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%I #13 Aug 29 2019 18:00:25
%S 1,1,1,0,1,0,1,1,0,0,1,1,1,0,1,0,1,0,0,0,0,1,0,1,0,1,0,0,1,0,0,0,1,0,
%T 0,0,0,0,0,1,1,1,1,1,1,0,1,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
%U 1,0,1,0,1,1,0,1,0,1,0,0,1,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0
%N Characteristic array marking partitions of m whose parts are exponents of partitions of n into m parts.
%C With N=A000217(n-1) + m, where A000217(n-1) is the largest triangular number less than N, a(N,k)=1 if there is at least one partition of n into m parts which has the parts of the k-th partition of m (in Abramowitz-Stegun order) as exponents. Otherwise a(N,k)=0.
%C The sequence of row lengths of this array is p(m)= A000041(m) (number of partitions of m) and m is determined from N (the row index) as explained above. It is [1,1,2,1,2,3,1,2,3,5,1,2,3,5,7,1,2,3,5,7,11,...]=A092080(N), N>=1.
%C One can find the (n,m; k) numbers for the p-th entry (p>2) of the sequence as follows: p= a(n-1) + b(m-1) + k, where a(n-1) := A085360(n-1) is the largest number from the numbers A085360 less than p and b(m-1)=A026905(m-1) is the largest number from the numbers A026905 less than p-a(n-1). p=1 belongs to (1,1;1) and p=2 to (2,1;1).
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 831-2.
%H W. Lang, <a href="/A092079/a092079.txt">First 36 rows and more comments</a>.
%e N=13 = 10 + 3 with 10=A000217(4), hence n=5 and m=3.
%e N=10 = 6 + 4 with 6=A000217(3), hence n=4 and m=4.
%e The sequence entry nr. p=16, which is 0, belongs to (n=4,m=3; k=3) because 16 = 10 + 3 + 3 with 10=A085360(3), hence n=4 and 3=A026905(2), hence m=3.
%e a(N=13,k=3)=0: There is no partition of 5 into 3 parts which has as exponents 1,1,1, the parts of the third (k=3) partition of 3.
%e a(N=13,k=2)=1, n=5, m=3; there is a partition of 5 into 3 parts, which has the parts of the second (k=2) partitions of 3, i.e. 1,2, as exponents. In fact there are two such partitions, namely [1^2, 3^1] and [1^1, 2^2].
%Y Cf. A092078 (with multiplicities).
%K nonn,easy,tabf
%O 1,1
%A _Wolfdieter Lang_, Mar 19 2004
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