A000389 -id:A000389 - cp's OEIS frontend

A085439 a(n) = Sum_{i=1..n} binomial(i+1,2)^4.

1, 82, 1378, 11378, 62003, 256484, 871140, 2550756, 6651381, 15802006, 34776742, 71791798, 140366759, 261917384, 469277384, 811379400, 1359360681, 2214396762, 3517606762, 5462416762, 8309813083, 12406965164, 18209748140, 26309748140, 37466388765, 52644875166

Offset: 1

André F. Labossière, Jul 03 2003

			a(15) = (2520*(15^9) +22680*(15^8) +79920*(15^7) +136080*(15^6) +107352*(15^5) +22680*(15^4) -10080*(15^3) +1728*15)/9! = 469277384.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Column k=4 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(2520*n^9 +22680*n^8 +79920*n^7 +136080*n^6 +107352*n^5 +22680*n^4 -10080*n^3 +1728*n)/Factorial(9): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(2520*(n^9) + 22680*(n^8) + 79920*(n^7) + 136080*(n^6) + 107352*(n^5) + 22680*(n^4) - 10080*(n^3) + 1728*n)/9!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)

Vec(x*(x^6+72*x^5+603*x^4+1168*x^3+603*x^2+72*x+1)/(x-1)^10 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = sum(i=1, n, binomial(i+1, 2)^4); \\ Michel Marcus, Nov 22 2017

a(n) = (2520*n^9 +22680*n^8 +79920*n^7 +136080*n^6 +107352*n^5 +22680*n^4 -10080*n^3 +1728*n)/9!.

G.f.: x*(x^6+72*x^5+603*x^4+1168*x^3+603*x^2+72*x+1) / (x-1)^10. - Colin Barker, May 02 2014

More terms from Colin Barker, May 02 2014

Typo in example fixed by Colin Barker, May 02 2014

A085440 a(n) = Sum_{i=1..n} binomial(i+1,2)^5.

Original entry on oeis.org

1, 244, 8020, 108020, 867395, 4951496, 22161864, 82628040, 267156165, 770440540, 2022773116, 4909947484, 11150268935, 23913084560, 48796284560, 95322158736, 179163294729, 325374464580, 572984364580, 981394464580, 1639143014731, 2675722491224, 4277290592600

Offset: 1

André F. Labossière, Jun 30 2003

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Column k=5 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n )/Factorial(11): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!, {n,1,50}] (* G. C. Greubel, Nov 22 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^5), ", ")) \\ G. C. Greubel, Nov 22 2017

a(n) = (113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!.

G.f.: x*(x^8+232*x^7+5158*x^6+27664*x^5+47290*x^4+27664*x^3+5158*x^2+232*x+1) / (x-1)^12. - Colin Barker, May 02 2014

Formula edited by Colin Barker, May 02 2014

A085441 a(n) = Sum_{i=1..n} binomial(i+1,2)^6.

Original entry on oeis.org

1, 730, 47386, 1047386, 12438011, 98204132, 580094436, 2756876772, 11060642397, 38741283022, 121395233038, 346594833742, 914464085783, 2254559726408, 5240543726408, 11568062614344, 24395756421273, 49397866465794, 96443747465794, 182209868465794

Offset: 1

André F. Labossière, Jul 07 2003

			a(5) = C(7,3)*[191*106 + 450*(18*C(14,10) + 3851*C(13,10) + 61839*C(12,10) + 225352*C(11,10) + 225352*C(10,10))]/10010 = 12438011.

T. D. Noe, Table of n, a(n) for n = 1..1000
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Column k=6 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030, A234253.

