A085437 Duplicate of A024166.
0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088
Offset: 0
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.
For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27. For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - _Patrick J. McNab_, Mar 28 2016
a000578 = (^ 3) a000578_list = 0 : 1 : 8 : zipWith (+) (map (+ 6) a000578_list) (map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list) -- Reinhard Zumkeller, Sep 05 2015, May 24 2012, Oct 22 2011
[ n^3 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014
I:=[0,1,8,27]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jul 05 2014
A000578 := n->n^3; seq(A000578(n), n=0..50); isA000578 := proc(r) local p; if r = 0 or r =1 then true; else for p in ifactors(r)[2] do if op(2, p) mod 3 <> 0 then return false; end if; end do: true ; end if; end proc: # R. J. Mathar, Oct 08 2013
Table[n^3, {n, 0, 30}] (* Stefan Steinerberger, Apr 01 2006 *) CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *) Accumulate[Table[3n^2+3n+1,{n,0,20}]] (* or *) LinearRecurrence[{4,-6,4,-1},{1,8,27,64},20](* Harvey P. Dale, Aug 18 2018 *)
A000578(n):=n^3$ makelist(A000578(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
A000578(n)=n^3 \\ M. F. Hasler, Apr 12 2008
is(n)=ispower(n,3) \\ Charles R Greathouse IV, Feb 20 2012
A000578_list, m = [], [6, -6, 1, 0] for _ in range(10**2): A000578_list.append(m[-1]) for i in range(3): m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015
(define (A000578 n) (* n n n)) ;; Antti Karttunen, Oct 06 2017
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
G.f. = x + 9*x^2 + 36*x^3 + 100*x^4 + 225*x^5 + 441*x^6 + ... - _Michael Somos_, Aug 29 2022
List([0..40],n->(n*(n+1)/2)^2); # Muniru A Asiru, Dec 05 2018
a000537 = a000290 . a000217 -- Reinhard Zumkeller, Mar 26 2015
[(n*(n+1)/2)^2: n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
a:= n-> (n*(n+1)/2)^2: seq(a(n), n=0..40);
Accumulate[Range[0, 50]^3] (* Harvey P. Dale, Mar 01 2011 *) f[n_] := n^2 (n + 1)^2/4; Array[f, 39, 0] (* Robert G. Wilson v, Nov 16 2012 *) Table[CycleIndex[{{1, 2, 3, 4}, {3, 2, 1, 4}, {1, 4, 3, 2}, {3, 4, 1, 2}}, s] /. Table[s[i] -> n, {i, 1, 2}], {n, 0, 30}] (* Geoffrey Critzer, Jun 18 2014 *) Accumulate @ Range[0, 50]^2 (* Waldemar Puszkarz, Jan 24 2015 *) Binomial[Range[20], 2]^2 (* Eric W. Weisstein, Jan 02 2018 *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 9, 36, 100}, 20] (* Eric W. Weisstein, Jan 02 2018 *) CoefficientList[Series[-((x (1 + 4 x + x^2))/(-1 + x)^5), {x, 0, 20}], x] (* Eric W. Weisstein, Jan 02 2018 *)
a(n)=(n*(n+1)/2)^2
def A000537(n): return (n*(n+1)>>1)**2 # Chai Wah Wu, Oct 20 2023
Northwest corner (the array is read by southwest falling antidiagonals): 1, 4, 10, 20, 35, 56, 84, ... 2, 7, 16, 30, 50, 77, 112, ... 3, 10, 22, 40, 65, 98, 140, ... 4, 13, 28, 50, 80, 119, 168, ... 5, 16, 34, 60, 95, 140, 196, ... 6, 19, 40, 70, 110, 161, 224, ... T(6,1) = (1)**(6) = 6; T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19; T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
b[n_] := n; c[n_] := n t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}] TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]] Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]] r[n_] := Table[t[n, k], {k, 1, 60}] (* A213500 *)
t(n,k) = sum(i=0, k - 1, (k - i) * (n + i)); tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););}; tabl(12) \\ Indranil Ghosh, Mar 26 2017
def t(n, k): return sum((k - i) * (n + i) for i in range(k)) for n in range(1, 13): print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017
a001715 = (flip div 6) . a000142 -- Reinhard Zumkeller, Aug 31 2014
[Factorial(n)/6: n in [3..30]]; // Vincenzo Librandi, Jun 20 2011
