A000389 -id:A000389 - cp's OEIS frontend

A334581 Number of ways to choose 3 points that form an equilateral triangle from the A000292(n) points in a regular tetrahedral grid of side length n.

0, 0, 4, 24, 84, 224, 516, 1068, 2016, 3528, 5832, 9256, 14208, 21180, 30728, 43488, 60192, 81660, 108828, 142764, 184708, 236088, 298476, 373652, 463524, 570228, 696012, 843312, 1014720, 1213096, 1441512, 1703352, 2002196, 2341848, 2726400, 3160272, 3648180

Offset: 0

Peter Kagey, May 06 2020

nonn

a(n) >= 4 * A269747(n).
a(n) >= 4 * A000389(n+3) = A210569(n+2).
a(n) >= 4 * (n-1) + 4 * a(n-1) - 6 * a(n-2) + 4 * a(n-3) - a(n-4) for n >= 4.

Peter Kagey, Table of n, a(n) for n = 0..1000 (Computed via Anders Kaseorg's program in the Stack Exchange link.)
Code Golf Stack Exchange, Triangles in a tetrahedron

Cf. A000292, A000389, A210569.
Cf. A000332 (equilateral triangles in triangular grid), A269747 (regular tetrahedra in a tetrahedral grid), A102698 (equilateral triangles in cube), A103158 (regular tetrahedra in cube).
Cf. A077435, A085582, A187452, A189412-A189418, A271910.

A346893 Expansion of e.g.f. 1 / (1 - x^5 * exp(x) / 5!).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 6, 21, 56, 126, 504, 6006, 67320, 577863, 4038034, 24975951, 165481680, 1553590220, 19495772856, 249507077436, 2910465717648, 31103684847837, 326286335505438, 3766644374319673, 51399738264984648, 785038533451101930

Offset: 0

Ilya Gutkovskiy, Aug 06 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..498

Column k=5 of A351703.
Cf. A006153, A346888, A346889, A346890.
Cf. A000389, A145455.

nmax = 25; CoefficientList[Series[1/(1 - x^5 Exp[x]/5!), {x, 0, nmax}], x] Range[0, nmax]!
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] Binomial[k, 5] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 25}]

my(x='x+O('x^25)); Vec(serlaplace(1/(1-x^5*exp(x)/5!))) \\ Michel Marcus, Aug 06 2021

a(n) = n!*sum(k=0, n\5, k^(n-5*k)/(120^k*(n-5*k)!)); \\ Seiichi Manyama, May 13 2022

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * binomial(k,5) * a(n-k).
a(n) ~ n! / ((1 + LambertW(2^(3/5)*3^(1/5)/5^(4/5))) * 5^(n+1) * LambertW(2^(3/5)*3^(1/5)/5^(4/5))^n). - Vaclav Kotesovec, Aug 08 2021
a(n) = n! * Sum_{k=0..floor(n/5)} k^(n-5*k)/(120^k * (n-5*k)!). - Seiichi Manyama, May 13 2022

A017764 a(n) = binomial coefficient C(n,100).

Original entry on oeis.org

1, 101, 5151, 176851, 4598126, 96560646, 1705904746, 26075972546, 352025629371, 4263421511271, 46897636623981, 473239787751081, 4416904685676756, 38393094575497956, 312629484400483356, 2396826047070372396, 17376988841260199871, 119594570260437846171

Offset: 100

N. J. A. Sloane

More generally, the ordinary generating function for the binomial coefficients C(n,k) is x^k/(1 - x)^(k+1). - Ilya Gutkovskiy, Mar 21 2016

G. C. Greubel, Table of n, a(n) for n = 100..1100

Cf. similar sequences of the binomial coefficients C(n,k): A000012 (k = 0), A001477 (k = 1), A000217 (k = 2), A000292 (k = 3), A000332 (k = 4), A000389 (k = 5), A000579-A000582 (k = 6..9) A001287 (k = 10), A001288 (k = 11), A010965-A011001 (k = 12..48), A017713-A017763 (k = 49..99), this sequence (k = 100).

Cf. A001787, A242091.

