A024852 Duplicate of A023856.
2, 3, 10, 13, 28, 34, 60, 70, 110, 125, 182, 203, 280, 308, 408, 444, 570, 615, 770, 825
Offset: 2
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(2) = 3*4*5/6 = 10, the number of balls in a pyramid of 3 layers of balls, 6 in a triangle at the bottom, 3 in the middle layer and 1 on top. Consider the square array 1 2 3 4 5 6 ... 2 4 6 8 10 12 ... 3 6 9 12 16 20 ... 4 8 12 16 20 24 ... 5 10 15 20 25 30 ... ... then a(n) = sum of n-th antidiagonal. - _Amarnath Murthy_, Apr 06 2003 G.f. = x + 4*x^2 + 10*x^3 + 20*x^4 + 35*x^5 + 56*x^6 + 84*x^7 + 120*x^8 + 165*x^9 + ... Example for a(3+1) = 20 nondecreasing 3-letter words over {1,2,3,4}: 111, 222, 333; 444, 112, 113, 114, 223, 224, 122, 224, 133, 233, 144, 244, 344; 123, 124, 134, 234. 4 + 4*3 + 4 = 20. - _Wolfdieter Lang_, Jul 29 2014 Example for a(4-2) = 4 independent components of a rank 3 antisymmetric tensor A of dimension 4: A(1,2,3), A(1,2,4), A(1,3,4) and A(2,3,4). - _Wolfdieter Lang_, Dec 10 2015
a:=n->Binomial(n+2,3);; A000292:=List([0..50],n->a(n)); # Muniru A Asiru, Feb 28 2018
a000292 n = n * (n + 1) * (n + 2) `div` 6 a000292_list = scanl1 (+) a000217_list -- Reinhard Zumkeller, Jun 16 2013, Feb 09 2012, Nov 21 2011
[n*(n+1)*(n+2)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 03 2014
a:=n->n*(n+1)*(n+2)/6; seq(a(n), n=0..50); A000292 := n->binomial(n+2,3); seq(A000292(n), n=0..50); isA000292 := proc(n) option remember; local a,i ; for i from iroot(6*n,3)-1 do a := A000292(i) ; if a > n then return false; elif a = n then return true; end if; end do: end proc: # R. J. Mathar, Aug 14 2024
Table[Binomial[n + 2, 3], {n, 0, 20}] (* Zerinvary Lajos, Jan 31 2010 *) Accumulate[Accumulate[Range[0, 50]]] (* Harvey P. Dale, Dec 10 2011 *) Table[n (n + 1)(n + 2)/6, {n,0,100}] (* Wesley Ivan Hurt, Sep 25 2013 *) Nest[Accumulate, Range[0, 50], 2] (* Harvey P. Dale, May 24 2017 *) Binomial[Range[20] + 1, 3] (* Eric W. Weisstein, Sep 08 2017 *) LinearRecurrence[{4, -6, 4, -1}, {0, 1, 4, 10}, 20] (* Eric W. Weisstein, Sep 08 2017 *) CoefficientList[Series[x/(-1 + x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *) Table[Range[n].Range[n,1,-1],{n,0,50}] (* Harvey P. Dale, Mar 02 2024 *)
A000292(n):=n*(n+1)*(n+2)/6$ makelist(A000292(n),n,0,60); /* Martin Ettl, Oct 24 2012 */
