A050494 -id:A050494 - cp's OEIS frontend

A053367 Partial sums of A050494.

1, 11, 63, 255, 825, 2277, 5577, 12441, 25740, 50050, 92378, 163098, 277134, 455430, 726750, 1129854, 1716099, 2552517, 3725425, 5344625, 7548255, 10508355, 14437215, 19594575, 26295750, 34920756, 45924516, 59848228, 77331980, 99128700, 126119532, 159330732, 199952181, 249357615

Offset: 0

Barry E. Williams, Jan 06 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A050494.

Cf. A093560 ((3, 1) Pascal, column m=8).

[(3*n+8)*Binomial(n+7,7)/8: n in [0..30]]; // G. C. Greubel, May 25 2018

LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {1, 11, 63, 255, 825, 2277, 5577, 12441, 25740}, 30] (* or *) Table[(3*n+8)* Binomial[n+7,7]/8, {n,0,30}] (* G. C. Greubel, May 25 2018 *)

a(n)=binomial(n+7, 7)*(3*n+8)/8 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = binomial(n+7, 7)*(3n+8)/8.

G.f.: (1+2*x)/(1-x)^9.

Terms a(24) onward added by G. C. Greubel, May 25 2018

A093560 (3,1) Pascal triangle.

Original entry on oeis.org

1, 3, 1, 3, 4, 1, 3, 7, 5, 1, 3, 10, 12, 6, 1, 3, 13, 22, 18, 7, 1, 3, 16, 35, 40, 25, 8, 1, 3, 19, 51, 75, 65, 33, 9, 1, 3, 22, 70, 126, 140, 98, 42, 10, 1, 3, 25, 92, 196, 266, 238, 140, 52, 11, 1, 3, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1, 3, 31, 145, 405, 750, 966, 882, 570, 255, 75, 13, 1

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(3;n,m) gives in the columns m >= 1 the figurate numbers based on A016777, including the pentagonal numbers A000326 (see the W. Lang link).
This is the third member, d=3, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal (d=1), A029653 (d=2).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(1+2*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(1+2*z)/(1-(1+x)*z).
The SW-NE diagonals give the Lucas numbers A000032: L(n) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with L(0)=2. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by [3,-2,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
From Wolfdieter Lang, Jan 09 2015: (Start)
The signed lower triangular matrix (-1)^(n-1)*a(n,m) is the inverse of the Riordan matrix A106516; that is Riordan ((1-2*x)/(1+x),x/(1+x)).
See the Peter Bala comment from Dec 23 2014 in A106516 for general Riordan triangles of the type (g(x), x/(1-x)): exp(x)*r(n,x) = d(n,x) with the e.g.f. r(n,x) of row n and the e.g.f. of diagonal n.
Similarly, for general Riordan triangles of the type (g(x), x/(1+x)): exp(x)*r(n,-x) = d(n,x). (End)
The n-th row polynomial is (3 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018
Binomial(n-2,k)+2*Binomial(n-3,k) is also the number of permutations avoiding both 123 and 132 with k double descents, i.e., positions with w[i]>w[i+1]>w[i+2]. - Lara Pudwell, Dec 19 2018

			Triangle begins
  1,
  3,  1,
  3,  4,  1,
  3,  7,  5,   1,
  3, 10, 12,   6,   1,
  3, 13, 22,  18,   7,   1,
  3, 16, 35,  40,  25,   8,   1,
  3, 19, 51,  75,  65,  33,   9,  1,
  3, 22, 70, 126, 140,  98,  42, 10,  1,
  3, 25, 92, 196, 266, 238, 140, 52, 11, 1,

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, A note on the diagonals of a proper Riordan Array
M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
W. Lang, First 10 rows and array of figurate numbers .

Cf. Column sequences for m=1..9: A016777, A000326 (pentagonal), A002411, A001296, A051836, A051923, A050494, A053367, A053310;

A007318 (Pascal's triangle), A029653 ((2,1) Pascal triangle), A093561 ((4,1) Pascal triangle), A228196, A228576.