[(n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12): n in [1..30]]; // G. C. Greubel, Nov 22 2017

f:= sum(binomial(1+i,2)^6,i=1..n):
seq(f, n=1..30); # Robert Israel, Nov 22 2017

Table[Sum[Binomial[i+1,2]^6,{i,n}],{n,20}] (* or *) LinearRecurrence[ {14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1},{1,730,47386,1047386,12438011, 98204132,580094436, 2756876772,11060642397, 38741283022,121395233038, 346594833742, 914464085783, 2254559726408},20] (* Harvey P. Dale, Jun 05 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^6), ", ")) \\ G. C. Greubel, Nov 22 2017

G.f.: x*(x^10 +716*x^9 +37257*x^8 +450048*x^7 +1822014*x^6 +2864328*x^5 +1822014*x^4 +450048*x^3 +37257*x^2 +716*x +1) / (x -1)^14. - Colin Barker, May 02 2014

a(n) = (n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12). - G. C. Greubel, Nov 22 2017

A119900 Triangle read by rows: T(n,k) is the number of binary words of length n with k strictly increasing runs, for 0<=k<=n.

Original entry on oeis.org

1, 0, 2, 0, 1, 3, 0, 0, 4, 4, 0, 0, 1, 10, 5, 0, 0, 0, 6, 20, 6, 0, 0, 0, 1, 21, 35, 7, 0, 0, 0, 0, 8, 56, 56, 8, 0, 0, 0, 0, 1, 36, 126, 84, 9, 0, 0, 0, 0, 0, 10, 120, 252, 120, 10, 0, 0, 0, 0, 0, 1, 55, 330, 462, 165, 11, 0, 0, 0, 0, 0, 0, 12, 220, 792, 792, 220, 12, 0, 0, 0, 0, 0, 0, 1, 78

Offset: 0

Emeric Deutsch, May 27 2006

Sum of terms in row n is 2^n (A000079). Sum of terms in column k is A001906(k+1) (the even-indexed Fibonacci numbers). Row n contains 1+floor(n/2) nonzero terms. Sum_{k=0..n} k*T(n,k) = (3n+1)*2^(n-2) = A066373(n+1) for n>=1.
Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1/2,-1/2,0,0,0,0,0, 0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 02 2008
From R. Bagula's comment in A053122 (cf. Damianou link), the columns of this array give the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014
Odd rows contain the Pascal triangle numbers A091042. See A034867 and A034839 for some relations to tan(x). - Tom Copeland, Oct 15 2014

			The binary word 1/0/01/01/1/1/01 has 7 strictly increasing runs.
T(5,3)=6 because we have 0/01/01, 01/0/01, 01/01/0, 01/1/01, 01/01/1 and 1/01/01 (the runs are separated by /).
Triangle starts:
  1;
  0,2;
  0,1,3;
  0,0,4,4;
  0,0,1,10,5;
  0,0,0,6,20,6;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
A. Collins et al., Binary words, n-color compositions and bisection of the Fibonacci numbers, Fib. Quarterly, 51 (2013), 130-136.
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv:1110.6620 [math.RT], 2014.
R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.

Cf. A000079, A001906, A066373, A034867.
Cf. A098158. - Philippe Deléham, Dec 02 2008
Cf. A000292, A000389, A000580, A001263, A049310, A053122, A134264, A207543.

/* triangle */ [[Binomial(n+1, 2*k-n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Oct 22 2017

T:=(n,k)->binomial(n+1,2*k-n): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Table[Binomial[n + 1, 2 k - n], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 21 2016 *)

for(n=0,10, for(k=0,n, print1(binomial(n+1, 2*k-n), ", "))) \\ G. C. Greubel, Oct 22 2017

