A092288 Triangle read by rows: T(n,k) = count of parts k in all plane partitions of n.
1, 4, 1, 11, 2, 1, 28, 7, 2, 1, 62, 15, 5, 2, 1, 137, 38, 13, 5, 2, 1, 278, 76, 28, 11, 5, 2, 1, 561, 164, 60, 26, 11, 5, 2, 1, 1080, 316, 124, 52, 24, 11, 5, 2, 1, 2051, 623, 244, 108, 50, 24, 11, 5, 2, 1, 3778, 1156, 469, 208, 100, 48, 24, 11, 5, 2, 1, 6885, 2160, 886, 404, 194, 98, 48, 24, 11, 5, 2, 1
Offset: 1
Examples
Triangle begins:
1;
4, 1;
11, 2, 1;
28, 7, 2, 1;
62, 15, 5, 2, 1;
137, 38, 13, 5, 2, 1;
...
Links
- Alois P. Heinz, Rows n = 1..50, flattened
- M. F. Hasler, A092288, rows 1 - 18.
Crossrefs
Programs
-
Mathematica
Table[Length /@ Split[Sort[Flatten[planepartitions[k]]]], {k, 12}] -
PARI
A092288_row(n, c=vector(n), m, k)={for(i=1, #n=PlanePartitions(n), for(j=1,#m=n[i], for(i=1,#k=m[j], c[k[i]]++))); c} \\ See A091298 for PlanePartitions(). See below for more efficient code. M92288=[]; A092288(n,k,L=0)={n>1||return(if(L,[n,n==k],n==k)); if(#L&& #L<3, my(j=setsearch(M92288,[[n,k,L],[]],1)); j<=#M92288&& M92288[j][1]==[n,k,L]&& return(M92288[j][2])); my(c(p)=sum(i=1,#p,p[i]==k),S=[0,0],t); for(m=1,n,my(P=if(L,select(p->vecmin(L-Vecrev(p,#L))>=0, partitions(m,L[1],#L)), partitions(m))); if(m
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