A000580 -id:A000580 - cp's OEIS frontend

A347056 Triangle read by rows: T(n,k) = (n+1)(n+2)(k+3)*binomial(n,k)/6, 0 <= k <= n.

1, 3, 4, 6, 16, 10, 10, 40, 50, 20, 15, 80, 150, 120, 35, 21, 140, 350, 420, 245, 56, 28, 224, 700, 1120, 980, 448, 84, 36, 336, 1260, 2520, 2940, 2016, 756, 120, 45, 480, 2100, 5040, 7350, 6720, 3780, 1200, 165, 55, 660, 3300, 9240, 16170, 18480, 13860, 6600, 1815, 220

Offset: 0

Luc Rousseau, Aug 14 2021

This triangle is T[3] in the sequence (T[p]) of triangles defined by: T[p](n,k) = (k+p)*(n+p-1)!/(k!*(n-k)!*p!) and T[0](0,0)=1.

Riordan triangle (1/(1-x)^3, x/(1-x)) with column k scaled with A000292(k+1) = binomial(k+3, 3), for k >= 0. - Wolfdieter Lang, Sep 30 2021

			T(6,2) = (6+1)*(6+2)*(2+3)*binomial(6,2)/6 = 7*8*5*15/6 = 700.
The triangle T begins:
n \ k  0   1    2     3     4     5     6     7     8    9  10 ...
0:     1
1:     3   4
2:     6  16   10
3:    10  40   50    20
4:    15  80  150   120    35
5:    21 140  350   420   245    56
6:    28 224  700  1120   980   448    84
7:    36 336 1260  2520  2940  2016   756   120
8:    45 480 2100  5040  7350  6720  3780  1200   165
9:    55 660 3300  9240 16170 18480 13860  6600  1815  220
10:   66 880 4950 15840 32340 44352 41580 26400 10890 2640 286
... - _Wolfdieter Lang_, Sep 30 2021

Luc Rousseau, Illustration: the A347056 triangle in a sequence of contiguous triangles.

Cf. A097805 (p=0), A103406 (p=1), A124932 (essentially p=2).
From Wolfdieter Lang, Sep 30 2021: (Start)
Columns (with leading zeros): A000217(n+1), 4*A000294, 10*A000332(n+2), 20*A000389(n+2), 35*A000579(n+2), 56*A000580(n+2), 84*A000581(n+2), 120*A000582(n+2), ...
Diagonals: A000292(k+1), A004320(k+1), 2*A006411(k+1), 10*A040977, ... (End)

T(p,n,k)=if(n==0&&p==0,1,((k+p)*(n+p-1)!)/(k!*(n-k)!*p!))
for(n=0,9,for(k=0,n,print1(T(3,n,k),", ")))

T(n,k) = (n+1)*(n+2)*(k+3)*binomial(n,k)/6.

G.f. column k: x^k*binomial(k+3, 3)/(1 - x)^(k+3), for k >= 0. - Wolfdieter Lang, Sep 30 2021

A059251 A sequence related to numeric partitions and Fermat Coefficients.

Original entry on oeis.org

1, 1, 5, 15, 44, 99, 217, 429, 811, 1430, 2438, 3978, 6312, 9690, 14550, 21318, 30669, 43263, 60115, 82225, 111044, 148005, 195143, 254475, 328759, 420732, 534076, 672452, 840656, 1043460, 1287036, 1577532, 1922745, 2330445, 2810385, 3372291

Offset: 1

Alford Arnold, Jan 22 2001

The sequences m1^8, m2^4 and 6*m4^2 correspond to eight elements of a finite group of order eight belonging to the appropriate partition class.

			a(5)= 44 because (1/8)*( 330 + 10 + 12) = 352/8; a(9)= 811 because (1/8)*(6435 + 35 + 18) = 6488/8.

Cf. A000041, A000292, A000580, A000973, A058936.

Let m1^8 = A000580, m2^4 = 1 0 4 0 10 0 20 ... and let m4^2 = 1 0 0 0 2 0 0 0 3 0 0 0 4 ... Then a(n) = (1/8)*(m1^8 + m2^4 + 6*m4^2).

Empirical g.f.: x*(1 - 3*x + 5*x^2 + 3*x^3 - 4*x^4 + 3*x^5 + 5*x^6 - 3*x^7 + x^8) / ((1 - x)^8*(1 + x)^4*(1 + x^2)^2). - Colin Barker, Mar 30 2017

More terms from David Wasserman, Jun 07 2002

A095670 Eighth column (m=7) of (1,4)-Pascal triangle A095666.

Original entry on oeis.org

4, 29, 120, 372, 960, 2178, 4488, 8580, 15444, 26455, 43472, 68952, 106080, 158916, 232560, 333336, 468996, 648945, 884488, 1189100, 1578720, 2072070, 2691000, 3460860, 4410900, 5574699, 6990624, 8702320, 10759232, 13217160, 16138848

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X then, for n>=10, a(n-10) is the number of 7-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

G.f.: (4-3*x)/(1-x)^8.

a(n) = 4*b(n)-3*b(n-1) =(n+28)*binomial(n+6, 6)/7, with b(n):=binomial(n+7, 7)= A000580(n+7, 7).

