A008969
Triangle of differences of reciprocals of unity.
Original entry on oeis.org
1, 1, 3, 1, 11, 7, 1, 50, 85, 15, 1, 274, 1660, 575, 31, 1, 1764, 48076, 46760, 3661, 63, 1, 13068, 1942416, 6998824, 1217776, 22631, 127, 1, 109584, 104587344, 1744835904, 929081776, 30480800, 137845, 255, 1, 1026576, 7245893376, 673781602752, 1413470290176, 117550462624, 747497920, 833375, 511
Offset: 1
Triangle T(n,k) begins:
1;
1, 3;
1, 11, 7;
1, 50, 85, 15;
1, 274, 1660, 575, 31;
1, 1764, 48076, 46760, 3661, 63;
1, 13068, 1942416, 6998824, 1217776, 22631, 127;
1, 109584, 104587344, 1744835904, 929081776, 30480800, 137845, 255;
...
- F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 228.
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T:= (n,k)-> `if`(k<=n, (n-k+2)!^k *
add((-1)^(j+1)*binomial(n-k+2, j)/ j^k, j=1..n-k+2), 0):
seq(seq(T(n,k), k=0..n), n=0..7); # Alois P. Heinz, Sep 05 2008
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T[n_, k_] := If[k <= n, (n-k+2)!^k*Sum[(-1)^(j+1)*Binomial[n-k+2, j]/j^k, {j, 1, n-k+2}], 0]; Table[Table[T[n, k], {k, 0, n}], {n, 0, 7}] // Flatten (* Jean-François Alcover, Mar 10 2014, after Alois P. Heinz *)
A112492
Triangle from inverse scaled Pochhammer symbols.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 7, 11, 1, 1, 15, 85, 50, 1, 1, 31, 575, 1660, 274, 1, 1, 63, 3661, 46760, 48076, 1764, 1, 1, 127, 22631, 1217776, 6998824, 1942416, 13068, 1, 1, 255, 137845, 30480800, 929081776, 1744835904, 104587344, 109584, 1, 1, 511, 833375, 747497920, 117550462624, 1413470290176, 673781602752, 7245893376, 1026576, 1
Offset: 0
Triangle begins:
1;
1, 1;
1, 3, 1;
1, 7, 11, 1;
1, 15, 85, 50, 1;
1, 31, 575, 1660, 274, 1;
1, 63, 3661, 46760, 48076, 1764, 1;
1, 127, 22631, 1217776, 6998824, 1942416, 13068, 1; ...
The g.f.s for the rows are illustrated by:
Sum_{n>=0} (n+1)^(n-1)*exp((n+1)*x)*(-x)^n/n! = 1;
Sum_{n>=0} (n+1)^(n-2)*exp((n+1)*x)*(-x)^n/n! = 1 + 1*x/2!;
Sum_{n>=0} (n+1)^(n-3)*exp((n+1)*x)*(-x)^n/n! = 1 + 3*x/2!^2 + 1*x^2/3!;
Sum_{n>=0} (n+1)^(n-4)*exp((n+1)*x)*(-x)^n/n! = 1 + 7*x/2!^3 + 11*x^2/3!^2 + 1*x^3/4!;
Sum_{n>=0} (n+1)^(n-5)*exp((n+1)*x)*(-x)^n/n! = 1 + 15*x/2!^4 + 85*x^2/3!^3 + 50*x^3/4!^2 + 1*x^4/5!; ...
which are derived from a LambertW() identity. - _Paul D. Hanna_, Oct 20 2012
- Charles Jordan, Calculus of Finite Differences, Chelsea, 1965.
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function T(n,k) // T = A112492
if k eq 0 or k eq n then return 1;
else return (k+1)^(n-k)*T(n-1,k-1) + Factorial(k)*T(n-1,k);
end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 24 2023
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T[, 0]=1; T[n, m_]:= -m!^(n-m+1)*Sum[(-1)^j*Binomial[m, j]/j^(n-m+ 1), {j,m}]; Table[T[n, m], {n,10}, {m,0,n}]//Flatten (* Jean-François Alcover, Jul 09 2013, from 2nd formula *)
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{h(n,recurse=1) = if(recurse == 0, return(1)); ;
return( sum(k=0,n, h(k,recurse-1) / (1+k) )); }
a(r,c) = h(r-1,c-r) * r!^(c-r) \\ Gottfried Helms, Dec 11 2001
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/* From g.f. for column k: */
T(n,k) = (k+1)!^(n-k+1)*polcoeff(prod(j=0,k,1/(j+1-x +x*O(x^(n-k)))),n-k)
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Oct 20 2012
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/* From g.f. for row n: */
T(n,k) = (k+1)!^(n-k+1)*polcoeff(sum(j=0,k,(j+1)^(j-n-1)*exp((j+1)*x +x*O(x^k))*(-x)^j/j!),k)
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Oct 20 2012
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def T(n,k): # T = A112492
if (k==0 or k==n): return 1
else: return (k+1)^(n-k)*T(n-1,k-1) + factorial(k)*T(n-1,k)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 24 2023
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