T(n,k) = binomial(n+1,2k-n).
G.f.: 1/(1 - 2*t*z - t*(1-t)*z^2).
T(n,k) = A034867(n,n-k)
From Tom Copeland, Sep 30 2011: (Start)
With K(x,t) = 1/{d/dx{x/[t-1+1/(1-x)]}} = [t-1+1/(1-x)]^2/{t-[x/(1-x)]^2}, the g.f. of A119900 = K(x*t,t)-t+1.
From formulas in A134264: K(x,t)d/dx is a generator for A001263. A refinement of A119900 to partition polynomials is given by umbralizing
  K(x,t) roughly as K(h.x,h_0) and precisely as in A134264 as
  W(x)= 1/{d/dx[f(x)]}=1/{d/dx[x/h(x)]}. (End)
T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2). - Philippe Deléham, Oct 02 2011
From Tom Copeland, Dec 07 2015: (Start)
An alternate o.g.f. is (1/(x*t)) {-1 + 1 / [1 - (1/t)[x*t/(1-x*t)]^2]} = Sum_{n>0} x^(2(n-1)+1) t^(n-1) / (1-t*x)^(2n) = x + 2t x^2 + (t+3t^2) x^3 + ... .
The n-th diagonal has elements binomial(2n+1+k,k), starting with k=0 for the first non-vanishing element, with o.g.f. (1-x)^(-2(n+1)). The first few subdiagonals are shifted versions of A000292, A000389, and A000580. Cf. A049310.
See A034867 for the matrix representation for the infinitesimal generator K(x,t) d/dx for the Narayana polynomials. (End)
From Peter Bala, Aug 17 2016: (Start)
Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:
Sum_{k = 0..p} T(p,k)*S(2*k + 1,n) = (n*(n + 1)/2)^(p+1).
For example, for row 6 we find S(7,n) + 21*S(9,n) + 35*S(11,n) + 7*S(13,n) = (n*(n + 1)/2)^7.
There appears to be a similar result for the even power sums S(2*k,n) involving A207543. (End)

Keyword tabl added by Philippe Deléham, Jan 26 2010

A052787 Product of 5 consecutive integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 120, 720, 2520, 6720, 15120, 30240, 55440, 95040, 154440, 240240, 360360, 524160, 742560, 1028160, 1395360, 1860480, 2441880, 3160080, 4037880, 5100480, 6375600, 7893600, 9687600, 11793600, 14250600, 17100720, 20389320, 24165120, 28480320, 33390720

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Appears in Harriot along with the formula (for a different offset) a(n) = n^5 + 10n^4 + 35n^3 + 50n^2 + 24n, see links. - Charles R Greathouse IV, Oct 22 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Thomas Harriot, Manuscript 6782, p. 77, c. 1599.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 744.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A002378, A007531, A052762, A000389, A054559, A173333.

[n*(n-1)*(n-2)*(n-3)*(n-4): n in [0..35]]; // Vincenzo Librandi, May 26 2011

spec := [S,{B=Set(Z),S=Prod(Z,Z,Z,Z,Z,B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
seq(numbperm (n,5), n=0..31); # Zerinvary Lajos, Apr 26 2007
G(x):=x^5*exp(x): f[0]:=G(x): for n from 1 to 31 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..31); # Zerinvary Lajos, Apr 05 2009

Times@@@(Partition[Range[-4,35],5,1])  (* Harvey P. Dale, Feb 04 2011 *)

a(n)=120*binomial(n,5) \\ Charles R Greathouse IV, Nov 20 2011

a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)=n!/(n-5)!. [Corrected by Philippe Deléham, Dec 12 2003]
a(n) = 120*A000389(n) = 4*A054559(n).
E.g.f.: x^5*exp(x).
Recurrence: {a(1)=0, a(2)=0, a(4)=0, a(3)=0, (-1-n)*a(n)+(-4+n)*a(n+1), a(5)=120}.
O.g.f.: 120*x^5/(-1+x)^6. - R. J. Mathar, Nov 16 2007
For n>5: a(n) = A173333(n,n-5). - Reinhard Zumkeller, Feb 19 2010
a(n) = a(n-1) + 5*A052762(n). - J. M. Bergot, May 30 2012
From Amiram Eldar, Mar 08 2022: (Start)
Sum_{n>=5} 1/a(n) = 1/96.
Sum_{n>=5} (-1)^(n+1)/a(n) = 2*log(2)/3 - 131/288. (End)

More terms from Henry Bottomley, Mar 20 2000

A061336 Smallest number of triangular numbers which sum to n.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 2, 2, 3, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 2, 3, 2, 1, 2, 2, 2, 3, 3, 2, 3, 1, 2, 2, 2, 3, 3, 2, 2, 3, 1, 2, 3, 2, 2, 3, 2, 3, 3, 3, 1, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 1, 2, 3, 2, 2, 3, 2, 2, 3, 3, 2, 3, 1, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 1, 2, 2, 2, 3, 3, 2, 3, 2, 2, 2, 2, 3, 3