A127157 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and 2k nodes of odd degree (not outdegree; 1 <= k <= ceiling(n/2)).

Original entry on oeis.org

1, 2, 3, 2, 4, 10, 5, 30, 7, 6, 70, 56, 7, 140, 252, 30, 8, 252, 840, 330, 9, 420, 2310, 1980, 143, 10, 660, 5544, 8580, 2002, 11, 990, 12012, 30030, 15015, 728, 12, 1430, 24024, 90090, 80080, 12376, 13, 2002, 45045, 240240, 340340, 111384, 3876, 14, 2730

Offset: 1

Emeric Deutsch, Feb 27 2007

Row n has ceiling(n/2) terms.
Row sums are the Catalan numbers (A000108).
T(n,1) = n;
T(n,2) = 2*binomial(n+1, 4) = 2*A000332(n+1);
T(n,3) = 7*binomial(n+2, 7) = 7*A000580(n+2);
T(n,4) = 30*binomial(n+3, 10) = 30*A001287(n+3);
T(n,5) = 143*binomial(n+4, 13) = 143*A010966(n+4);
T(2n-1,n) = A006013(n-1).
T(n,k) is the number of ordered trees (A000108) with n edges, exactly k of whose vertices possess at least one leaf child. [David Callan, Aug 22 2014]

			Triangle starts:
  1;
  2;
  3,  2;
  4, 10;
  5, 30,  7;
  6, 70, 56;

J.-C. Aval, A. Boussicault, B. Delcroix-Oger, F. Hivert, et al., Non-ambiguous trees: new results and generalization, arXiv preprint arXiv:1511.09455 [math.CO], 2015.
Bérénice Delcroix-Oger, Florent Hivert, Patxi Laborde-Zubieta, Jean-Christophe Aval, and Adrien Boussicault, Non-ambiguous trees: new results and generalisation, hal-03165269v2 [cs.DM] [math.CO], 2021.
Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See p. 20.

Cf. A000108, A000332, A000580, A001287, A010966, A006013.

T:=(n,k)->2*binomial(3*k-1,2*k)*binomial(n-1+k,3*k-2)/(3*k-1): for n from 1 to 15 do seq(T(n,k),k=1..ceil(n/2)) od;

m = 14(*rows*); G = 0; Do[G = Series[(1 + t^2 z - G^3 z^2 + G^2 z (2+z))/ (1+2z), {t, 0, m}, {z, 0, m}] // Normal // Expand, m]; Rest[ CoefficientList[#, t^2]]& /@ Rest[CoefficientList[G-1, z] ] // Flatten (* Jean-François Alcover, Jan 23 2019 *)

T(n,k) = 2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1) (formula obtained only by inspection).

G.f.: G-1, where G=G(t,z) satisfies z^2*G^3 - z(z+2)G^2 + (1+2z)*G - t^2*z - 1 = 0.

A356037 Conjecturally, a(n) is the smallest number m such that every natural number is a sum of at most m n-simplex numbers.

Original entry on oeis.org

1, 3, 5, 8, 10, 13, 15, 15, 19, 24

Offset: 1

Mohammed Yaseen, Jul 24 2022

n-simplex numbers are {binomial(k,n); k>=n}.
This problem is the simplex number analog of Waring's problem.
a(2) = 3 was proposed by Fermat and proved by Gauss, see A061336.
Pollock conjectures that a(3) = 5. Salzer and Levine prove this for numbers up to 452479659. See A104246 and A000797.
Kim gives a(4)=8, a(5)=10, a(6)=13 and a(7)=15 (not proved).

			2-simplex numbers are {binomial(k,2); k>=2} = {1,3,6,10,...}, the triangular numbers. 3 is the smallest number m such that every natural number is a sum of at most m triangular numbers. So a(2)=3.
3-simplex numbers are {binomial(k,3); k>=3} = {1,4,10,20,...}, the tetrahedral numbers. 5 is presumed to be the smallest number m such that every natural number is a sum of at most m tetrahedral numbers. So a(3)=5.

Hyun Kwang Kim, On regular polytope numbers, Proc. Amer. Math. Soc. 131 (2003), p. 65-75.

Cf. A002804, A079611.
Minimal number of x-simplex numbers whose sum equals n: A061336 (x=2), A104246 (x=3), A283365 (x=4), A283370 (x=5).
x-simplex numbers: A000217 (x=2), A000292 (x=3), A000332 (x=4), A000389 (x=5), A000579 (x=6), A000580 (x=7), A000581 (x=8), A000582 (x=9).

cp's OEIS Frontend

A347056 Triangle read by rows: T(n,k) = (n+1)*(n+2)*(k+3)*binomial(n,k)/6, 0 <= k <= n.

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A059251 A sequence related to numeric partitions and Fermat Coefficients.

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A095670 Eighth column (m=7) of (1,4)-Pascal triangle A095666.

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A127157 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and 2k nodes of odd degree (not outdegree; 1 <= k <= ceiling(n/2)).

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A356037 Conjecturally, a(n) is the smallest number m such that every natural number is a sum of at most m n-simplex numbers.

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A347056 Triangle read by rows: T(n,k) = (n+1)(n+2)(k+3)*binomial(n,k)/6, 0 <= k <= n.