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A(m, n) = lcm_{p in Partitions(n)} (Product_{r in p}(r + m)). Array read by ascending antidiagonals, for m, n >= 0.
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%I #12 Dec 14 2023 16:28:26

%S 1,1,1,1,2,2,1,3,12,6,1,4,36,24,12,1,5,80,540,720,60,1,6,150,960,6480,

%T 1440,360,1,7,252,5250,134400,136080,60480,2520,1,8,392,1512,315000,

%U 537600,8164800,120960,5040,1,9,576,24696,63504,1575000,32256000,24494400,3628800,15120

%N A(m, n) = lcm_{p in Partitions(n)} (Product_{r in p}(r + m)). Array read by ascending antidiagonals, for m, n >= 0.

%C We say q is a 'm-shifted partition of n' if there is a partition p of n, p = (t1, t2, ..., tk) and q = (t1 + m, t2 + m, ..., tk + m), where m is a nonnegative integer. q is a partition of n + k*m.

%C Let P(n) denote the partitions of n and P_{m}(n) the m-shifted partitions of n. The product of a partition is the product of its parts, Prod(p) = p1*p2*...*pk if p = (p1, p2, ..., pk). Using this terminology, the definition can be written as A(m, n) = lcm_{p in P_{m}(n)} Prod(p).

%C With m = 0 the cumulative radical A048803 is computed, and with m = 1 the Hirzebruch numbers A091137.

%e Array A(m, n) begins:

%e [0] 1, 1, 2, 6, 12, 60, 360, ... A048803

%e [1] 1, 2, 12, 24, 720, 1440, 60480, ... A091137

%e [2] 1, 3, 36, 540, 6480, 136080, 8164800, ... A368048

%e [3] 1, 4, 80, 960, 134400, 537600, 32256000, ...

%e [4] 1, 5, 150, 5250, 315000, 1575000, 330750000, ...

%e [5] 1, 6, 252, 1512, 63504, 1905120, 880165440, ...

%e [6] 1, 7, 392, 24696, 6914880, 532445760, 268352663040, ...

%e [7] 1, 8, 576, 23040, 18247680, 145981440, 683193139200, ...

%e [8] 1, 9, 810, 80190, 7217100, 844400700, 5851696851000, ...

%e .

%e Let m = 2 and n = 4. The partitions of 4 are [(4), (3,1), (2,2), (2,1,1), (1, 1, 1, 1)]. Thus A(2, 4) = lcm([6, 5*3, 4*4, 4*3*3, 3*3*3*3]) = 6480.

%o (SageMath)

%o def A(m, n): return lcm(product(r + m for r in p) for p in Partitions(n))

%o for m in range(9): print([A(m, n) for n in range(7)])

%Y Cf. A048803 (m=0), A091137 (m=1), A368048 (m=2).

%Y Columns include: A000027, A011379.

%K nonn,tabl

%O 0,5

%A _Peter Luschny_, Dec 12 2023