Offset: 0

Henry Bottomley, Apr 25 2001

nonn

a(n)=3 if n=5 or 8 mod 9, since triangular numbers are {0,1,3,6} mod 9.
From Bernard Schott, Jul 16 2022: (Start)
In September 1636, Fermat, in a letter to Mersenne, made the statement that every number is a sum of at most three triangular numbers. This was proved by Gauss, who noted this event in his diary on July 10 1796 with the notation:
EYPHKA! num = DELTA + DELTA + DELTA (where Y is in fact the Greek letter Upsilon and DELTA is the Greek letter of that name).
This proof was published in his book Disquisitiones Arithmeticae, Leipzig, 1801. (End)

			a(3)=1 since 3=3, a(4)=2 since 4=1+3, a(5)=3 since 5=1+1+3, with 1 and 3 being triangular.

Elena Deza and Michel Marie Deza, Fermat's polygonal number theorem, Figurate numbers, World Scientific Publishing (2012), Chapter 5, pp. 313-377.
C. F. Gauss, Disquisitiones Arithmeticae, Yale University Press, 1966, New Haven and London, p. 342, art. 293.

Giovanni Resta, Table of n, a(n) for n = 0..10000
George E. Andrews, EYPHKA! num = Delta + Delta + Delta, J. Number Theory 23 (1986), 285-293.
Eric Weisstein's World of Mathematics, Fermat's Polygonal Number Theorem.

Cf. A000217, A007294, A057945, A061337, A051533, A020757.

Cf. A100878 (analog for A000326), A104246 (analog for A000292), A283365 (analog for A000332), A283370 (analog for A000389).

t[n_]:=n*(n+1)/2; a[0]=0; a[n_]:=Block[ {k=1, tt= t/@ Range[Sqrt[2*n]]}, Off[IntegerPartitions::take]; While[{} == IntegerPartitions[n, {k}, tt, 1], k++]; k]; a/@ Range[0, 104] (* Giovanni Resta, Jun 09 2015 *)

\\ see A283370 for generic code, working but not optimized for this case of triangular numbers. - M. F. Hasler, Mar 06 2017

a(n)=my(m=n%9,f); if(m==5 || m==8, return(3)); f=factor(4*n+1); for(i=1,#f~, if(f[i,2]%2 && f[i,1]%4==3, return(3))); if(ispolygonal(n,3), n>0, 2) \\ Charles R Greathouse IV, Mar 17 2022

a(n) = 0 if n=0, otherwise 1 if n is in A000217, otherwise 2 if n is in A051533, otherwise 3 in which case n is in A020757.

a(n) <= 3 (proposed by Fermat and proved by Gauss). - Bernard Schott, Jul 16 2022

A001753 Expansion of 1/((1+x)*(1-x)^6).

Original entry on oeis.org

1, 5, 16, 40, 86, 166, 296, 496, 791, 1211, 1792, 2576, 3612, 4956, 6672, 8832, 11517, 14817, 18832, 23672, 29458, 36322, 44408, 53872, 64883, 77623, 92288, 109088, 128248, 150008, 174624, 202368, 233529

Offset: 0

N. J. A. Sloane

Number of symmetric nonnegative integer 5 X 5 matrices with sum of elements equal to 4*n under action of dihedral group D_4.

a(n) = A108561(n+6,n) for n>0. - Reinhard Zumkeller, Jun 10 2005

			There are 5 symmetric nonnegative integer 5 X 5 matrices with sum of elements equal to 4 under action of D_4:
[1 0 0 0 1] [0 0 1 0 0] [0 0 0 0 0] [0 0 0 0 0] [0 0 0 0 0]
[0 0 0 0 0] [0 0 0 0 0] [0 1 0 1 0] [0 0 1 0 0] [0 0 0 0 0]
[0 0 0 0 0] [1 0 0 0 1] [0 0 0 0 0] [0 1 0 1 0] [0 0 4 0 0]
[0 0 0 0 0] [0 0 0 0 0] [0 1 0 1 0] [0 0 1 0 0] [0 0 0 0 0]
[1 0 0 0 1] [0 0 1 0 0] [0 0 0 0 0] [0 0 0 0 0] [0 0 0 0 0].

