A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers.

1, 1926, 850137, 2564464982, 1132138928657, 3415133918621062, 1507685261236261801, 4547981651299964079126, 2007805569980855008712097, 6056610836775865229750164742, 2673822786819976661810784866297, 8065673443881586606920210924732502

Offset

1,2

Comments

We solve the equation P(p) = P(r) + P(r+1) with unknowns p and r, equivalent to (6*p-1)^2 = 2*(6*r+2)^2 + 17. The Diophantine equation X^2 = 2*Y^2 + 17 whose solutions give p and r are obtained by (x(n), y(n)) such that:

x(1)=5, x(2)=215, x(3)=4517, x(4)=248087 and the same recurrence relation on the odd and even indices x(n+2) = 1154*x(n+1) - x(n);

y(1)=2, y(2)=152, y(3)=3194, y(4)=175424 and the same recurrence relation on the odd and even indices y(n+2) = 1154*y(n+1) - y(n).

The solutions (p,r) are given by the (u(n),v(n)) such that

u(1)=1, u(2)=36, u(3)=753, u(4)=41348 and the same recurrence relation on the odd and even indices u(n+2) = 1154*u(n+1) - u(n) - 192 or u(n+1) = 577*u(n) - 96 + 68*sqrt(72*u(n)^2 - 24*u(n) - 32);

v(1)=0, v(2)=25, v(3)=532, v(4)=29237 and the same recurrence relation on the odd and even indices v(n+2) = 1154*v(n+1) - v(n) + 384 or v(n+1) = 577*v(n) + 192 + 68*sqrt(72*u(n)^2 + 48*u(n) + 15).

Links

Colin Barker, Table of n, a(n) for n = 1..300

Index entries for linear recurrences with constant coefficients, signature (1,1331714,-1331714,-1,1).

Formula

For odd and even indices respectively: a(n+2) = 1331714*a(n+1) - a(n) - 416160; on the odd and the even indices respectively we also have a(n+1) = 665857*a(n) - 208080 + 19618*sqrt(1152*a(n)^2 - 720*a(n) - 32).

The g.f., h(z), is such that h(z) = a(1)*z + a(2)*z^2 + ... and is given by h(z) = z*(1 + 1925*z - 483503*z^2 + 65395*z^3 + 22*z^4)/((1-z)*(1 - 1331714*z^2 + z^4)).

Example

With P(m) = m*(3*m-1)/2,
a(1)=1 because a(1) = P(1) = P(0) + P(1);
a(2)=1926 because P(36) = 1926 = P(25) + P(26) = 925 + 1001;
a(3)=850137 because P(753) = 850137 = P(532) + P(533) = 424270 + 425867 ...

Maple

a:=proc(m) if type (sqrt(18*m^2-6*m-8)/6-1/3), integer=true then m*(3*m-1)/2 else fi end : seq(a(m), m=1..100000)od; # Emeric Deutsch

Mathematica

# (3 # - 1)/2 &@ Select[Range[10^6], IntegerQ[Sqrt[18 #^2 - 6 # - 8]/6 - 1/3] &] (* or *)
Rest@ CoefficientList[Series[x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/( (1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)), {x, 0, 12}], x] (* Michael De Vlieger, Jul 14 2016 *)

Prog

(PARI) Vec(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) + O(x^20)) \\ Colin Barker, Oct 20 2014
(Magma) R<x>:=PowerSeriesRing(Integers(), 20); Coefficients(R!( x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2)) )); // G. C. Greubel, Mar 16 2019
(Sage) a=(x*(1+1925*x-483503*x^2+65395*x^3+22*x^4)/((1-x)*(1 -1154*x + x^2)*(1 +1154*x + x^2))).series(x, 20).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 16 2019

Crossrefs

Cf. A046173, A144796 (x(n)).

Sequence in context: A326018 A202051 A283949 * A258841 A099482 A253337

Adjacent sequences: A133298 A133299 A133300 * A133302 A133303 A133304

Keyword

nonn,easy

Author

Richard Choulet, Dec 20 2007

Extensions

Fixed typo in g.f. in formula, and more terms from Colin Barker, Oct 20 2014

Status

approved