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A026820
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Euler's table: triangular array T read by rows, where T(n,k) = number of partitions in which every part is <= k for 1 <= k <= n. Also number of partitions of n into at most k parts.
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41
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1, 1, 2, 1, 2, 3, 1, 3, 4, 5, 1, 3, 5, 6, 7, 1, 4, 7, 9, 10, 11, 1, 4, 8, 11, 13, 14, 15, 1, 5, 10, 15, 18, 20, 21, 22, 1, 5, 12, 18, 23, 26, 28, 29, 30, 1, 6, 14, 23, 30, 35, 38, 40, 41, 42, 1, 6, 16, 27, 37, 44, 49, 52, 54, 55, 56, 1, 7, 19, 34, 47, 58, 65, 70, 73, 75, 76, 77
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history;
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internal format)
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OFFSET
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1,3
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REFERENCES
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G. Chrystal, Algebra, Vol. II, p. 558.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XIV.2, p. 493.
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, p. 831. [scanned copy]
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FORMULA
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T(n,k) = Sum_{i=1..min(k,floor(n/2))} T(n-i,i) + Sum_{j=1+floor(n/2)..k} A000041(n-j). - Bob Selcoe, Aug 22 2014 [corrected by Álvar Ibeas, Mar 15 2018]
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EXAMPLE
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Triangle starts:
1;
1, 2;
1, 2, 3;
1, 3, 4, 5;
1, 3, 5, 6, 7;
1, 4, 7, 9, 10, 11;
1, 4, 8, 11, 13, 14, 15;
1, 5, 10, 15, 18, 20, 21, 22;
1, 5, 12, 18, 23, 26, 28, 29, 30;
1, 6, 14, 23, 30, 35, 38, 40, 41, 42;
1, 6, 16, 27, 37, 44, 49, 52, 54, 55, 56;
...
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MAPLE
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T:= proc(n, k) option remember;
`if`(n=0 or k=1, 1, T(n, k-1) + `if`(k>n, 0, T(n-k, k)))
end:
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MATHEMATICA
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t[n_, k_] := Length@ IntegerPartitions[n, k]; Table[ t[n, k], {n, 12}, {k, n}] // Flatten
(* Second program: *)
T[n_, k_] := T[n, k] = If[n==0 || k==1, 1, T[n, k-1] + If[k>n, 0, T[n-k, k]]]; Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 22 2015, after Alois P. Heinz *)
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PROG
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(Haskell)
import Data.List (inits)
a026820 n k = a026820_tabl !! (n-1) !! (k-1)
a026820_row n = a026820_tabl !! (n-1)
a026820_tabl = zipWith
(\x -> map (p x) . tail . inits) [1..] $ tail $ inits [1..] where
p 0 _ = 1
p _ [] = 0
p m ks'@(k:ks) = if m < k then 0 else p (m - k) ks' + p m ks
(SageMath)
from sage.combinat.partition import number_of_partitions_length
from itertools import accumulate
for n in (1..11):
print(list(accumulate([number_of_partitions_length(n, k) for k in (1..n)])))
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CROSSREFS
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T(n,n) = A000041(n), T(n,1) = A000012(n), T(n,2) = A008619(n) for n>1, T(n,3) = A001399(n) for n>2, T(n,4) = A001400(n) for n>3, T(n,5) = A001401(n) for n>4, T(n,6) = A001402(n) for n>5, T(n,7) = A008636(n) for n>6, T(n,8) = A008637(n) for n>7, T(n,9) = A008638(n) for n>8, T(n,10) = A008639(n) for n>9, T(n,11) = A008640(n) for n>10, T(n,12) = A008641(n) for n>11, T(n,n-2) = A007042(n-1) for n>2, T(n,n-1) = A000065(n) for n>1.
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KEYWORD
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AUTHOR
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STATUS
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approved
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