Concatenation([1],Flat(List([1..11],n->List([0..n],k->Binomial(n,k)+2*Binomial(n-1,k))))); # Muniru A Asiru, Dec 20 2018

a093560 n k = a093560_tabl !! n !! k
a093560_row n = a093560_tabl !! n
a093560_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [3, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093560(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<1))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(3;n-m, m) for 0<= m <= n, otherwise 0, with F(3;0, 0)=1, F(3;n, 0)=3 if n>=1 and F(3;n, m):=(3*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+2*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=3 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 2*C(n-1, k). - Philippe Deléham, Aug 28 2005
Equals M * A007318, where M = an infinite triangular matrix with all 1's in the main diagonal and all 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Sum_{k=0..n} T(n,k) = A151821(n+1). - Philippe Deléham, Sep 17 2009
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(3 + 7*x + 5*x^2/2! + x^3/3!) = 3 + 10*x + 22*x^2/2! + 40*x^3/3! + 65*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-2*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009

A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

Original entry on oeis.org

1, 4, 1, 7, 5, 1, 10, 12, 6, 1, 13, 22, 18, 7, 1, 16, 35, 40, 25, 8, 1, 19, 51, 75, 65, 33, 9, 1, 22, 70, 126, 140, 98, 42, 10, 1, 25, 92, 196, 266, 238, 140, 52, 11, 1, 28, 117, 288, 462, 504, 378, 192, 63, 12, 1

Offset: 0

Gary W. Adamson, Dec 01 2007

A007318 * a bidiagonal matrix with all 1's in the main diagonal and all 3's in the subdiagonal.
Row sums give A036563(n+2), n >= 0.
From Wolfdieter Lang, Mar 23 2015: (Start)
This is the triangle of iterated partial sums of A016777. Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given).
This is therefore the Riordan triangle ((1+2*x)/(1-x)^2, x/(1-x)) with o.g.f. of the columns ((1+2*x)/(1-x)^2)*(x/(1-x))^k, k >= 0.
The column sequences are A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310, for k = 0..8.
The alternating row sums are A122553(n) = {1, repeat(3)}.
The Riordan A-sequence is A(y) = 1 + y (implying the Pascal triangle recurrence for k >= 1).
The Riordan Z-sequence is A256096, leading to a recurrence for T(n,0) given in the formula section. See the link "Sheffer a- and z-sequences" under A006232 also for Riordan A- and Z-sequences with references. (End)
When the first column (k = 0) is removed from this triangle, the result is A125232. - Georg Fischer, Jul 26 2023

			The triangle T(n, k) begins:
n\k  0   1   2    3    4    5    6   7   8  9 10 11
0:   1
1:   4   1
2:   7   5   1
3:  10  12   6    1
4:  13  22  18    7    1
5:  16  35  40   25    8    1
6:  19  51  75   65   33    9    1
7:  22  70 126  140   98   42   10   1
8:  25  92 196  266  238  140   52  11   1
9:  28 117 288  462  504  378  192  63  12  1
10: 31 145 405  750  966  882  570 255  75 13  1
11: 34 176 550 1155 1716 1848 1452 825 330 88 14  1
... reformatted and extended by _Wolfdieter Lang_, Mar 23 2015
From _Wolfdieter Lang_, Mar 23 2015: (Start)
T(3, 1) = T(2, 0) + T(2, 1) = 7 + 5 = 12 (Pascal, from the A-sequence given above).
T(4, 0) = 4*T(3, 0) - 9*T(3, 1) + 27*T(3, 2) - 81* T(3, 3) = 4*10 - 9*12 + 27*6 - 81*1 = 13, from the Z-sequence given above and in A256096.
T(4, 0) = 2*T(3, 0) - T(2, 0) = 2*10 - 7 = 13.
(End)

Columns k=0-8 give: A016777, A000326, A002411, A001296, A051836, A051923, A050494, A053367, A053310.
Cf. A036563, A125232.
Cf. A110813. - Philippe Deléham, Feb 08 2009

Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal and all 3's in the subdiagonal; i.e., by columns - every column = (1, 3, 0, 0, 0, ...).
T(n,k) = (3n-2k+1)*binomial(n+1,k+1)/(n+1). - Philippe Deléham, Feb 08 2009
From Wolfdieter Lang, Mar 23 2015: (Start)
O.g.f. for row polynomials: (1 + 2*z)/((1- z*(1 + x))*(1 - z)) (see the Riordan property from the comment).
O.g.f. for column k (without leading zeros): (1 + 2*x)/(1-x)^(2+k), k >= 0, (Riordan property).
T(n, k) = T(n-1, k-1) + T(n-1, k) for k >= 1. From the Riordan A-sequence given above in a comment.
T(n, 0) = Sum_{j=0..n} Z(j)*T(n-1, j), for n >= 1, from the Riordan Z-sequence A256096 mentioned above in a comment. Of course, T(n, 0) = 2*T(n-1, 0) - T(n-2, 0) for n >= 2 (see A016777).
(End)

Edited. Offset is 0 from the old name and the Philippe Deléham formula. New name, old name as first comment. - Wolfdieter Lang, Mar 23 2015

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

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A053367 Partial sums of A050494.

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A093560 (3,1) Pascal triangle.

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A135857 Partial sums triangle based on A016777. Riordan convolution triangle ((1 + 2*x)/(1-x)^2, x/(1-x)).

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