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A262705
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Triangle: Newton expansion of C(n,m)^4, read by rows.
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2
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1, 0, 1, 0, 14, 1, 0, 36, 78, 1, 0, 24, 978, 252, 1, 0, 0, 4320, 8730, 620, 1, 0, 0, 8460, 103820, 46890, 1290, 1, 0, 0, 7560, 581700, 1159340, 185430, 2394, 1, 0, 0, 2520, 1767360, 13387570, 8314880, 595476, 4088, 1, 0, 0, 0, 3087000, 85806000, 170429490, 44341584, 1642788, 6552, 1
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OFFSET
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0,5
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COMMENTS
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Triangle here T_4(n,m) is such that C(n,m)^4 = Sum_{j=0..n} C(n,j)*T_4(j,m).
Equivalently, lower triangular matrix T_4 such that
T_4(n,m) = 0 for n < m and for 4*m < n.
Example:
C(x,2)^4 = x^4*(x-1)^4 /16 = 1*C(x,2) + 78*C(x,3) + 978*C(x,4) + 4320*C(x,5) + 8460*C(x,6) + 7560*C(x,7) + 2520*C(x,8);
C(5,2)^4 = C(5,3)^4 = 10000 = 1*C(5,2) + 78*C(5,3) + 978*C(5,4) + 4320*C(5,5) = 1*C(5,3) + 252*C(5,4) + 8730*C(5,5).
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LINKS
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FORMULA
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T_4(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^4.
Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above, then T_4(n,m) = n! / (m!)^4 * S(m,m)(4,n).
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EXAMPLE
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Triangle starts:
[1];
[0, 1];
[0, 14, 1];
[0, 36, 78, 1];
[0, 24, 978, 252, 1];
[0, 0, 4320, 8730, 620, 1];
[0, 0, 8460, 103820, 46890, 1290, 1];
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MATHEMATICA
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T4[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^4, {j, 0, n}]; Table[T4[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)
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PROG
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(MuPAD)
// as a function
T_4:=(n, m)->_plus((-1)^(n-j)*binomial(n, j)*binomial(j, m)^4 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n, m) $m=0..h]$n=0..h]):
_P_4:=h->matrix([[binomial(n, m)^4 $m=0..h]$n=0..h]):
_T_4:=h->_P(h)^-1*_P_4(h):
(Magma) [&+[(-1)^(n-j)*Binomial(n, j)*Binomial(j, m)^4: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015
(PARI) T_4(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n, j)*binomial(j, m)^4), ", ")); print())} \\ Colin Barker, Oct 01 2015
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CROSSREFS
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Row sums are, by definition, the inverse binomial transform of A005260.
Second diagonal (T_4(n+1,n)) is A058895(n+1).
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KEYWORD
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AUTHOR
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STATUS
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approved
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