[Binomial(n,100): n in [100..130]]; // G. C. Greubel, Nov 24 2017

Table[Binomial[n, 100], {n, 100, 5!}] (* Vladimir Joseph Stephan Orlovsky, Sep 25 2008 *)

a(n)=binomial(n,100) \\ Charles R Greathouse IV, Jun 28 2012

A017764_list, m = [], [1]*101
for _ in range(10**2):
    A017764_list.append(m[-1])
    for i in range(100):
        m[i+1] += m[i] # Chai Wah Wu, Jan 24 2016

[binomial(n, 100) for n in range(100,115)] # Zerinvary Lajos, May 23 2009

G.f.: x^100/(1 - x)^101. - Ilya Gutkovskiy, Mar 21 2016
E.g.f.: x^100 * exp(x)/(100)!. - G. C. Greubel, Nov 24 2017
From Amiram Eldar, Dec 20 2020: (Start)
Sum_{n>=100} 1/a(n) = 100/99.
Sum_{n>=100} (-1)^n/a(n) = A001787(100)*log(2) - A242091(100)/99! = 63382530011411470074835160268800*log(2) - 1914409165727592211172313915606932788039791776845041612575266508424929 / 43575234518570298227833630584570189723 = 0.9902877001... (End)

A051747 a(n) = n(n+1)(n+2)(n^2+7n+32)/120.

Original entry on oeis.org

2, 10, 31, 76, 161, 308, 546, 912, 1452, 2222, 3289, 4732, 6643, 9128, 12308, 16320, 21318, 27474, 34979, 44044, 54901, 67804, 83030, 100880, 121680, 145782, 173565, 205436, 241831, 283216, 330088, 382976, 442442, 509082, 583527, 666444, 758537

Offset: 1

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 07 1999

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000217, A000389, A005583.

[n*(n+1)*(n+2)*(n^2+7*n+32)/120: n in [1..40]]; // Vincenzo Librandi, Jun 15 2011

Table[(1/120)*n*(n + 1)*(n + 2)*(n^2 + 7*n + 32), {n, 60}] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{2,10,31,76,161,308},60] (* Harvey P. Dale, Oct 03 2012 *)

conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w; t(n)=n*(n+1)/2; u=vector(10,i,t(i)); v=vector(10,i,t(i)+1); conv(u,v)

Vec(x*(x^2-2*x+2)/(x-1)^6 + O(x^100)) \\ Colin Barker, Mar 18 2015

a(n) = binomial(n+4, n-1)+binomial(n+2, n-1).
Convolution of triangular numbers with triangular numbers + 1, i.e. [1, 3, 6, 10, 15, 21, ...] with [2, 4, 7, 11, 16, 22, ...].
a(1)=2, a(2)=10, a(3)=31, a(4)=76, a(5)=161, a(6)=308, 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). - Harvey P. Dale, Oct 03 2012
G.f.: x*(x^2-2*x+2) / (x-1)^6. - Colin Barker, Mar 18 2015

A053643 a(n) = ceiling(binomial(n,6)/n).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 4, 10, 21, 42, 77, 132, 215, 334, 501, 728, 1032, 1428, 1938, 2584, 3392, 4389, 5609, 7084, 8855, 10964, 13455, 16380, 19793, 23751, 28319, 33563, 39556, 46376, 54106, 62832, 72650, 83657, 95960, 109668, 124900

Offset: 1

N. J. A. Sloane, Mar 25 2000

Robert Israel, Table of n, a(n) for n = 1..10000
R. L. Graham and N. J. A. Sloane, Lower bounds for constant weight codes, IEEE Trans. Inform. Theory, 26 (1980), 37-43.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1,0,0,0,0,0,0,0,0,0,0,0,0,1,-5,10,-10,5,-1,0,0,0,0,0,0,0,0,0,0,0,0,-1,5,-10,10,-5,1,0,0,0,0,0,0,0,0,0,0,0,0,1,-5,10,-10,5,-1).

Cf. A000389, A000579, A053618.