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-9,5,5,-9,5,-1)

Cf. A000217, A002620, A008804, A038163, A054343, A001769 (partial sums), A001752 (first differences), A169793 (binomial transf).

[(4*n^5+70*n^4+460*n^3+1400*n^2+1936*n+945)/960+(-1)^n/64: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011

CoefficientList[Series[1/((1+x)*(1-x)^6), {x, 0, 50}], x] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{5,-9,5,5,-9,5,-1},{1,5,16,40,86,166,296},40] (* Harvey P. Dale, Jun 05 2021 *)

a(n)=(4*n^5+70*n^4+460*n^3+1400*n^2+1936*n)\/960+1 \\ Charles R Greathouse IV, Apr 17 2012

a(n) = Sum{k=0..n} (-1)^(n-k)*binomial(k+5, 5); a(n) = (4*n^5 + 70*n^4 + 460*n^3 + 1400*n^2 + 1936*n + 945)/960 + (-1)^n/64. - Paul Barry, Jul 01 2003
a(n) = a(n-2) + (n*(n + 1)*(n + 2)*(n - 1))/24, a(1) = 0, a(2) = 1; (15*(-1)^n - 15*(-1)^(2*n) + 96*n - 160*(-1)^(2*n)*n + 200*n^2 - 200*(-1)^(2*n)*n^2 + 140*n^3 - 80*(-1)^(2*n)*n^3 + 40*n^4 - 10*(-1)^(2*n)*n^4 + 4*n^5)/960. - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
a(n) + a(n+1) = A000389(n+6). - R. J. Mathar, Mar 14 2011

Comment and example from Vladeta Jovovic, May 14 2000

A280880 Number T(n,k) of set partitions of [n] into exactly k blocks where sizes of distinct blocks are coprime; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 4, 6, 1, 0, 1, 15, 10, 10, 1, 0, 1, 6, 75, 20, 15, 1, 0, 1, 63, 21, 245, 35, 21, 1, 0, 1, 64, 476, 56, 630, 56, 28, 1, 0, 1, 171, 540, 2100, 126, 1386, 84, 36, 1, 0, 1, 130, 4185, 2640, 6930, 252, 2730, 120, 45, 1

Offset: 0

Alois P. Heinz, Jan 09 2017

			T(5,1) = 1: 12345.
T(5,2) = 15: 1234|5, 1235|4, 123|45, 1245|3, 124|35, 125|34, 12|345, 1345|2, 134|25, 135|24, 13|245, 145|23, 14|235, 15|234, 1|2345.
T(5,3) = 10: 123|4|5, 124|3|5, 125|3|4, 134|2|5, 135|2|4, 1|234|5, 1|235|4, 145|2|3, 1|245|3, 1|2|345.
T(5,4) = 10: 12|3|4|5, 13|2|4|5, 1|23|4|5, 14|2|3|5, 1|24|3|5, 1|2|34|5, 15|2|3|4, 1|25|3|4, 1|2|35|4, 1|2|3|45.
T(5,5) = 1: 1|2|3|4|5.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,   1;
  0, 1,   3,    1;
  0, 1,   4,    6,    1;
  0, 1,  15,   10,   10,    1;
  0, 1,   6,   75,   20,   15,    1;
  0, 1,  63,   21,  245,   35,   21,    1;
  0, 1,  64,  476,   56,  630,   56,   28,   1;
  0, 1, 171,  540, 2100,  126, 1386,   84,  36,  1;
  0, 1, 130, 4185, 2640, 6930,  252, 2730, 120, 45, 1;

