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A105487
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Number of partitions of {1...n} containing 5 strings of 3 consecutive integers, where each string is counted within a block and a string of more than 3 consecutive integers are counted three at a time.
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4
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1, 2, 12, 56, 297, 1632, 9531, 58634, 379371, 2574254, 18276457, 135463074, 1046041114, 8399533370, 70013963418, 604840440328, 5407301690915, 49958478263502, 476403955991034, 4683463406478004, 47414166201239781, 493803423334040824, 5285548108715948453
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OFFSET
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7,2
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LINKS
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FORMULA
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a(n) = Sum_{k=1..n} c(n, k, 5), where c(n, k, 5) is the case r=5 of c(n, k, r) given by c(n, k, r)=c(n-1, k-1, r)+(k-1)c(n-1, k, r)+c(n-2, k-1, r)+(k-1)c(n-2, k, r)+c(n-1, k, r-1)-c(n-2, k-1, r-1)-(k-1)c(n-2, k, r-1), r=0, 1, .., n-k-1, k=1, 2, .., n-2r, c(n, k, 0) = Sum_{j= 0..floor(n/2)} binomial(n-j, j)*S2(n-j-1, k-1).
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EXAMPLE
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a(8) = 2 because the partitions of {1,...,8} with 5 strings of 3 consecutive integers are 1234567/8, 1/2345678.
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MAPLE
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c := proc(n, k, r) option remember ; local j ; if r =0 then add(binomial(n-j, j)*combinat[stirling2](n-j-1, k-1), j=0..floor(n/2)) ; else if r <0 or r > n-k-1 then RETURN(0) fi ; if n <1 then RETURN(0) fi ; if k <1 then RETURN(0) fi ; RETURN( c(n-1, k-1, r)+(k-1)*c(n-1, k, r)+c(n-2, k-1, r)+(k-1)*c(n-2, k, r) +c(n-1, k, r-1)-c(n-2, k-1, r-1)-(k-1)*c(n-2, k, r-1) ) ; fi ; end: A105487 := proc(n) local k ; add(c(n, k, 5), k=1..n) ; end: for n from 7 to 30 do printf("%d, ", A105487(n)) ; od ; # R. J. Mathar, Feb 20 2007
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MATHEMATICA
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S2[_, -1] = 0;
S2[n_, k_] = StirlingS2[n, k];
c[n_, k_, r_] := c[n, k, r] = Which[
r == 0, Sum[Binomial[n - j, j]*S2[n - j - 1, k - 1],
{j, 0, Floor[n/2]}],
r < 0 || r > n - k - 1, 0,
n < 1, 0,
k < 1, 0,
True, c[n - 1, k - 1, r] +
(k - 1)*c[n - 1, k, r] +
c[n - 2, k - 1, r] +
(k - 1)*c[n - 2, k, r] +
c[n - 1, k, r - 1] -
c[n - 2, k - 1, r - 1] -
(k - 1)*c[n - 2, k, r - 1]];
A105487[n_] := Sum[c[n, k, 5], {k, 1, n}];
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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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