cp's OEIS Frontend

Row sums form A000070, which is the partial sums of the partition numbers (A000041). Rows read backwards converge to the row sums (A000070).

From Alford Arnold, Feb 07 2010: (Start)

The table can also be generated by summing sequences embedded within Table A008284

For example,

1 1 1 1 ... yields 1 2 3 4 ...

1 1 2 2 3 3 ... yields 1 2 4 6 9 12 ...

1 1 2 3 4 5 7 ... yields 1 2 4 7 11 16 ...

(End)

T(n,k) is also count of all 'replacable' cells in the (Ferrers plots of) the partitions on n in exactly k parts. [Wouter Meeussen, Sep 16 2010]

From Wolfdieter Lang, Dec 03 2012: (Start)

The triangle entry T(n,k) is obtained from triangle A072233 by summing the entries of column k up to n (see the partial sum type o.g.f. given by Vladeta Jovovic in the formula section).

Therefore, the o.g.f. for the sequence in column k is x^k/((1-x)* product(1-x^j,j=1..k)).

The triangle with entry a(n,m) = T(n-1,m-1), n >= 1, m = 1, ..., n, is obtained from the partition array A103921 when in row n all entries belonging to part number m are summed (a conjecture). (End)

Is this a finite sequence?

It is possible to generalize this class of sequences by taking some integer-valued function f(j,k) decreasing in k such that f(j,1) = j and f(j,m) = c (for example, c=1 or c=2) for all sufficiently large m and considering those j that are divisible by all of f(j,1), f(j,2), ... If f(j,k) is slowly decreasing in k, then the set of corresponding j's is likely to have a very small number (if any) of terms, while if f(j,k) decreases rapidly, then there will be too many suitable j's. I believe the balance is achieved at functions like f(j,k) = floor(j^(1/k)) so that f(j,k) stabilizes to c at k ~= log(j). - Max Alekseyev, Aug 16 2015

If it exists, a(20) > 10^35. - Jon E. Schoenfield, Oct 17 2015

If one replaces the equilateral triangle by a rectangle isoscele triangle with hypotenuse [0,n], one gets A087811. - Michel Marcus, Aug 05 2013

It appears that the row sums give the positive integers of A001629. - Omar E. Pol, Jul 09 2016

The formula for this sequence can be derived by separating the conformed manifolds into three sets. The first set consists of those conformations where the handles of the manifold are pinched at the boundary, the second set have two or more handles pinched at the interior of the manifold, and the third set are pinched at the boundary and may or may not have handles drawn into this pinch. The order of the first set is n, the order of the second is n - 1, and the order of the third set is given by the following series: (Sum_{k mod 2 = 0..n} (k/2)*(n - k + 1) + (2*(n - k) + (-1)^(n - k) + 3)/4) + (Sum_{j mod 2 = 1..n} ((j + 1)/2)*(n - j + 1)). These can then be combined into a single expression, Sum_{i = 0..n} ((2*i + (-1)^(i + 1) + 1)/4)*(n - i + 1) + ((2*(n - i) + (-1)^(n - i) + 3)/4)*(((-1)^i + 1)/2). The i in this series can be thought of as the number of handles drawn into the central pinch. If one factors out the expressions in the series and simplifies each term individually, the resulting functions can then be combined into a single formula. However, when we add 2n - 1 to this we find that for n = 0 the formula also equals zero. This cannot be, because there is one way to pinch a compact 2-manifold with 0 handles. Therefore, ((-1)^(2^n - 1) + 1)/2 is added as a corrective term for this one case.

This sequence can be calculated by a recursive algorithm:

Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').

The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).

There exist some transformation pairs of infinite sequences in the database:

A000027 <--> A000142; A000079 <--> A000079; A000045 <--> A001654;

A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;

A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;

A098558 <--> A004277; A010551 <--> A087811; A100538 <--> A329227;

A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;

A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;

A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;

A279312 <--> A------; A031923 <--> A------.

These transformation pairs are conjectured:

A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;

A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;

A349079 <--> A------.

("A------" means not yet in the database.)

Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.

If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.