A082535 Duplicate of A007326.
0, 0, 0, 0, 0, 0, 1, 3, 8, 19, 40, 83, 176, 365, 775, 1643, 3483
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.
with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d,j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:= etr(n-> n*(n+1)/2): seq(a(n), n=0..30); # Alois P. Heinz, Sep 08 2008
a[0] = 1; a[n_] := a[n] = 1/(2*n)*Sum[(DivisorSigma[2, k]+DivisorSigma[3, k])*a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 05 2014, after Vladeta Jovovic *) nmax=50; CoefficientList[Series[Product[1/(1-x^k)^(k*(k+1)/2),{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Mar 11 2015 *)
a(n)=if(n<0, 0, polcoeff(exp(sum(k=1, n, x^k/(1-x^k)^3/k, x*O(x^n))), n)) \\ Joerg Arndt, Apr 16 2010
# uses[EulerTransform from A166861] b = EulerTransform(lambda n: binomial(n+1, 2)) print([b(n) for n in range(37)]) # Peter Luschny, Nov 11 2020
Examples for n=2 and n=3. a(2) = 4: 2; 11 where the first 1 is at the origin and the second 1 is in the x, y or z direction. a(3) = 10: 3; 21 where the 2 is at the origin and the 1 is on the x, y or z axis; 111 (a row of 3 ones on the x, y or z axes); and three 1's with one 1 at the origin and the other two 1's on two of the three axes. From _Gus Wiseman_, Jan 22 2019: (Start) The a(1) = 1 through a(4) = 26 solid partitions, represented as chains of chains of integer partitions: ((1)) ((2)) ((3)) ((4)) ((11)) ((21)) ((22)) ((1)(1)) ((111)) ((31)) ((1))((1)) ((2)(1)) ((211)) ((11)(1)) ((1111)) ((2))((1)) ((2)(2)) ((1)(1)(1)) ((3)(1)) ((11))((1)) ((21)(1)) ((1)(1))((1)) ((11)(11)) ((1))((1))((1)) ((111)(1)) ((2))((2)) ((3))((1)) ((2)(1)(1)) ((21))((1)) ((11))((11)) ((11)(1)(1)) ((111))((1)) ((2)(1))((1)) ((1)(1)(1)(1)) ((11)(1))((1)) ((2))((1))((1)) ((1)(1))((1)(1)) ((1)(1)(1))((1)) ((11))((1))((1)) ((1)(1))((1))((1)) ((1))((1))((1))((1)) (End)
planePtns[n_]:=Join@@Table[Select[Tuples[IntegerPartitions/@ptn],And@@(GreaterEqual@@@Transpose[PadRight[#]])&],{ptn,IntegerPartitions[n]}]; solidPtns[n_]:=Join@@Table[Select[Tuples[planePtns/@y],And@@(GreaterEqual@@@Transpose[Join@@@(PadRight[#,{n,n}]&/@#)])&],{y,IntegerPartitions[n]}]; Table[Length[solidPtns[n]],{n,10}] (* Gus Wiseman, Jan 23 2019 *)
List([0..40], n-> (120 + 404*n + 490*n^2 + 255*n^3 + 50*n^4 + n^5)/120); # G. C. Greubel, Sep 11 2019
R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1+5*x-3*x^2-2*x^3)/(1-x)^6 )); // G. C. Greubel, Sep 11 2019
seq(1+10*n+27*binomial(n,2)+29*binomial(n,3)+12*binomial(n,4)+binomial(n,5), n=0..40);
Table[1+10n+27Binomial[n,2]+29Binomial[n,3]+12Binomial[n,4]+ Binomial[n,5], {n,0,40}] (* Harvey P. Dale, Jul 27 2011 *) CoefficientList[Series[(1+5x-3x^2-2x^3)/(1-x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 17 2013 *) LinearRecurrence[{6,-15,20,-15,6,-1},{1,11,48,141,331,672},40] (* Harvey P. Dale, Aug 28 2019 *)
my(x='x+O('x^40)); Vec((1+5*x-3*x^2-2*x^3)/(1-x)^6) \\ G. C. Greubel, Sep 11 2019
[(120 + 404*n + 490*n^2 + 255*n^3 + 50*n^4 + n^5)/120 for n in (0..40)] # G. C. Greubel, Sep 11 2019
List([0..40],n->(n+1)*(n+4)*(n^3+40*n^2+61*n+30)/120); # Muniru A Asiru, Feb 17 2019
[1 + 10*n + 27*Binomial(n,2) + 28*Binomial(n,3) + 11*Binomial(n,4) + Binomial(n,5): n in [0..40]]; // Vincenzo Librandi, Oct 27 2013
a:= n-> 1+10*n+27*binomial(n, 2)+28*binomial(n, 3) +11*binomial(n, 4)+binomial(n, 5): seq(a(n), n=0..34);
LinearRecurrence[{6,-15,20,-15,6,-1},{1,11,48,140,326,657},40] (* Harvey P. Dale, Jan 27 2013 *) CoefficientList[Series[(x^4 -3x^3 -3x^2 +5x +1)/(x-1)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Oct 27 2013 *)
my(x='x+O('x^40)); Vec((x^4-3*x^3-3*x^2+5*x+1)/(x-1)^6) \\ G. C. Greubel, Feb 17 2019
((x^4-3*x^3-3*x^2+5*x+1)/(x-1)^6).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019
A116672 begins 1; 1,1; 1,2,1; 1,4,4,1; 1,6,11,7,1; 1,10,27,29,12,1; 1,14,57,96,72,21,1; 1,21,117,277,319,176,38,1; . . . so A116673 begins 1 2 4 10 26 80 262 950 ...
Comments