OFFSET
1,1
COMMENTS
Indices m: 4, 16, 61, 229, 856, 3196, 11929, with recurrence m(i) = 5(m(i-1) - m(i-2)) + m(i-3) (see A133161).
If first term is omitted, same sequence as A128862. - R. J. Mathar, Jun 13 2008
REFERENCES
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), page 427.
LINKS
Colin Barker, Table of n, a(n) for n = 1..850
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).
FORMULA
a(n) = tr(m) = tr(k) + tr(k+1) + tr(k+2), where tr(k) = k(k+1)/2 = A000217(k).
From Richard Choulet, Oct 06 2007: (Start)
a(n+2) = 14*a(n+1) - a(n) - 3.
a(n+1) = 7*a(n) - 3/2 + 1/2*sqrt(192*a(n)^2 - 96*a(n) - 15).
G.f.: x*(10-14*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)
a(n) = (4-3*(7-4*sqrt(3))^n*(-2+sqrt(3))+3*(2+sqrt(3))*(7+4*sqrt(3))^n)/16. - Colin Barker, Mar 05 2016
E.g.f.: 3*exp(7*x)*(2*cosh(4*sqrt(3)*x) + sqrt(3)*sinh(4*sqrt(3)*x))/8 - 1 + exp(x)/4. - Stefano Spezia, Nov 08 2025
EXAMPLE
With tr(k) = k(k+1)/2 = A000217(k):
10 = tr(4) = tr(1) + tr(2) + tr(3) = 1 + 3 + 6,
136 = tr(16) = tr(8) + tr(9) + tr(10) = 36 + 45 + 55,
1891 = tr(61) = tr(34) + tr(35) + tr(36) = 595 + 630 + 666,
26335 = tr(229) = tr(131) + tr(132) + tr(133) = 8646 + 8778 + 8911,
366796 = tr(856) = tr(493) + tr(494) + tr(495) = 121771 + 122265 + 122760.
MATHEMATICA
LinearRecurrence[{15, -15, 1}, {10, 136, 1891}, 20] (* Harvey P. Dale, Oct 31 2024 *)
PROG
(PARI) Vec((10*z - 14*z^2 + z^3)/((1-z)*(1 - 14*z + z^2)) + O(z^30)) \\ Michel Marcus, Sep 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Zak Seidov, May 18 2007
STATUS
approved