f := proc(n) n!/6; end; BB:= [S, {S = Prod(Z,Z,C), C = Union(B,Z,Z), B = Prod(Z,C)}, labelled]: seq(combstruct[count](BB, size=n)/12, n=3..20); # Zerinvary Lajos, Jun 19 2008 G(x):=1/(1-x)^4: f[0]:=G(x): for n from 1 to 18 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..16); # Zerinvary Lajos, Apr 01 2009
a[n_]:=n!/6; (*Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *) Range[3,30]!/6 (* Harvey P. Dale, Aug 12 2012 *)
a(n)=n!/6 \\ Charles R Greathouse IV, Jan 12 2012
Row 3 contains 1,8,19,18,6, so Sum_{i=1..n} C(i+1,2)^3 = (n+2) * C(n+1,2) * [ a(1,3)/3 + a(2,3)*C(n-1,1)/4 + a(3,3)*C(n-1,2)/5 + a(4,3)*C(n-1,3)/6 + a(5,3)*C(n-1,4)/7 ] = [ (n+2)*(n+1)*n/2 ] * [ 1/3 + (8/4)*C(n-1,1) + (19/5)*C(n-1,2) + (18/6)*C(n-1,3) + (6/7)*C(n-1,4). Cf. A085438 for more details. From _Peter Bala_, Mar 08 2018: (Start) Table begins n=0 |1 n=1 |1 2 1 n=2 |1 8 19 18 6 n=3 |1 26 163 432 564 360 90 n=4 |1 80 1135 6354 18078 28800 26100 12600 2520 ... Row 2: C(i+2,2)^2 = C(i,0) + 8*C(i,1) + 19*C(i,2) + 18*C(i,3) + 6*C(i,4). Hence, Sum_{i = 0..n-1} C(i+2,2)^2 = C(n,1) + 8*C(n,2) + 19*C(n,3) + 18*C(n,4) + 6*C(n,5). (End)
Flat(List([0..6],n->List([0..2*n],k->Sum([0..k],i->(-1)^(k-i)*Binomial(k,i)*Binomial(i+2,2)^n)))); # Muniru A Asiru, Mar 22 2018
seq(seq(add( (-1)^(k-i)*binomial(k,i)*binomial(i+2,2)^n, i = 0..k), k = 0..2*n), n = 0..8); # Peter Bala, Mar 08 2018
a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 3, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 2, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 2*p - 1}]//Flatten (* G. C. Greubel, Nov 23 2017 *) a[i_,p_]:=(-1)^i HypergeometricPFQ[ConstantArray[3,p]~Join~{2-i},ConstantArray[1,p],1];Table[a[i,p],{p,0,10},{i,2,2 p+2}]//Flatten (* Jonathan Burns, Mar 20 2018 *)
{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 3, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 2, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 2*p-1, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017
a(8) = Sum_{i=1..8} binomial(i+2,3)^2 = (20*(8^7) + 210*(8^6) + 854*(8^5) + 1680*(8^4) + 1610*(8^3) + 630*(8^2) + 36*8)/7! = 26334.
[n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520: n in [1..30]]; // G. C. Greubel, Nov 22 2017
a:=n->n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520: seq(a(n),n=1..31); # Emeric Deutsch
Accumulate[Binomial[Range[30]+2,3]^2] (* Harvey P. Dale, Mar 24 2011 *) LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{1,17,117,517,1742,4878, 11934, 26334},30] (* Harvey P. Dale, Aug 17 2014 *)
a(n)=n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520 \\ Charles R Greathouse IV, May 18 2015
a(10) = (90*(10^7)+630*(10^6)+1638*(10^5)+1890*(10^4)+840*(10^3)-48*(10))/5040 = 339988.
[(90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/ Factorial(7): n in [1..30]]; // G. C. Greubel, Nov 22 2017
Table[(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n)/7!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)
Vec(x*(x^4+20*x^3+48*x^2+20*x+1)/(x-1)^8 + O(x^100)) \\ Colin Barker, May 02 2014
a(n) = sum(i=1, n, binomial(i+1, 2)^3); \\ Michel Marcus, Nov 22 2017
[(1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568): n in [1..30]]; // G. C. Greubel, Nov 22 2017
Table[Sum[Binomial[k+1,2]^7, {k,1,n}], {n,1,30}] (* G. C. Greubel, Nov 22 2017 *) LinearRecurrence[{16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1},{1,2188,282124,10282124,181141499,1982230040,15475158552,93839322648,467508775773,1989944010148,7445104711204,25010673566116,76686775501847,217396817767472,575714897767472,1436257466526768},20] (* Harvey P. Dale, May 11 2022 *)
for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^7), ", ")) \\ G. C. Greubel, Nov 22 2017
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