Alois P. Heinz, Rows n = 0..200, flattened
Wikipedia, Coprime integers
Wikipedia, Partition of a set

Columns k=0-10 give: A000007, A057427, A194924, A280881, A280882, A280883, A280884, A280885, A280886, A280887, A280888.
T(n+k,n) for k=0-4 give: A000012, A000217, A000292, A051880(n-1) if n>0, A000389(n+4).
Row sums give A280275.
T(2n,n) gives A280889.

with(numtheory):
b:= proc(n, i, s) option remember; expand(
      `if`(n=0 or i=1, x^n, b(n, i-1, select(x->x<=i-1, s))+
      `if`(i>n or factorset(i) intersect s<>{}, 0, x*b(n-i, i-1,
      select(x->x<=i-1, s union factorset(i)))*binomial(n, i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, {})):
seq(T(n), n=0..12);

b[n_, i_, s_] := b[n, i, s] = Expand[If[n == 0 || i == 1, x^n, b[n, i - 1, Select[s, # <= i - 1 &]] + If[i > n || FactorInteger[i][[All, 1]]  ~Intersection~ s != {}, 0, x*b[n - i, i - 1, Select[ s ~Union~ FactorInteger[i][[All, 1]], # <= i - 1 &]]*Binomial[n, i]]]]; T[n_] := Function [p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n, {}]]; Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Jan 20 2017, after Alois P. Heinz *)

A325000 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1

Offset: 1

Robert A. Russell, Mar 23 2019

For n=1, the figure is a line segment with two vertices. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with four triangular faces. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Each of its n+1 facets is a regular (n-1)-dimensional simplex. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
Note that antidiagonals are part of rows of the Pascal triangle.
T(n,k-n) is the number of chiral pairs of colorings of the facets (or vertices) of a regular n-dimensional simplex using k or fewer colors. - Robert A. Russell, Sep 28 2020

			The array begins with T(1,1):
  1  3  6  10  15   21   28    36    45    55    66     78     91    105 ...
  1  4 10  20  35   56   84   120   165   220   286    364    455    560 ...
  1  5 15  35  70  126  210   330   495   715  1001   1365   1820   2380 ...
  1  6 21  56 126  252  462   792  1287  2002  3003   4368   6188   8568 ...
  1  7 28  84 210  462  924  1716  3003  5005  8008  12376  18564  27132 ...
  1  8 36 120 330  792 1716  3432  6435 11440 19448  31824  50388  77520 ...
  1  9 45 165 495 1287 3003  6435 12870 24310 43758  75582 125970 203490 ...
  1 10 55 220 715 2002 5005 11440 24310 48620 92378 167960 293930 497420 ...
  ...
For T(1,2) = 3, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses two colors. For T(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.

Robert A. Russell, Table of n, a(n) for n = 1..1275

Cf. A324999 (oriented), A325001 (achiral).
Unoriented: A007318(n,k-1) (exactly k colors), A327084 (edges, ridges), A337884 (faces, peaks), A325005 (orthotope facets, orthoplex vertices), A325013 (orthoplex facets, orthotope vertices).
Chiral: A327085 (edges, ridges), A337885 (faces, peaks), A325006 (orthotope facets, orthoplex vertices), A325014 (orthoplex facets, orthotope vertices).
Cf. A104712 (same sequence for a triangle; same sequence apart from offset).
Rows 1-4 are A000217, A000292, A000332(n+3), A000389(n+4). - Robert A. Russell, Sep 28 2020

Table[Binomial[d+1,n+1], {d,1,15}, {n,1,d}] // Flatten

T(n,k) = binomial(n+k,n+1) = A007318(n+k,n+1).
T(n,k) = Sum_{j=1..n+1} A007318(n,j-1) * binomial(k,j).
T(n,k) = A324999(n,k) + T(n,k-n) = (A324999(n,k) - A325001(n,k)) / 2 = T(n,k-n) + A325001(n,k). - Robert A. Russell, Sep 28 2020
G.f. for row n: x / (1-x)^(n+2).
Linear recurrence for row n: T(n,k) = Sum_{j=1..n+2} -binomial(j-n-3,j) * T(n,k-j).
G.f. for column k: (1 - (1-x)^k) / (x * (1-x)^k) - k.
T(n,k-n) = A324999(n,k) - T(n,k) = (A324999(n,k) - A325001(n,k)) / 2 = T(n,k) - A325001(n,k). - Robert A. Russell, Oct 10 2020

A338965 Number of unoriented colorings of the 120 dodecahedral facets of the 4-D 120-cell (or 120 vertices of the 4-D 600-cell) using subsets of a set of n colors.