a(n) = (n) * (n+1) * (n+2) / 6 \\ corrected by Harry J. Smith, Dec 22 2008
a=vector(10000);a[2]=1;for(i=3,#a,a[i]=a[i-2]+i*i); \\ Stanislav Sykora, Nov 07 2013
is(n)=my(k=sqrtnint(6*n,3)); k*(k+1)*(k+2)==6*n \\ Charles R Greathouse IV, Dec 13 2016
# Compare A000217. def A000292(): x, y, z = 1, 1, 1 yield 0 while True: yield x x, y, z = x + y + z + 1, y + z + 1, z + 1 a = A000292(); print([next(a) for i in range(45)]) # Peter Luschny, Aug 03 2019
a023855 n = sum $ zipWith (*) [1 .. div (n+1) 2] [n, n-1 ..] -- Reinhard Zumkeller, Jan 23 2012
[(4*n^3 +15*n^2 +14*n +3 -3*(n+1)^2*(-1)^n)/48: n in [1..60]]; // G. C. Greubel, Jul 12 2022
seq(-(1/3)*floor((k+1)/2)^3 + (k/2)*floor((k+1)/2)^2 + ((3*k+2)/6)*floor((k+1)/2), k=1..100); # Wesley Ivan Hurt, Sep 18 2013
LinearRecurrence[{1,3,-3,-3,3,1,-1}, {1,2,7,10,22,28,50}, 60] (* Vincenzo Librandi, Jan 23 2012 *) Table[-Ceiling[n/2] (Ceiling[n/2] + 1) (2 Ceiling[n/2] - 3 n - 2)/6, {n, 100}] (* Wesley Ivan Hurt, Sep 20 2013 *)
a(n)=if(n%2, (n+1)*(n+3)*(2*n+1)/24, n*(n+1)*(n+2)/12)
my(x='x+O('x^99)); Vec(x*(1+x+2*x^2)/((1-x)^4*(1+x)^3)) \\ Altug Alkan, Mar 03 2018
[(4*n^3 +15*n^2 +14*n +3 -3*(n+1)^2*(-1)^n)/48 for n in (1..60)] # G. C. Greubel, Jul 12 2022
List([1..60], n-> (4*n^3 +27*n^2 +50*n +21 -3*(n^2+6*n+7)*(-1)^n)/48) # G. C. Greubel, Jun 12 2019
[(4*n^3 +27*n^2 +50*n +21 -3*(n^2+6*n+7)*(-1)^n)/48: n in [1..60]]; // G. C. Greubel, Jun 12 2019
seq(sum(i*(n-i+3), i=1..ceil(n/2)), n=1..60); # Wesley Ivan Hurt, Sep 20 2013
Table[-Ceiling[n/2]*(Ceiling[n/2]+1)*(2*Ceiling[n/2]-3n-8)/6, {n,60}] (* Wesley Ivan Hurt, Sep 20 2013 *) LinearRecurrence[{1,3,-3,-3,3,1,-1},{3,4,13,16,34,40,70},60] (* Harvey P. Dale, Feb 13 2018 *)
a(n) = (4*n^3 +27*n^2 +50*n +21 -3*(n^2+6*n+7)*(-1)^n)/48; \\ G. C. Greubel, Jun 12 2019
[(4*n^3 +27*n^2 +50*n +21 -3*(n^2+6*n+7)*(-1)^n)/48 for n in (1..60)] # G. C. Greubel, Jun 12 2019
b:= func< n | (1-(-1)^n)/2 >; [(2*n^3 + 3*(6 +b(n))*n^2 + 2*(14 +9*b(n))*n + 27*b(n))/24 : n in [1..50]] // G. C. Greubel, Jul 12 2022
seq(sum((i+1)*(k-i+2), i=1..ceil(k/2)), k=1..70); # Wesley Ivan Hurt, Sep 20 2013
Table[Ceiling[n/2]*(-2*Ceiling[n/2]^2+3n*Ceiling[n/2]+9n+14)/6,{n,100}] (* Wesley Ivan Hurt, Sep 20 2013 *)
