A122172 Triangle read by rows relating A074139, A074141, A078436 and A079025.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 3, 2, 1, 1, 3, 4, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 3, 4, 4, 3, 1, 1, 3, 5, 5, 3, 1, 1, 4, 7, 7, 4, 1, 1, 5, 10, 10, 5, 1, 1, 1, 1, 1, 1, 1, 1

Offset

0,8

Comments

A proper definition is needed for this sequence.

Are the row sums A074139(n) and the row lengths A000041(n)? - R. J. Mathar, May 08 2019 [Not exactly: see below. - M. F. Hasler, Jan 07 2024]

From M. F. Hasler, Jan 06 2024: (Start)

I get this triangle as T(n,k) = # { v in S(p_n), |v| = k }, where p_n is the n-th partition as listed in A036036 or A036037 (which has a nice table of the p's), and S(p) = {0, ..., p[1]} x ... x {0, ..., p[#p]}, the set of vectors v with 0 <= v[i] <= p[i] for all indices i from 1 to #p = number of parts in p.

Then the row sums are indeed the total number of elements in S(p_n) which is equal to the product (p[1]+1)*...*(p[#p]+1) which is also the number of divisors of the Heinz number of p (cf. A185974).

The row lengths are 1 + |p| = 1 + sum of all parts of p (corresponding to the possible values of |v| ranging from 0 to |p|), repeated A000041(|p|) times: A000041(0) = 1 row of length 0+1 for the partition () of 0, A000041(1) = 1 row of length 1+1 for partition (1) of 1; A000041(2) = 2 rows of length 2+1 for the two partitions (2) and (1,1) of 2; A000041(3) = 3 rows of length 3+1 for the 3 partitions {(3), (2,1), (1,1,1)} of 3; etc. (End)

Links

M. F. Hasler, Table of n, a(n) for n = 0..9172 (up to partitions of 15), Jan 07 2024

Example

The triangle begins
1
1 1
1 1 1
1 2 1
1 1 1 1
1 2 2 1
1 3 3 1
1 1 1 1 1
1 2 2 2 1
1 2 3 2 1
1 3 4 3 1
1 4 6 4 1
1 1 1 1 1 1
1 2 2 2 2 1
1 2 3 3 2 1
1 3 4 4 3 1
1 3 5 5 3 1
1 4 7 7 4 1
1 5 10 10 5 1

Prog

From M. F. Hasler, Jan 06 2024: (Start)
(PARI) A122172_row(n, p=part(n))={my(c=Vec(0, vecsum(p)+1)); forvec(v=[[0, k]| k<-p], c[vecsum(v)+1]++); c} \\ instead of n one can directly give p as 2nd arg
/* helper function: n-th partition as listed in A036036, A036037 or A185974 */
part(n)={my(c, r=0); while(n >= c = numbpart(r), n -= c; r++); partitions(r)[n+1]
for(n=0, 5, forpart(p=n, print(A122172_row(, Vec(p))) )) \\ Illustration. (End)

Crossrefs

Cf. A036036 (partitions in A-S order), A036037 (the same, parts reversed), A185974 (corresponding Heinz numbers).

Sequence in context: A279061 A206491 A359997 * A030613 A025910 A002637

Adjacent sequences: A122169 A122170 A122171 * A122173 A122174 A122175

Keyword

obsc,nonn,tabf,more

Author

Alford Arnold, Aug 23 2006

Extensions

More terms from M. F. Hasler, Jan 07 2024

Status

approved