(0)=0, a(1)=1.
G.f.: x*(1+x)^2/((1-x)*(1-10*x^2+x^4)).
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(n) = 10*a(n-2) - a(n-4) + 4, n > 4. - Vladimir Pletser, Feb 29 2020
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
2*a(n) + 1 = A080806(n+1). - R. J. Mathar, Oct 01 2021
EXAMPLE
a(3) = 2*3 - 1 + 9 = 14, a(4) = 2*14 - 3 + 9 = 34, etc.
G.f. = x + 3*x^2 + 14*x^3 + 34*x^4 + 143*x^5 + 341*x^6 + 1420*x^7 + 3380*x^8 + ... - Michael Somos, Jul 15 2018
MAPLE
f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 100)]); - Vladimir Pletser, Jul 24 2020
MATHEMATICA
CoefficientList[Series[x*(1 + x)^2/((1 - x)*(1 - 10 x^2 + x^4)), {x, 0, 40}], x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 3, 14, 34}, 50] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ x (1 + x)^2 / ((1 - x) (1 - 10 x^2 + x^4)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)
PROG
(PARI) my(x='x+O('x^30)); concat([0], Vec(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))) \\ G. C. Greubel, Jul 15 2018
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))); // G. C. Greubel, Jul 15 2018
CROSSREFS
The k values are in A077291
Cf. A053141.
Sequence in context: A081269 A140064 A064226 * A094627 A009394 A076533
KEYWORD
easy,nonn
AUTHOR
Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002
STATUS
approved