A005585 5-dimensional pyramidal numbers: a(n) = n*(n+1)*(n+2)*(n+3)*(2n+3)/5!.

1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289, 5005, 7371, 10556, 14756, 20196, 27132, 35853, 46683, 59983, 76153, 95634, 118910, 146510, 179010, 217035, 261261, 312417, 371287, 438712, 515592, 602888, 701624, 812889, 937839, 1077699, 1233765, 1407406

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Convolution of triangular numbers (A000217) and squares (A000290) (n>=1). - Graeme McRae, Jun 07 2006

p^k divides a(p^k-3), a(p^k-2), a(p^k-1) and a(p^k) for prime p > 5 and integer k > 0. p^k divides a((p^k-3)/2) for prime p > 5 and integer k > 0. - Alexander Adamchuk, May 08 2007

If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-5) is the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

5-dimensional square numbers, fourth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+4, i+4)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

Antidiagonal sums of the convolution array A213550. - Clark Kimberling, Jun 17 2012

Binomial transform of (1, 6, 14, 16, 9, 2, 0, 0, 0, ...). - Gary W. Adamson, Jul 28 2015

2*a(n) is number of ways to place 4 queens on an (n+3) X (n+3) chessboard so that they diagonally attack each other exactly 6 times. The maximal possible attack number, p=binomial(k,2)=6 for k=4 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015

While adjusting for offsets, add A000389 to find the next in series A000389, A005585, A051836, A034263, A027800, A051843, A051877, A051878, A051879, A051880, A056118, A271567. (See Bruno Berselli's comments in A271567.) - Bruce J. Nicholson, Jun 21 2018

Coefficients in the terminating series identity 1 - 7*n/(n + 6) + 27*n*(n - 1)/((n + 6)*(n + 7)) - 77*n*(n - 1)*(n - 2)/((n + 6)*(n + 7)*(n + 8)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A040977. - Peter Bala, Feb 18 2019

Examples

			G.f. = x + 7*x^2 + 27*x^3 + 77*x^4 + 182*x^5 + 378*x^6 + 714*x^7 + 1254*x^8 + ... - _Michael Somos_, Jun 24 2018
		

References

• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 797.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000 (first 121 terms from Alexander Adamchuk)
• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
• Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
• R. K. Guy, Letter to N. J. A. Sloane, Feb 1988
• Milan Janjic, Two Enumerative Functions
• C. H. Karlson and N. J. A. Sloane, Correspondence, 1974
• Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
• R. P. Stanley and F. Zanello, The Catalan case of Armstrong's conjecture on core partitions, arXiv preprint arXiv:1312.4352 [math.CO], 2013.
• Index entries for sequences related to Chebyshev polynomials.
• Index to sequences related to pyramidal numbers
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

a(n) = ((-1)^(n+1))*A053120(2*n+3, 5)/16, (1/16 of sixth unsigned column of Chebyshev T-triangle, zeros omitted).

Partial sums of A002415.

Cf. A006542, A040977, A047819, A111125 (third column).

Cf. a(n) = ((-1)^(n+1))*A084960(n+1, 2)/16 (compare with the first line). - Wolfdieter Lang, Aug 04 2014

Cf. A000389, A051836, A034263, A027800, A051843, A051877, A051878, A051879.

Programs

Magma

I:=[1, 7, 27, 77, 182, 378]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Jun 09 2013

Maple

[seq(binomial(n+2,6)-binomial(n,6), n=4..45)]; # Zerinvary Lajos, Jul 21 2006
A005585:=(1+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation

Mathematica

With[{c=5!},Table[n(n+1)(n+2)(n+3)(2n+3)/c,{n,40}]] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{1,7,27,77,182,378},40] (* Harvey P. Dale, Oct 04 2011 *)
CoefficientList[Series[(1 + x) / (1 - x)^6, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

PARI

a(n)=binomial(n+3,4)*(2*n+3)/5 \\ Charles R Greathouse IV, Jul 28 2015

Formula

G.f.: x*(1+x)/(1-x)^6.

a(n) = 2*C(n+4, 5) - C(n+3, 4). - Paul Barry, Mar 04 2003

a(n) = C(n+3, 5) + C(n+4, 5). - Paul Barry, Mar 17 2003

a(n) = C(n+2, 6) - C(n, 6), n >= 4. - Zerinvary Lajos, Jul 21 2006

a(n) = Sum_{k=1..n} T(k)*T(k+1)/3, where T(n) = n(n+1)/2 is a triangular number. - Alexander Adamchuk, May 08 2007

a(n-1) = (1/4)*Sum_{1 <= x_1, x_2 <= n} |x_1*x_2*det V(x_1,x_2)| = (1/4)*Sum_{1 <= i,j <= n} i*j*|i-j|, where V(x_1,x_2) is the Vandermonde matrix of order 2. First differences of A040977. - Peter Bala, Sep 21 2007

a(n) = C(n+4,4) + 2*C(n+4,5). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

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(1)=1, a(2)=7, a(3)=27, a(4)=77, a(5)=182, a(6)=378. - Harvey P. Dale, Oct 04 2011

a(n) = (1/6)*Sum_{i=1..n+1} (i*Sum_{k=1..i} (i-1)*k). - Wesley Ivan Hurt, Nov 19 2014

E.g.f.: x*(2*x^4 + 35*x^3 + 180*x^2 + 300*x + 120)*exp(x)/120. - Robert Israel, Nov 19 2014

a(n) = A000389(n+3) + A000389(n+4). - Bruce J. Nicholson, Jun 21 2018

a(n) = -a(-3-n) for all n in Z. - Michael Somos, Jun 24 2018

From Amiram Eldar, Jun 28 2020: (Start)

Sum_{n>=1} 1/a(n) = 40*(16*log(2) - 11)/3.

Sum_{n>=1} (-1)^(n+1)/a(n) = 20*(8*Pi - 25)/3. (End)

a(n) = A004302(n+1) - A207361(n+1). - J. M. Bergot, May 20 2022

a(n) = Sum_{i=0..n+1} Sum_{j=i..n+1} i*j*(j-i)/2. - Darío Clavijo, Oct 11 2023

a(n) = (A000538(n+1) - A000330(n+1))/12. - Yasser Arath Chavez Reyes, Feb 21 2024