A050483 -id:A050483 - cp's OEIS frontend

A052181 Partial sums of A050483.

1, 12, 72, 300, 990, 2772, 6864, 15444, 32175, 62920, 116688, 206856, 352716, 581400, 930240, 1449624, 2206413, 3287988, 4807000, 6906900, 9768330, 13616460, 18729360, 25447500, 34184475, 45439056, 59808672, 78004432, 100867800, 129389040, 164727552, 208234224

Offset: 0

Barry E. Williams, Jan 26 2000

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

Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A027819, A050483.

Cf. A093561 ((4, 1) Pascal, column m=8).

a:=n->(sum((numbcomp(n,8)), j=7..n))/2:seq(a(n), n=8..31); # Zerinvary Lajos, Aug 26 2008

Table[(n + 2)*Binomial[n + 7, 7]/2, {n, 0, 40}] (* Amiram Eldar, Feb 11 2022 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,12,72,300,990,2772,6864,15444,32175},40] (* Harvey P. Dale, Sep 06 2024 *)

a(n) = A027819(n+1)/7.
a(n) = (n+2)*C(n+7, 7)/2.
G.f.: (1+3*x)/(1-x)^9.
a(n) = C(n+2, 2)*C(n+7, 6)/7. - Zerinvary Lajos, Jul 29 2005
From Amiram Eldar, Feb 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 41783/300 - 14*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 7*Pi^2 - 2688*log(2)/5 + 91343/300. (End)

A093561 (4,1) Pascal triangle.

Original entry on oeis.org

1, 4, 1, 4, 5, 1, 4, 9, 6, 1, 4, 13, 15, 7, 1, 4, 17, 28, 22, 8, 1, 4, 21, 45, 50, 30, 9, 1, 4, 25, 66, 95, 80, 39, 10, 1, 4, 29, 91, 161, 175, 119, 49, 11, 1, 4, 33, 120, 252, 336, 294, 168, 60, 12, 1, 4, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1, 4, 41, 190, 525, 960, 1218

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). 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) = (1+3*z)/(1-(1+x)*z).
The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + 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

			Triangle begins
  [1];
  [4, 1];
  [4, 5, 1];
  [4, 9, 6, 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
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.
Cf. A007318, A093562 (d=5), A228196, A228576.

a093561 n k = a093561_tabl !! n !! k
a093561_row n = a093561_tabl !! n
a093561_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093561(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+3*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 3*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*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

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A052181 Partial sums of A050483.

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A093561 (4,1) Pascal triangle.

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