[Ceiling(Binomial(n,6)/n) : n in [1..50]]; // Wesley Ivan Hurt, Nov 01 2015

seq(ceil(binomial(n,5)/6), n=0..50); # Zerinvary Lajos, Jan 12 2009

Table[Ceiling[Binomial[n, 6]/n], {n, 50}] (* Michael De Vlieger, Nov 01 2015 *)

vector(50, n, ceil(binomial(n, 6)/n)) \\ Altug Alkan, Nov 01 2015

[ceil(binomial(n,6)/n) for n in (1..50)] # G. C. Greubel, May 17 2019

From Robert Israel, Nov 01 2015: (Start)
a(n) = ceiling(A000389(n-1)/6).
G.f.: (x^52 -2*x^51 +4*x^50 -4*x^49 +2*x^48 +x^47 -x^46 +2*x^44 -2*x^43 +x^42 -x^41 +4*x^40 -5*x^39 +4*x^38 -2*x^37 +3*x^36 -5*x^35 +5*x^34 -2*x^33 -2*x^32 +5*x^31 -5*x^30 +2*x^29 +2*x^28 -5*x^27 +5*x^26 -2*x^25 -2*x^24 +5*x^23 -5*x^22 +2*x^21 +2*x^20 -5*x^19 +5*x^18 -3*x^17 +4*x^16 -7*x^15 +9*x^14 -7*x^13 +5*x^12 -4*x^11 +4*x^10 -2*x^9 +3*x^7 -4*x^6 +x^5 +6*x^4 -10*x^3 +9*x^2 -4*x +1)*x^6/((x -1)^6*(x +1)*(x^4 +1)*(x^2 +x +1)*(x^2 -x +1)*(x^6 +x^3 +1)*(x^6 -x^3 +1)*(x^8 -x^4 +1)*(x^24 -x^12 +1)).
(End)

A095704 Triangle read by rows giving coefficients of the trigonometric expansion of sin(n*x).

Original entry on oeis.org

1, 2, 0, 3, 0, -1, 4, 0, -4, 0, 5, 0, -10, 0, 1, 6, 0, -20, 0, 6, 0, 7, 0, -35, 0, 21, 0, -1, 8, 0, -56, 0, 56, 0, -8, 0, 9, 0, -84, 0, 126, 0, -36, 0, 1, 10, 0, -120, 0, 252, 0, -120, 0, 10, 0, 11, 0, -165, 0, 462, 0, -330, 0, 55, 0, -1, 12, 0, -220, 0, 792, 0, -792, 0, 220, 0, -12, 0, 13, 0, -286, 0, 1287, 0

Offset: 1

Robert G. Wilson v, Jul 06 2004

			The trigonometric expansion of sin(4x) is 4*cos(x)^3*sin(x) - 4*cos(x)*sin(x)^3, so the fourth row is 4, 0, -4, 0.
Triangle begins:
1
2 0
3 0 -1
4 0 -4 0
5 0 -10 0 1
6 0 -20 0 6 0
7 0 -35 0 21 0 -1
8 0 -56 0 56 0 -8 0

Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4, section 5.

First column is A000027 = C(n, 1), third column is A000292 = C(n, 3), fifth column is A000389 = C(n, 5), seventh column is A000580 = C(n, 7), ninth column is A000582 = C(n, 9).
A001288 = C(n, 11), A010966 = C(n, 13), A010968 = C(n, 15), A010970 = C(n, 17), A010972 = C(n, 19),
A010974 = C(n, 21), A010976 = C(n, 23), A010978 = C(n, 25), A010980 = C(n, 27), A010982 = C(n, 29),
A010984 = C(n, 31), A010986 = C(n, 33), A010988 = C(n, 35), A010990 = C(n, 37), A010992 = C(n, 39),
A010994 = C(n, 41), A010996 = C(n, 43), A010998 = C(n, 45), A011000 = C(n, 47), A017713 = C(n, 49)
Another version of the triangle in A034867. Cf. A096754.
A017715 = C(n, 51), A017717 = C(n, 53), A017719 = C(n, 55), A017721 = C(n, 57), etc.

Flatten[ Table[ Plus @@ CoefficientList[ TrigExpand[ Sin[n*x]], {Sin[x], Cos[x]}], {n, 13}]]

T(n,k) = C(n+1,k+1)*sin(Pi*(k+1)/2). - Paul Barry, May 21 2006

A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 9, 10, 5, 1, 1, 6, 10, 16, 15, 6, 1, 1, 5, 18, 20, 25, 21, 7, 1, 1, 8, 15, 40, 35, 36, 28, 8, 1, 1, 9, 27, 35, 75, 56, 49, 36, 9, 1

Offset: 1

Alford Arnold, Mar 29 2007

The array can be constructed by beginning with A007318 (Pascal's triangle) placing each diagonal on a prime row. The other rows are filled in by mapping the prime factorization of the row number to the known sequences on the prime rows and multiplying term by term.

