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A014820
a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.
31
1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481, 350208, 396033, 446216, 501025, 560736, 625633, 696008, 772161
OFFSET
0,2
COMMENTS
a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009
If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008
REFERENCES
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), pages 189, 201.
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
LINKS
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
Maya Ahmed, Jesús De Loera and Raymond Hemmecke, Polyhedral cones of magic cubes and squares, in Discrete and Computational Geometry, Springer, Berlin, 2003, pp. 25-41.
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric Weisstein's World of Mathematics, 16-Cell.
FORMULA
Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009
From Vladeta Jovovic, Apr 03 2002: (Start)
G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013
From Bruce J. Nicholson, Jan 23 2019: (Start)
Sum_{i=0..n} a(i) = A061927(n+1).
a(n) = 4*A002415(n+1) + A000290(n+1) = A039623(n+1) + A002415(n+1). (End)
E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024
a(n) = A006325(n+1) + A006325(n+2). - Bruce Nye, Feb 12 2026
Sum_{n>=0} (-1)^n/a(n) = (Pi^2 - 3 + 3*sqrt(2)*Pi*cosech(sqrt(2)*Pi))/8. - Amiram Eldar, Feb 15 2026
MAPLE
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(4, n), n=0..100)];
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)
PROG
(Magma) [(1/3)*(n^2+2*n+3)*(n+1)^2: n in [0..40]]; // Vincenzo Librandi, May 22 2011
(PARI) a(n)=(n+1)^2*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Apr 17 2012
(R)
a <- c(1, 8, 33, 96, 225)
for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
a # Yosu Yurramendi, Sep 03 2013
(SageMath) [((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019
(GAP) List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019
KEYWORD
nonn,easy
EXTENSIONS
Formula index corrected by R. J. Mathar, Jul 18 2009
STATUS
approved