A117409 Number of partitions of n into odd parts in which the largest part occurs only once.
1, 0, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 18, 22, 27, 32, 38, 46, 54, 64, 76, 89, 104, 122, 142, 165, 192, 222, 256, 296, 340, 390, 448, 512, 585, 668, 760, 864, 982, 1113, 1260, 1426, 1610, 1816, 2048, 2304, 2590, 2910, 3264, 3658, 4097, 4582, 5120, 5718, 6378
Offset: 1
Keywords
Examples
a(9)=5 because we have [9],[7,1,1],[5,3,1],[5,1,1,1,1] and [3,1,1,1,1,1,1].
Crossrefs
Cf. A117408.
Programs
-
Maple
f:=sum(x^(2*k-1)/product(1-x^(2*i-1),i=1..k-1),k=1..40): fser:=series(f,x=0,70): seq(coeff(fser,x^n),n=1..65);
-
Mathematica
Table[SeriesCoefficient[Sum[x^(2 k - 1)/Product[1 - x^(2 i - 1), {i, k - 1}], {k, 0, n}] , {x, 0, n}], {n, 57}] (* Michael De Vlieger, Sep 16 2016 *) -
PARI
{a(n)=if(n<3, n==1, n-=2; polcoeff( prod(k=1, n, 1+x^k, 1+x*O(x^n)), n))} /* Michael Somos, May 28 2006 */
Formula
G.f.: Sum_{k>0} x^(2k-1)/(Product_{0
a(n) = A000009(n-2), n>2. - Michael Somos, May 28 2006
a(n) = A117408(n,1).
a(n) ~ exp(Pi*sqrt(n/3)) / (4*3^(1/4)*n^(3/4)). - Vaclav Kotesovec, Sep 27 2016
A303138 Regular triangle where T(n,k) is the number of strict integer partitions of n with greatest common divisor k.
1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 2, 0, 0, 0, 1, 2, 1, 0, 0, 0, 1, 4, 0, 0, 0, 0, 0, 1, 4, 1, 0, 0, 0, 0, 0, 1, 6, 0, 1, 0, 0, 0, 0, 0, 1, 7, 2, 0, 0, 0, 0, 0, 0, 0, 1, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 10, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 17, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 17, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 23, 0, 2, 0, 1
Offset: 1
Examples
Triangle begins: 01: 1 02: 0 1 03: 1 0 1 04: 1 0 0 1 05: 2 0 0 0 1 06: 2 1 0 0 0 1 07: 4 0 0 0 0 0 1 08: 4 1 0 0 0 0 0 1 09: 6 0 1 0 0 0 0 0 1 10: 7 2 0 0 0 0 0 0 0 1 11: 11 0 0 0 0 0 0 0 0 0 1 12: 10 2 1 1 0 0 0 0 0 0 0 1 13: 17 0 0 0 0 0 0 0 0 0 0 0 1 14: 17 4 0 0 0 0 0 0 0 0 0 0 0 1 15: 23 0 2 0 1 0 0 0 0 0 0 0 0 0 1 The strict partitions counted in row 12 are the following. T(12,1) = 10: (11,1) (9,2,1) (8,3,1) (7,5) (7,4,1) (7,3,2) (6,5,1) (6,3,2,1) (5,4,3) (5,4,2,1) T(12,2) = 2: (10,2) (6,4,2) T(12,3) = 1: (9,3) T(12,4) = 1: (8,4) T(12,12) = 1: (12)
Crossrefs
Programs
-
Mathematica
Table[Length[Select[IntegerPartitions[n],UnsameQ@@#&&GCD@@#===k&]],{n,15},{k,n}]
Formula
If k divides n, T(n,k) = A078374(n/k); otherwise T(n,k) = 0.
A115604 Triangle read by rows: T(n,k) is the number of partitions of n into odd parts in which the smallest part occurs k times (1<=k<=n).
1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 2, 1, 0, 1, 0, 0, 1, 2, 1, 1, 0, 1, 0, 0, 1, 2, 1, 2, 1, 0, 1, 0, 0, 1, 3, 2, 1, 1, 1, 0, 1, 0, 0, 1, 3, 3, 1, 1, 1, 1, 0, 1, 0, 0, 1, 4, 2, 2, 2, 1, 1, 1, 0, 1, 0, 0, 1, 5, 3, 2, 2, 1, 1, 1, 1, 0, 1, 0, 0, 1, 5, 4, 3, 2, 2, 1, 1, 1, 1, 0, 1, 0, 0, 1
Offset: 1
Examples
T(14,2)=4 because we have [9,3,1,1],[7,7],[7,5,1,1] and [3,3,3,3,1,1]. Triangle starts: 1; 0,1; 1,0,1; 1,0,0,1; 1,1,0,0,1; 1,1,1,0,0,1; 2,1,0,1,0,0,1;
Programs
-
Maple
g:=sum(t*x^(2*k-1)/(1-t*x^(2*k-1))/product(1-x^(2*i-1),i=k+1..40),k=1..40): gser:=simplify(series(g,x=0,55)): for n from 1 to 15 do P[n]:=expand(coeff(gser,x^n)) od: for n from 1 to 15 do seq(coeff(P[n],t^j),j=1..n) od; # yields sequence in triangular form
Formula
G.f.=G(t,x)=sum(tx^(2k-1)/[(1-tx^(2k-1))product(1-x^(2i-1), i=k+1..infinity)], k=1..infinity).
Comments