			Row six begins 1 6 18 40 75 126 ... because rows two and three are
1 2 3 4 5 6 ...
1 3 6 10 15 21 ...
The array begins
1 1 1 1 1 1 1 1 1 A000012
1 2 3 4 5 6 7 8 9 A000027
1 3 6 10 15 21 28 36 45 A000217
1 4 9 16 25 36 49 64 81 A000290
1 4 10 20 35 56 84 120 165 A000292
1 6 18 40 75 126 196 288 405 A002411
1 5 15 35 70 126 210 330 495 A000332
1 8 27 64 125 216 343 512 729 A000578
1 9 36 100 225 441 784 1296 2025 A000537
1 8 30 80 175 336 588 960 1485 A002417
1 6 21 56 126 252 462 792 1287 A000389
1 12 54 160 375 756 1372 2304 3645 A019582
1 7 28 84 210 462 924 1716 3003 A000579
1 10 45 140 350 756 1470 2640 4455 A027800
1 12 60 200 525 1176 2352 4320 7425 A004302
1 16 81 256 625 1296 2401 4096 6561 A000583
1 8 36 120 330 792 1716 3432 6435 A000580
1 18 108 400 1125 2646 5488 10368 18225 A019584
1 9 45 165 495 1287 3003 6435 12870 A000581
1 16 90 320 875 2016 4116 7680 13365 A119771
1 15 90 350 1050 2646 5880 11880 22275 A001297
1 12 63 224 630 1512 3234 6336 11583 A027810
1 10 55 220 715 2002 5005 11440 24310 A000582
1 24 162 640 1875 4536 9604 18432 32805 A019583
1 16 100 400 1225 3136 7056 14400 27225 A001249
1 14 84 336 1050 2772 6468 13728 27027 A027818
1 27 216 1000 3375 9261 21952 46656 91125 A059827
1 20 135 560 1750 4536 10290 21120 40095 A085284

Cf. A000040 A007318.

Cf. A064553 (second diagonal), A080688 (second diagonal resorted).

A128629 := proc(n,m) if n = 1 then 1; elif isprime(n) then p := numtheory[pi](n) ; binomial(p+m-1,p) ; else a := 1 ; for p in ifactors(n)[2] do a := a* procname(op(1,p),m)^ op(2,p) ; od: fi; end: # R. J. Mathar, Sep 09 2009

A-number added to each row of the examples by R. J. Mathar, Sep 09 2009

A167566 The third left hand column of triangle A167565.

Original entry on oeis.org

2, 16, 67, 202, 497, 1064, 2058, 3684, 6204, 9944, 15301, 22750, 32851, 46256, 63716, 86088, 114342, 149568, 192983, 245938, 309925, 386584, 477710, 585260, 711360, 858312, 1028601, 1224902, 1450087, 1707232, 1999624, 2330768, 2704394

Offset: 3

Johannes W. Meijer, Nov 10 2009

G. C. Greubel, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Equals the third left hand column of triangle A167565.

Other left hand columns are A000027, A000292, A167567 and A168304.

Table[(7*n^5 - 30*n^4 + 45*n^3 - 30*n^2 + 8*n)/5!, {n,3,100}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1}, {2, 16, 67, 202, 497, 1064}, 100] (* G. C. Greubel, Jun 16 2016 *)

Vec((1*z^2 + 4*z + 2)/(1-z)^6 + O(z^50)) \\ Michel Marcus, Jul 05 2017

a(n) = n*(7*n^4 - 30*n^3 + 45*n^2 - 30*n + 8)/120 \\ Charles R Greathouse IV, Jul 14 2017

From Johannes W. Meijer, Nov 23 2009: (Start)
a(n) = (7*n^5 - 30*n^4 + 45*n^3 - 30*n^2 + 8*n)/5!.
G.f.: (1*z^2 + 4*z + 2)/(1-z)^6.
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).
a(n) - 5*a(n-1) + 10*a(n-2) - 10*a(n-3) + 5*a(n-4) - a(n-5) = 7. (End)
a(n) = A024166(n-2) + A000389(n+2). - J. M. Bergot, Jul 04 2017

A175114 First differences of A175113.

