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 A092905 Triangle, read by rows, such that the partial sums of the n-th row form the n-th diagonal, for n>=0, where each row begins with 1. 6
 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 6, 4, 2, 1, 1, 6, 9, 7, 4, 2, 1, 1, 7, 12, 11, 7, 4, 2, 1, 1, 8, 16, 16, 12, 7, 4, 2, 1, 1, 9, 20, 23, 18, 12, 7, 4, 2, 1, 1, 10, 25, 31, 27, 19, 12, 7, 4, 2, 1, 1, 11, 30, 41, 38, 29, 19, 12, 7, 4, 2, 1, 1, 12, 36, 53, 53, 42, 30, 19, 12, 7, 4, 2, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS Row sums form A000070, which is the partial sums of the partition numbers (A000041). Rows read backwards converge to the row sums (A000070). Contribution 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. [From 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) LINKS R. J. Mathar, Size of the set of residues of integer powers of fixed exponent, (2017), Table 11. FORMULA T(n, k) = sum_{j=0..k} T(n-k, j), with T(n, 0) = 1 for all n>=0. A000070(n) = sum_{k=0..n} T(n, k). O.g.f.: (1/(1-y))*(1/Product(1-x*y^k, k=1..infinity)). - Vladeta Jovovic, Jan 29 2005 EXAMPLE The fourth row (n=3) is {1,3,2,1} and the fourth diagonal is the partial sums of the fourth row: {1,4,6,7,7,7,7,7,...}. The triangle T(n,k) begins: n\k 0  1  2  3  4  5  6  7  8  9 10 11 12  ... 0   1 1   1  1 2   1  2  1 3   1  3  2  1 4   1  4  4  2  1 5   1  5  6  4  2  1 6   1  6  9  7  4  2  1 7   1  7 12 11  7  4  2  1 8   1  8 16 16 12  7  4  2  1 9   1  9 20 23 18 12  7  4  2  1 10  1 10 25 31 27 19 12  7  4  2  1 11  1 11 30 41 38 29 19 12  7  4  2  1 12  1 12 36 53 53 42 30 19 12  7  4  2  1 ... Reformatted by Wolfdieter Lang, Dec 03 2012 T(5,3)=4 because the partitions of 5 in exactly 3 parts are 221 and 311, and they give rise to partitions of 4 in four ways: 221->22 and 211, 311->211 and 31, since both their Ferrers plots have 2 'mobile cells' each. [From Wouter Meeussen, Sep 16 2010] T(5,3) = a(6,4) = 4 because the partitions of 6 with 4 parts are 1113 and 1122, with the number of distinct parts 2 and 2, respectively, summing to 4 (see the array A103921). An example for the conjecture given as comment above.  - Wolfdieter Lang, Dec 03 2012 MAPLE T(n, k)=if(n

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Last modified January 23 04:40 EST 2019. Contains 319370 sequences. (Running on oeis4.)