def b(n): return (1-(-1)^n)/2 [(2*n^3 + 3*(6 +b(n))*n^2 + 2*(14 +9*b(n))*n + 27*b(n))/24 for n in (1..50)] # G. C. Greubel, Jul 12 2022
[(4*n^3+21*n^2+14*n-9+3*(n^2+6*n+3)*(-1)^n)/48: n in [2..60]]; // Vincenzo Librandi, Oct 31 2014
seq(sum(i*(k-i+3), i=1..floor(k/2)), k=2..70); # Wesley Ivan Hurt, Sep 20 2013
Table[-Floor[n/2] * (Floor[n/2] + 1) * (2 * Floor[n/2] - 3n - 8)/6, {n, 2, 100}] (* Wesley Ivan Hurt, Sep 20 2013 *) CoefficientList[Series[- (- 4 - x + x^2)/((1 + x)^3 (x - 1)^4), {x, 0, 60}], x] (* Vincenzo Librandi, Oct 31 2014 *)
[(4*n^3+21*n^2+14*n-9+3*(n^2+6*n+3)*(-1)^n)/48 for n in (2..60)] # G. C. Greubel, Jul 13 2022
List([2..50], n-> (8*n^3 +6*n^2 -2*n -3 +3*(-1)^n*(2*n+1))/48); # G. C. Greubel, Dec 12 2019
a = @(n) 1/4*(n*(n+1)*(4*n-1)/6 + (-1)^n*floor((n+1)/2)); % Néstor Jofré, Apr 24 2017
[n^3/6+n^2/8-n/24-1/16+(-1)^n*(n/8+1/16): n in [2..50]]; // Vincenzo Librandi, Apr 25 2017
seq( (8*n^3 +6*n^2 -2*n -3 +3*(-1)^n*(2*n+1))/48, n=2..50); # G. C. Greubel, Dec 12 2019
T[m_, n_]:=(m+n)(m-n+1)/2; Table[Sum[T[n, 2k], {k, Floor[n/2]}], {n, 2, 50}] (* Indranil Ghosh, Apr 24 2017 *) LinearRecurrence[{2, 1, -4, 1, 2, -1}, {2, 5, 13, 23, 41, 62}, 50] (* Vincenzo Librandi, Apr 25 2017 *)
a(n)=(4*n^3 + 3*n^2 + 2*n - if(n%2,6*n+3))/24 \\ Charles R Greathouse IV, Jun 08 2015
[(8*n^3 +6*n^2 -2*n -3 +3*(-1)^n*(2*n+1))/48 for n in (2..50)] # G. C. Greubel, Dec 12 2019
[19*n/24-9/16+n^3/12+11*n^2/16+(-1)^n*(3*n/8 +9/16+n^2/16): n in [2..50]]; // Vincenzo Librandi, Sep 26 2013
seq(sum((i+1)*(k-i+2), i=1..floor(k/2)), k=2..70); # Wesley Ivan Hurt, Sep 20 2013
Table[Floor[n/2] (-2Floor[n/2]^2 +3n*Floor[n/2] +9n +14)/6, {n, 2, 100}] (* Wesley Ivan Hurt, Sep 20 2013 *) CoefficientList[Series[(6 +2x -4x^2 -x^3 +x^4)/((1+x)^3 (1-x)^4), {x, 0, 60}], x] (* Vincenzo Librandi, Sep 26 2013 *) LinearRecurrence[{1,3,-3,-3,3,1,-1},{6,8,22,27,52,61,100},50] (* Harvey P. Dale, Aug 11 2023 *)
[(1/48)*(4*n^3 +33*n^2 +38*n -27 +3*(-1)^n*(n+3)^2) for n in (2..60)] # G. C. Greubel, Jul 13 2022
Irregular triangle: 2, 3, 6, 4, 8, 5, 12, 10, 6, 15, 12, 7, 20, 18, 14, 8, 24, 21, 16, 9, 30, 28, 24, 18, 10, 35, 32, 27, 20, 11, etc.
(Table[n^2 - k^2, {n, 3, 18}, {k, n}] /. m_ /; Or[OddQ@ m, m == 0] -> Nothing)/4 // Flatten (* Michael De Vlieger, Apr 14 2016 *)
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