login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A133301 a(n) is the n-th pentagonal number which is the sum of two consecutive pentagonal numbers. 9
1, 1926, 850137, 2564464982, 1132138928657, 3415133918621062, 1507685261236261801, 4547981651299964079126, 2007805569980855008712097, 6056610836775865229750164742, 2673822786819976661810784866297, 8065673443881586606920210924732502 (list; graph; refs; listen; history; text; internal format)
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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified May 25 02:01 EDT 2020. Contains 334581 sequences. (Running on oeis4.)