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A304011
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Number of same-sized pairs of subsets of set of n numbers that might have the same sum.
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1
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0, 0, 0, 1, 5, 20, 70, 231, 735, 2289, 7029, 21384, 64636, 194480, 583232, 1744847, 5210687, 15540023, 46299143, 137837666, 410127806, 1219804541, 3626853647, 10781440394, 32045015650, 95236794600, 283027305300, 841096898745, 2499595030581, 7428627412260
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OFFSET
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1,5
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COMMENTS
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Given a set with n different numbers, you only need to check a(n) pairs of subsets of the same cardinality to prove that no pair of same-cardinality subsets have the same total sum. The others can be eliminated by noting the dominance of members of one totally-ordered subset over the corresponding elements of the other totally-ordered subset.
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LINKS
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FORMULA
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a(n) = Sum_{i=1..floor(n/2)} binomial(n, 2*i)*A002054(i-1).
D-finite with recurrence: (n-4)*(n+2)*a(n) = (3*n^2 - 7*n - 5)*a(n-1) + (n-3)*(n-1)*a(n-2) - 3*(n-2)*(n-1)*a(n-3) for n >= 5.
a(n) ~ 3^(n + 1/2) / (4*sqrt(Pi*n)). (End)
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MATHEMATICA
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Table[1/2 + Hypergeometric2F1[(1 - n)/2, -n/2, 1, 4]/2 - Hypergeometric2F1[(1 - n)/2, -n/2, 2, 4], {n, 1, 30}] (* Vaclav Kotesovec, Aug 04 2018 *)
Join[{0, 0, 0, 1}, RecurrenceTable[{(n-4)*(n+2)*a[n]==(3*n^2-7*n-5)*a[n-1]+ (n-3)*(n-1)*a[n-2]-3*(n-2)*(n-1)*a[n-3], a[2]==0, a[3]==0, a[4]==1}, a, {n, 5, 25}]] (* Georg Fischer, Dec 06 2019 *)
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PROG
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(APL- NARS200 dialect) +/{((2×⍵)!n)×(⍵-2)!1+2×⍵-1}¨1..n÷2
(PARI) a(n) = sum(i=1, n\2, binomial(n, 2*i)*binomial(2*i-1, i-2)); \\ Michel Marcus, Jul 04 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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