cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A080852 Square array of 4D pyramidal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 5, 10, 1, 6, 15, 20, 1, 7, 20, 35, 35, 1, 8, 25, 50, 70, 56, 1, 9, 30, 65, 105, 126, 84, 1, 10, 35, 80, 140, 196, 210, 120, 1, 11, 40, 95, 175, 266, 336, 330, 165, 1, 12, 45, 110, 210, 336, 462, 540, 495, 220, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 286
Offset: 0

Views

Author

Paul Barry, Feb 21 2003

Keywords

Comments

The first row contains the tetrahedral numbers, which are really three-dimensional, but can be regarded as degenerate 4D pyramidal numbers. - N. J. A. Sloane, Aug 28 2015

Examples

			Array, n >= 0, k >= 0, begins
1 4 10 20  35  56 ...
1 5 15 35  70 126 ...
1 6 20 50 105 196 ...
1 7 25 65 140 266 ...
1 8 30 80 175 336 ...

Links

Crossrefs

Rows include A000292, A000332, A002415, A001296, A002418, A002419, A051740, A051797.
Cf. A057145, A080851, A180266, A055796 (antidiagonal sums).
See A257200 for another version of the array.

Programs

  • Derive
    vector(vector(poly_coeff(Taylor((1+kx)/(1-x)^5,x,11),x,n),n,0,11),k,-1,10)
  • Derive
    VECTOR(VECTOR(comb(k+3,3)+comb(k+3,4)n, k, 0, 11), n, 0, 11)
  • Maple
    A080852 := proc(n,k)
        binomial(k+4,4)+(n-1)*binomial(k+3,4) ;
end proc:
seq( seq(A080852(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021
  • Mathematica
    T[n_, k_] := Binomial[k+3, 3] + Binomial[k+3, 4]n;
Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 05 2023 *)

Formula

T(n, k) = binomial(k + 4, 4) + (n-1)*binomial(k + 3, 4), corrected Oct 01 2021.
T(n, k) = T(n - 1, k) + C(k + 3, 4) = T(n - 1, k) + k(k + 1)(k + 2)(k + 3)/24.
G.f. for rows: (1 + nx)/(1 - x)^5, n >= -1.
T(n,k) = sum_{j=0..k} A080851(n,j). - R. J. Mathar, Jul 28 2016

A034274 a(n)=f(n,n-1) where f is given in A034261.

Original entry on oeis.org

1, 5, 25, 119, 546, 2442, 10725, 46475, 199342, 848198, 3585946, 15080870, 63146500, 263432340, 1095517485, 4543460595, 18798494550, 77616288750, 319874637390, 1316106144210, 5407045011420, 22184521682700, 90910797617250, 372137346502974, 1521789223654476, 6217349014923452
Offset: 1

Views

Author

Clark Kimberling

Keywords

Links

Crossrefs

Cf. A000984, A000108, A034261, A180266.

Programs

Formula

From Peter Bala, Aug 19 2025: (Start)
a(n) = (n^2 + 1)/2 * A000108(n).
a(n) = (1/2) * A180266(n+1).
a(n) = Sum_{k = 1..n} k^2/(n+k-1) * binomial(n+k-1, k). Cf. Sum_{k = 1..n} k/(n+k-1) * binomial(n+k-1, k) = 1/2 * binomial(2*n, n) = 1/2 * A000984(n).
a(n) = 2*(n^2 + 1)*(2*n - 1)/((n + 1)*(n^2 - 2*n + 2)) * a(n-1) with a(1) = 1. (End)
a(n) ~ 2^(2*n-1) * sqrt(n/Pi). - Amiram Eldar, Sep 04 2025

Extensions

Corrected and extended by N. J. A. Sloane, Apr 21 2000

A301972 a(n) = n*(n^2 - 2*n + 4)*binomial(2*n,n)/((n + 1)*(n + 2)).

Original entry on oeis.org

0, 1, 4, 21, 112, 570, 2772, 13013, 59488, 266526, 1175720, 5123426, 22108704, 94645460, 402503220, 1702300725, 7165821120, 30043474230, 125523450360, 522857438070, 2172127120800, 9002522512620, 37233403401480, 153704429299746, 633442159732032, 2606543487445100, 10710790748646352, 43957192722175908
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 29 2018

Keywords

Comments

For n > 2, a(n) is the n-th term of the main diagonal of iterated partial sums array of n-gonal numbers (in other words, a(n) is the n-th (n+2)-dimensional n-gonal number, see also example).

Examples

			For n = 5 we have:
----------------------------
0   1    2    3     4    [5]
----------------------------
0,  1,   5,  12,   22,   35,  ... A000326 (pentagonal numbers)
0,  1,   6,  18,   40,   75,  ... A002411 (pentagonal pyramidal numbers)
0,  1,   7,  25,   65,  140,  ... A001296 (4-dimensional pyramidal numbers)
0,  1,   8,  33,   98,  238,  ... A051836 (partial sums of A001296)
0,  1,   9,  42,  140,  378,  ... A051923 (partial sums of A051836)
0,  1,  10,  52,  192, [570], ... A050494 (partial sums of A051923)
----------------------------
therefore a(5) = 570.

Links

Crossrefs

Cf. A000984, A002457, A006484, A057145, A060354, A080851, A080852, A100119, A180266, A275490, A292551.

Programs

  • Mathematica
    Table[n (n^2 - 2 n + 4) Binomial[2 n, n]/((n + 1) (n + 2)), {n, 0, 27}]
nmax = 27; CoefficientList[Series[(-4 + 31 x - 66 x^2 + 28 x^3 + (4 - 7 x) (1 - 4 x)^(3/2))/(2 x^2 (1 - 4 x)^(3/2)), {x, 0, nmax}], x]
nmax = 27; CoefficientList[Series[Exp[2 x] (4 - x + 2 x^2) BesselI[1, 2 x]/x - 2 Exp[2 x] (2 - x) BesselI[0, 2 x], {x, 0, nmax}], x] Range[0, nmax]!
Table[SeriesCoefficient[x (1 - 3 x + n x)/(1 - x)^(n + 3), {x, 0, n}], {n, 0, 27}]

Formula

O.g.f.: (-4 + 31*x - 66*x^2 + 28*x^3 + (4 - 7*x)*(1 - 4*x)^(3/2))/(2*x^2*(1 - 4*x)^(3/2)).
E.g.f.: exp(2*x)*(4 - x + 2*x^2)*BesselI(1,2*x)/x - 2*exp(2*x)*(2 - x)*BesselI(0,2*x).
a(n) = [x^n] x*(1 - 3*x + n*x)/(1 - x)^(n+3).
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
D-finite with recurrence: -(n+2)*(961*n-3215)*a(n) +4*(2081*n^2-4414*n-4668)*a(n-1) -28*(320*n-389)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 27 2020
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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