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A276965
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Square row sums of the triangle of Lah numbers (A105278).
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1
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1, 1, 5, 73, 2017, 86801, 5289301, 430814665, 45052534913, 5868875082817, 930114039075301, 175964489469769001, 39125942325820605025, 10092849114680961297553, 2987365449592984040715317, 1005030253302269078318250601
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OFFSET
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0,3
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n} lah(n,k)^2.
a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(n-1,k-1)^2*((n-k)!)^2.
a(n) = hypergeometric([-n+1,-n+1,-n,-n],[1],1).
a(n) = (n!)^2 * hypergeometric([-n+1,-n+1],[1,2,2],1) for n > 0.
Recurrence: n*(16*n^3 - 96*n^2 + 185*n - 116)*a(n) = 2*(32*n^6 - 272*n^5 + 930*n^4 - 1668*n^3 + 1670*n^2 - 867*n + 164)*a(n-1) - (n-2)*(96*n^7 - 1056*n^6 + 4646*n^5 - 10500*n^4 + 12990*n^3 - 8644*n^2 + 2827*n - 364)*a(n-2) + 2*(n-3)*(n-2)^3*(32*n^6 - 336*n^5 + 1410*n^4 - 2978*n^3 + 3268*n^2 - 1731*n + 353)*a(n-3) - (n-4)^2*(n-3)^3*(n-2)^4*(16*n^3 - 48*n^2 + 41*n - 11)*a(n-4). - Vaclav Kotesovec, Sep 27 2016
a(n) ~ n^(2*n - 3/4) * exp(4*sqrt(n) - 2*n - 1) / (2^(3/2) * sqrt(Pi)) * (1 + 31/(96*sqrt(n)) + 937/(18432*n)). - Vaclav Kotesovec, Sep 27 2016
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MATHEMATICA
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Table[HypergeometricPFQ[{1-n, 1-n, -n, -n}, {1}, 1], {n, 0, 100}]
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PROG
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(Maxima) makelist(hypergeometric([-n+1, -n+1, -n, -n], [1], 1), n, 0, 12);
(Perl) use ntheory ":all"; for my $n (0..20) { say "$n ", vecsum(map{my $l=stirling($n, $_, 3); vecprod($l, $l); } 0..$n) } # Dana Jacobsen, Mar 16 2017
(PARI) concat([1], for(n=1, 25, print1(sum(k=0, n, binomial(n, k)^2*binomial(n-1, k-1)^2*((n-k)!)^2), ", "))) \\ G. C. Greubel, Jun 05 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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