A080852 -id:A080852 - cp's OEIS frontend

A261721 Fourth-dimensional figurate numbers.

1, 1, 5, 1, 6, 15, 1, 7, 20, 35, 1, 8, 25, 50, 70, 1, 9, 30, 65, 105, 126, 1, 10, 35, 80, 140, 196, 210, 1, 11, 40, 95, 175, 266, 336, 330, 1, 12, 45, 110, 210, 336, 462, 540, 495, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 1, 14, 55, 140, 280, 476, 714, 960, 1155, 1210, 1001, 1

Offset: 1

Gary W. Adamson, Aug 30 2015

Generating polygons for the sequences are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ... .
n-th row sequence is the binomial transform of the fourth row of Pascal's triangle (1,n) followed by zeros; and the fourth partial sum of (1, n, n, n, ...).
n-th row sequence is the binomial transform of:
((n-1) * (0, 1, 3, 3, 1, 0, 0, 0) + (1, 4, 6, 4, 1, 0, 0, 0)).
Given the n-th row of the array (1, b, c, d, ...), the next row of the array is (1, b, c, d, ...) + (0, 1, 5, 15, 35, ...)

			The array as shown in A257200:
  1,  5, 15,  35,  70, 126, 210,  330, ... A000332
  1,  6, 20,  50, 105, 196, 336,  540, ... A002415
  1,  7, 25,  65, 140, 266, 462,  750, ... A001296
  1,  8, 30,  80, 175, 336, 588,  960, ... A002417
  1,  9, 35,  95, 210, 406, 714, 1170, ... A002418
  1, 10, 40, 110, 245, 476, 840, 1380, ... A002419
  ...
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of the fourth row of Pascal's triangle (1,3) followed by zeros: (1, 6, 12, 10, 3, 0, 0, 0, ...); and the fourth partial sum of (1, 3, 3, 3, 0, 0, 0).
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of: (2 * (0, 1, 3, 3, 1, 0, 0, 0, ...) + (1, 4, 6, 4, 1, 0, 0, 0, ...)); that is, the binomial transform of (1, 6, 12, 10, 3, 0, 0, 0, ...).
Row 2 of the array is (1, 5, 15, 35, 70, ...) + (0, 1, 5, 15, 35, ...), = (1, 6, 20, 50, 105, ...).

Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 195 (Table 80).

Alois P. Heinz, Antidiagonals n = 1..141, flattened
Index to sequences related to pyramidal numbers

Cf. A257200, A261720 (pyramidal numbers), A000332, A002415, A001296, A002417, A002418, A002419.
Similar to A080852 but without row n=0.
Main diagonal gives A256859.

A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):
seq(seq(A(d-k, k), k=0..d-1), d=1..13);  # Alois P. Heinz, Aug 31 2015

row[1] = LinearRecurrence[{5, -10, 10, -5, 1}, {1, 5, 15, 35, 70}, m = 10];
row1 = Join[{0}, row[1] // Most]; row[n_] := row[n] = row[n-1] + row1;
Table[row[n-k+1][[k]], {n, 1, m}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 26 2016 *)

A(n, k) = binomial(k+3, 3) + n*binomial(k+3, 4)
table(n, k) = for(x=1, n, for(y=0, k-1, print1(A(x, y), ", ")); print(""))
/* Print initial 6 rows and 8 columns as follows: */
table(6, 8) \\ Felix Fröhlich, Jul 28 2016

G.f. for row n: (1 + (n-1)*x)/(1 - x)^5.
A(n,k) = C(k+3,3) + n * C(k+3,4) = A080852(n,k).
E.g.f. as array: exp(y)*(exp(x)*(24 + 24*(3 + x)*y + 36*(1 + x)*y^2 + 4*(1 + 3*x)*y^3 + x*y^4) - 4*(6 + 18*y + 9*y^2 + y^3))/24. - Stefano Spezia, Aug 15 2024

A275490 Square array of 5D pyramidal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 5, 1, 6, 15, 1, 7, 21, 35, 1, 8, 27, 56, 70, 1, 9, 33, 77, 126, 126, 1, 10, 39, 98, 182, 252, 210, 1, 11, 45, 119, 238, 378, 462, 330, 1, 12, 51, 140, 294, 504, 714, 792, 495, 1, 13, 57, 161, 350, 630, 966, 1254, 1287, 715, 1, 14, 63, 182, 406, 756, 1218, 1716, 2079, 2002, 1001

Offset: 2

R. J. Mathar, Jul 30 2016

Let F(r,n,d) = binomial(r+d-2,d-1)* ((r-1)*(n-2)+d) /d be the d-dimensional pyramidal numbers. Then A(n,k) = F(k,n,5).

