OFFSET
1,2
COMMENTS
Also the index of a pentagonal number which is equal to the sum of two consecutive pentagonal numbers. - Colin Barker, Dec 22 2014
LINKS
Colin Barker, Table of n, a(n) for n = 1..653
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Polygonal Balancing Numbers I, Integers 22 (2022), A54. See p. 8.
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).
FORMULA
For the odd and even indices respectively the same recurrence is obtained: a(n+2) = 1154*a(n+1) - a(n) - 192.
We also have a(n+2) = 577*a(n+1) - 96 + 68*sqrt((72*a(n)^2-24*a(n)-32)).
G.f.: x*(1 + 35*x - 437*x^2 + 205*x^3 + 4*x^4) / ((1-x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)). - R. J. Mathar, Nov 27 2011
EXAMPLE
a(1) = 1 because 6*1 - 1 = 5 = A144796(1).
MATHEMATICA
LinearRecurrence[{1, 1154, -1154, -1, 1}, {1, 36, 753, 41348, 868769}, 30] (* Harvey P. Dale, Dec 27 2018 *)
PROG
(PARI) Vec(-x*(1+35*x-437*x^2+205*x^3+4*x^4) / ((x-1)*(x^2-34*x+1)*(x^2+34*x+1)) + O(x^30)) \\ Colin Barker, Dec 22 2014
(Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1+ 35*x-437*x^2+205*x^3+4*x^4)/((1-x)*(1-34*x+x^2)*(1+34*x+x^2)) )); // G. C. Greubel, Mar 16 2019
(SageMath) a=(x*(1+ 35*x-437*x^2+205*x^3+4*x^4)/((1-x)*(1-34*x+x^2)*(1+34*x +x^2))).series(x, 30).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 16 2019
(GAP) a:=[1, 36, 753, 41348, 868769];; for n in [6..30] do a[n]:=a[n-1] +1154*a[n-2]-1154*a[n-3]-a[n-4]+a[n-5]; od; a; # G. C. Greubel, Mar 16 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Richard Choulet, Sep 26 2008
EXTENSIONS
a(6) corrected and sequence extended by R. J. Mathar, Nov 27 2011
STATUS
approved