0,3

Hyun Kwang Kim asserts that every nonnegative integer can be represented by the sum of no more than 14 of these numbers. - Jonathan Vos Post, Nov 16 2004

If Y_i (i=1,2,3,4) are 2-blocks of a (n+4)-set X then a(n-4) is the number of 9-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Oct 28 2007

Starting with 1 = binomial transform of [1, 9, 32, 56, 48, 16, 0, 0, 0, ...], where (1, 9, 32, 56, 48, 16) = row 5 of the Chebyshev triangle A081277. Also = row 5 of the array in A142978. - Gary W. Adamson, Jul 19 2008

Starting with the term 1 this is the self-convolution of A001844(n). - Anton Zakharov, Sep 02 2016

H. S. M. Coxeter, Regular Polytopes, New York: Dover Publications, 1973.

Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), pages 200-201.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Milan Janjic, Two Enumerative Functions

Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Recurrence: a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

a(n) = n*(2*n^4 + 10*n^2 + 3)/15. - Jonathan Vos Post, Nov 16 2004

a(n) = C(n+4,5) + 4*C(n+3,5) + 6*C(n+2,5) + 4*C(n+1,5) + C(n,5).

Sum_{n>=1} 1/((1/15)*n*(2*n^4 + 10*n^2 + 3)*n!) = hypergeom([1, 1, 1+i*sqrt(10-2*sqrt(19))*(1/2), 1-i*sqrt(10-2*sqrt(19))*(1/2), 1+i*sqrt(10+2*sqrt(19))*(1/2), 1-i*sqrt(10+2*sqrt(19))*(1/2)], [2, 2, 2+i*sqrt(10-2*sqrt(19))*(1/2), 2-i*sqrt(10-2*sqrt(19))*(1/2), 2+i*sqrt(10+2*sqrt(19))*(1/2), 2-i*sqrt(10+2*sqrt(19))*(1/2)], 1) = 1.05351734968093116819345664995829700099916... - Stephen Crowley, Jul 14 2009

a(n) = a(n-1) + A014820(n-1) + A014820(n-2). - Bruce J. Nicholson, Apr 18 2018

a(n) = 10*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

Euler transform of length 2 sequence [10, -4]. - Michael Somos, Jun 19 2018

Sum_{k >= 1} (-1)^k/(a(k)*a(k+1)) = 10*log(2) - 41/6 = 1/(10 + 2/(10 + 6/(10 + ... + n*(n-1)/(10 + ...)))). See A142983. Cf. A005900 and A014820. - Peter Bala, Mar 08 2024

E.g.f.: exp(x)*x*(15 + 60*x + 60*x^2 + 20*x^3 + 2*x^4)/15. - Stefano Spezia, Mar 10 2024

al:=proc(s, n) binomial(n+s-1, s); end; be:=proc(d, n) local r; add( (-1)^r*binomial(d-1, r)*2^(d-1-r)*al(d-r, n), r=0..d-1); end; [seq(be(5, n), n=0..100)];

CoefficientList[Series[x (1 + x)^4/(1 - x)^6, {x, 0, 32}], x] (* Michael De Vlieger, Sep 02 2016 *)

LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 10, 51, 180, 501}, 40] (* Harvey P. Dale, Jun 19 2021 *)

(Magma) [n*(2*n^4 + 10*n^2 + 3)/15: n in [0..40]]; // Vincenzo Librandi, May 22 2011

(PARI) concat(0, Vec(x*(1+x)^4/(1-x)^6 + O(x^99))) \\ Altug Alkan, Sep 02 2016

(PARI) {a(n) = n * (2*n^4 + 10*n^2 + 3) / 15}; /* Michael Somos, Jun 17 2018 */

Cf. A000332, A014820, A005900.

Cf. A142978, A081277.

Cf. A001844.

Sequence in context: A143855 A124162 A077044 * A213563 A030183 A224327

Adjacent sequences: A069035 A069036 A069037 * A069039 A069040 A069041

nonn,easy

Vladeta Jovovic, Apr 03 2002

approved