OFFSET
4,1
COMMENTS
Tree degree sequences of n nodes are in one-to-one correspondence with the partitions of n-2, as for instance set out in Myerson's collection of problems [Myerson]. For series-reduced trees, these partitions have no part 1.
Given a term t, the respective degree sequence D is determined by Decode(t). See second (PARI) entry.
LINKS
EXAMPLE
Triangle begins:
n \ k| 1 2 ... n \ k| 1 2 ...
-----+------------- -----+-----------------------------------
4 | 40; 4 | [3,1,1,1];
5 | 112; 5 | [4,1,1,1,1];
6 | 352, 400; <=> 6 | [5,1,1,1,1,1], [3,3,1,1,1,1];
7 | 832, 1120; 7 | [6,1,1,1,1,1,1], [4,3,1,1,1,1,1];
... ...
Row n = 7 follows from table
.
+---------------------+------------------+---------------------------+
| Partitions of n-2 = | | |
| 5 without parts 1 | Degree sequences | Encoding |
+---------------------+------------------+---------------------------+
| [5] | 6,1,1,1,1,1,1 | prime(6) * 2^6 |
| [2, 3] | 4,3,1,1,1,1,1 | prime(4) * prime(3) * 2^5 |
+---------------------+------------------+---------------------------+
PROG
(PARI) Row(n) = {my(j=0, V = vector(numbpart(n-2) - numbpart(n-3)));
forpart(P=n-2, V[j++] = prod(k=1, #P, prime(P[k]+1)) << (n-#P), [2, n-2]); V};
(PARI) Decode(t) = {my(V = [], i = 1, p); while(t > 1, p = prime(i); while(t % p == 0, t /= p; V = concat(V, Vec(i)) ); i++); vecsort(V, (x, y)->y-x) };
CROSSREFS
KEYWORD
AUTHOR
Washington Bomfim, Jul 01 2021
STATUS
approved