Original entry on oeis.org

1, 364, 7448, 51012, 206896, 620060, 1527624, 3281908, 6373472, 11454156, 19360120, 31134884, 48052368, 71639932, 103701416, 146340180, 201982144, 273398828, 363730392, 476508676, 615680240, 785629404, 991201288, 1237724852

Offset: 0

R. J. Mathar, Feb 13 2010

Convolution of the finite sequence 1,358,5279,11764,5279,358,1 with A000389. Number of points in the standard root system of the D_6 lattice having L_infinity norm n.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A110907, A117216, A175112.

I:=[1,364,7448,51012,206896,620060,1527624]; [n le 7 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Dec 20 2012

CoefficientList[Series[(358 x + 5279 x^2 + 11764 x^3 + 5279 x^4 + 358 x^5 + 1+x^6)/(x - 1)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 20 2012 *)

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), n>6.
a(n) = ((2*n+1)^6-(2*n-1)^6)/2 = 4*n*(12*n^2+1)*(4*n^2+3), n>0. - Bruno Berselli, Dec 27 2010
G.f.: (358*x+5279*x^2+11764*x^3+5279*x^4+358*x^5+1+x^6)/(x-1)^6. - R. J. Mathar, Jan 03 2011

A190049 Expansion of (16+24x+2x^2)/(x-1)^6.

Original entry on oeis.org

16, 120, 482, 1412, 3402, 7168, 13692, 24264, 40524, 64504, 98670, 145964, 209846, 294336, 404056, 544272, 720936, 940728, 1211098, 1540308, 1937474, 2412608, 2976660, 3641560, 4420260, 5326776, 6376230, 7584892

Offset: 0

Johannes W. Meijer, May 06 2011

Equals the sixth right hand column of A175136.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A175136, A162148, A190048.

Related to A000389 and A091894.

[(21*n^5+245*n^4+1105*n^3+2395*n^2+2474*n+960)/60: n in [0..50]]; // Vincenzo Librandi, May 07 2011

A190049 := proc(n) option remember; a(n):=(21*n^5 +245*n^4 +1105*n^3 +2395*n^2 +2474*n +960)/60 end: seq(A190049(n),n=0..27);

LinearRecurrence[{6,-15,20,-15,6,-1}, {16,120,482,1412,3402,7168}, 30] (* or *) CoefficientList[Series[(16 +24*x +2*x^2)/(1-x)^6, {x, 0, 50}], x] (* G. C. Greubel, Jan 10 2018 *)

x='x+O('x^30); Vec((16 +24*x +2*x^2)/(1-x)^6) \\ G. C. Greubel, Jan 10 2018

for(n=0,30, print1((21*n^5 +245*n^4 +1105*n^3 +2395*n^2 +2474*n +960)/60, ", ")) \\ G. C. Greubel, Jan 10 2018

G.f.: (16 +24*x +2*x^2)/(1-x)^6.
a(n) = 16*binomial(n+5,5) +24*binomial(n+4,5) +2*binomial(n+3,5).
a(n) = A091894(5,0)*binomial(n+5,5) + A091894(5,1)*binomial(n+4,5) + A091894(5,2)*binomial(n+3,5).
a(n) = (21*n^5 +245*n^4 +1105*n^3 +2395*n^2 +2474*n +960)/60.

cp's OEIS Frontend

A334581 Number of ways to choose 3 points that form an equilateral triangle from the A000292(n) points in a regular tetrahedral grid of side length n.

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A346893 Expansion of e.g.f. 1 / (1 - x^5 * exp(x) / 5!).

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A017764 a(n) = binomial coefficient C(n,100).

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A051747 a(n) = n*(n+1)*(n+2)*(n^2+7*n+32)/120.

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A053643 a(n) = ceiling(binomial(n,6)/n).

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A095704 Triangle read by rows giving coefficients of the trigonometric expansion of sin(n*x).

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A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

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A167566 The third left hand column of triangle A167565.

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A051747 a(n) = n(n+1)(n+2)(n^2+7n+32)/120.

A190049 Expansion of (16+24x+2x^2)/(x-1)^6.