Original entry on oeis.org

1, 92307499707443390526727850063504, 124792381938502167392338612231208163827413085862945471, 122697712831832245109951221276235414511846772206539032522116543043328

Offset: 1

Robert A. Russell, Dec 04 2020

Each chiral pair is counted as one when enumerating unoriented arrangements. The Schläfli symbols of the 120-cell and 600-cell are {5,3,3} and {3,3,5} respectively. They are mutually dual.
Sequences for other elements of the 120-cell and 600-cell are not suitable for the OEIS as the first significant datum is too big. We provide formulas here.
For the 600 facets of the 600-cell (vertices of the 120-cell), the formula is (960*n^20 + 1440*n^30 + 960*n^40 + 1200*n^50 + 2064*n^60 + 1440*n^66 + 40*n^100 + 1600*n^104 + 1200*n^114 + 624*n^120 + 60*n^150 + 1800*n^152 + 40*n^200 + 400*n^208 + 61*n^300 + 450*n^302 + 60*n^330 + n^600) / 14400.
For the 720 pentagonal faces of the 120-cell (edges of the 600-cell), the formula is (960 n^24 + 1440 n^36 + 960 n^48 + 1200 n^60 + 336 n^72 + 1728 n^76 + 1440 n^84 + 1640 n^120 + 1200 n^132 + 336 n^144 + 288 n^152 + 60 n^180 + 1800 n^182 + 440 n^240 + 61 n^360 + 450 n^364 + 60 n^396 + n^720) / 14400.
For the 1200 edges of the 120-cell (triangular faces of the 600-cell), the formula is (960*n^40 + 1440*n^60 + 960*n^80 + 1200*n^100 + 2064*n^120 + 1440*n^128 + 40*n^200 + 1600*n^202 + 1200*n^216 + 624*n^240 + 60*n^300 + 1800*n^302 + 40*n^400 + 400*n^404 + 61*n^600 + 450*n^604 + 60*n^640 + n^1200) / 14400.

Robert A. Russell, Table of n, a(n) for n = 1..30
Index entries for linear recurrences with constant coefficients, order 121.

Cf. A338964 (oriented), A338966 (chiral), A338967 (achiral), A338981 (exactly n colors), A000389 (5-cell), A128767 (8-cell vertices, 16-cell facets), A337957(16-cell vertices, 8-cell facets), A338949 (24-cell).

Table[(960n^4+1440n^6+960n^8+1200n^10+336n^12+288n^16+1440n^17+1440n^19+40n^20+400n^22+1200n^23+336n^24+1200n^27+60n^30+1800n^31+288n^32+40n^40+400n^44+n^60+60n^61+450n^62+60n^75+n^120)/14400,{n,10}]

a(n) = (960*n^4 + 1440*n^6 + 960*n^8 + 1200*n^10 + 336*n^12 + 288*n^16 + 1440*n^17 + 1440*n^19 + 40*n^20 + 400*n^22 + 1200*n^23 + 336*n^24 + 1200*n^27 + 60*n^30 + 1800*n^31 + 288*n^32 + 40*n^40 + 400*n^44 + n^60 + 60*n^61 + 450*n^62 + 60*n^75 +*n^120) / 14400.
a(n) = Sum_{j=1..Min(n,120)} A338981(n) * binomial(n,j).
a(n) = A338964(n) - A338966(n) =(A338964(n) + A338967(n)) / 2 = A338966(n) + A338967(n).

cp's OEIS Frontend

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