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A094216
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Triangle read by rows giving the coefficients of formulas generating each variety of S1(n,k) (unsigned Stirling numbers of first kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2*p, where T(i,p) satisfies Sum_{i=1..2*p} T(i,p) * C(n,i).
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22
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1, 1, 2, 7, 8, 3, 6, 38, 93, 111, 65, 15, 24, 226, 874, 1821, 2224, 1600, 630, 105, 120, 1524, 8200, 24860, 47185, 58465, 47474, 24430, 7245, 945, 720, 11628, 81080, 326712, 852690, 1522375, 1905168, 1676325, 1018682, 407925, 97020, 10395, 5040
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OFFSET
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1,3
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COMMENTS
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The formulas S1(n+p,n) obtained are those of S1(n+2,n) { A000914 }, S1(n+3,n) { A001303 }, S1(n+4,n) { A000915 }, S1(n+5,n) { A053567 } and so on.
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REFERENCES
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Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964, 9th Printing (1970), pp. 833-834.
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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 [alternative scanned copy].
Francis L. Miksa (1901-1975), Stirling numbers of the first kind, "27 leaves reproduced from typewritten manuscript on deposit in the UMT File", Mathematical Tables and Other Aids to Computation, vol. 10, no. 53, January 1956, pp. 37-38 (Reviews and Descriptions of Tables and Books, 7[I]).
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FORMULA
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a(1,k) = k!
...
a(2*k-5,k) = a(2*k,k) * (175000*k^8 -2117500*k^7 +10856650*k^6 -30743377*k^5 +52511770*k^4 -55386931*k^3 +35321832*k^2 -12560580*k+1944000) / (1632960*k^3 -7348320*k^2 +9389520*k -3061800).
a(2*k-4,k) = a(2*k,k) * (2500*k^6 -17400*k^5 +48511*k^4 -69378*k^3 +53929*k^2 -21906*k +3744) / (7776*k^2-15552*k+5832).
a(2*k-3,k) = a(2*k,k) * (1250*k^4-4225*k^3+5023*k^2-2600*k+528) / (1620*k-810).
a(2*k-2,k) = a(2*k,k) * (50*k^3-93*k^2+55*k-12) / (36*k-18).
a(2*k-1,k) = a(2*k,k) * (5*k-2) / 3.
a(2*k,k) = (2*k)! / (k!*2^k).
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EXAMPLE
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Row 5 contains 120,1524,8200,24860,47185,58465,47474,24430,7245,945, so the formula generating S1(n+5,n) numbers { A053567 } will be the following : 120*n +1524*C(n,2) +8200*C(n,3) +24860*C(n,4) +47185*C(n,5) +58465*C(n,6) +47474*C(n,7) +24430*C(n,8) +7245*C(n,9) +945*C(n,10). And then substituting for the 10th number of such a S1(n+p,n) gives S1(15,10) = 37312275.
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MATHEMATICA
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row[m_] := Module[{eq, t}, eq[n_] := Array[t, 2 m].Table[Binomial[n, k], {k, 1, 2 m}] == Abs[StirlingS1[n + m, n]]; Array[t, 2 m] /. Solve[ Array[ eq, 2 m]] // First];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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