Sum of the n-th antidiagonal is binomial(n+4,7) + binomial(n+4,5) = A055797(n-1). - Mathew Englander, Oct 27 2020

			The array starts in rows n>=2 and columns k>=1 as
   1    5   15   35   70  126  210  330  495   715  1001  1365  1820
   1    6   21   56  126  252  462  792 1287  2002  3003  4368  6188
   1    7   27   77  182  378  714 1254 2079  3289  5005  7371 10556
   1    8   33   98  238  504  966 1716 2871  4576  7007 10374 14924
   1    9   39  119  294  630 1218 2178 3663  5863  9009 13377 19292
   1   10   45  140  350  756 1470 2640 4455  7150 11011 16380 23660
   1   11   51  161  406  882 1722 3102 5247  8437 13013 19383 28028
   1   12   57  182  462 1008 1974 3564 6039  9724 15015 22386 32396
   1   13   63  203  518 1134 2226 4026 6831 11011 17017 25389 36764

Michael De Vlieger, Table of n, a(n) for n = 2..11176 (rows 2 <= n <= 150, flattened)
Index to sequences related to polygonal numbers

Cf. Row sums of A080852 (4D), A080851 (3D), A057145 (2D), A077028 (1D).

Cf. A055797.

Table[Binomial[k + 3, 4] + (# - 2)*Binomial[k + 3, 5] &[m - k + 1], {m, 2, 12}, {k, m - 1}] // Flatten (* Michael De Vlieger, Nov 05 2020 *)

A(n+2,k) = Sum_{j=0..k-1} A080852(n,j).

A(n,k) = binomial(k+3,4) + (n-2)*binomial(k+3,5). - Mathew Englander, Oct 27 2020

A301972 a(n) = n(n^2 - 2n + 4)binomial(2n,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

Ilya Gutkovskiy, Mar 29 2018

nonn

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).

			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.

Index to sequences related to polygonal numbers

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

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}]

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

A301973 a(n) = (n^2 - 3n + 6)binomial(n+2,3)/4.

Original entry on oeis.org

0, 1, 4, 15, 50, 140, 336, 714, 1380, 2475, 4180, 6721, 10374, 15470, 22400, 31620, 43656, 59109, 78660, 103075, 133210, 170016, 214544, 267950, 331500, 406575, 494676, 597429, 716590, 854050, 1011840, 1192136, 1397264, 1629705, 1892100, 2187255, 2518146, 2887924, 3299920, 3757650, 4264820

Offset: 0

Ilya Gutkovskiy, Mar 29 2018

For n > 2, a(n) is the n-th term of the partial sums of n-gonal pyramidal numbers (in other words, a(n) is the n-th 4-dimensional n-gonal number).

Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000292, A000332, A001296, A002415, A006484, A060354, A080852, A256859, A291288 (partial sums), A292551.

Table[(n^2 - 3 n + 6) Binomial[n + 2, 3]/4, {n, 0, 40}]
nmax = 40; CoefficientList[Series[x (1 - 2 x + 6 x^2)/(1 - x)^6, {x, 0, nmax}], x]
nmax = 40; CoefficientList[Series[Exp[x] x (24 + 24 x + 24 x^2 + 10 x^3 + x^4)/24, {x, 0, nmax}], x] Range[0, nmax]!
Table[SeriesCoefficient[x (1 - 3 x + n x)/(1 - x)^5, {x, 0, n}], {n, 0, 40}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 4, 15, 50, 140}, 41]

O.g.f.: x*(1 - 2*x + 6*x^2)/(1 - x)^6.
E.g.f.: exp(x)*x*(24 + 24*x + 24*x^2 + 10*x^3 + x^4)/24.
a(n) = [x^n] x*(1 - 3*x + n*x)/(1 - x)^5.
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).

A333932 a(n) is the least integer that is 4-dimensional pyramidal in exactly n ways.

Original entry on oeis.org

5, 15, 35, 140, 1820, 11375, 820820, 19019000, 10790015600, 1568726956160, 7278234628665, 7271181889157550

Offset: 1

Ilya Gutkovskiy, Apr 10 2020

a(n) has exactly n representations as an 4-dimensional pyramidal number P(m, k) = binomial(k + 2, 3)*(k*(m - 2) - m + 6) / 4, with m > 2, k > 1.

			a(3) = 35 because 35 is the least integer which is 4-dimensional pyramidal in 3 ways (35 = P(3, 4) = P(7, 3) = P(33, 2)).

Index to sequences related to pyramidal numbers

Cf. A063778, A080852, A261721, A333916.

a(9) from Giovanni Resta, Apr 11 2020

a(9) corrected and a(10)-a(12) from Bert Dobbelaere, Apr 14 2020

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A261721 Fourth-dimensional figurate numbers.

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A275490 Square array of 5D pyramidal numbers, read by antidiagonals.

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

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A301973 a(n) = (n^2 - 3*n + 6)*binomial(n+2,3)/4.

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A333932 a(n) is the least integer that is 4-dimensional pyramidal in exactly n ways.

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

A301973 a(n) = (n^2 - 3n + 6)binomial(n+2,3)/4.