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A282628
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Triangle T(n, k) read by rows: row n gives for n >= 0 the coefficients of the exponential numerator polynomial used for the exponential generating function of {Sum_{j=1..m} (1 + 2*j)^n}_{m>=0}.
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1
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1, 1, 1, 1, 3, 2, 1, 9, 16, 8, 1, 27, 98, 120, 48, 1, 81, 544, 1232, 1152, 384, 1, 243, 2882, 10800, 17760, 13440, 3840, 1, 729, 14896, 87128, 224640, 289920, 184320, 46080, 1, 2187, 75938, 669480, 2544528, 4986240, 5295360, 2903040, 645120, 1, 6561, 384064, 4990112, 26917632, 75204864, 118702080, 107089920, 51609600, 10321920
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OFFSET
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-1,5
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COMMENTS
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The n = -1 row with T(-1, 0) = 1 has been added in order to have a triangle, and is not used.
For n >= 0 the exponential row polynomials are R(n, t) = Sum_{k=0..n+1} T(n, k)*t^k/k!.
The e.g.f. Eodd(n, t) = Sum_{m >= 0} Sodd(n, m)*t^m/m! with Sodd(n, m) = Sum_{j=0..m} (1+2*j)^n is R(n, t)*exp(t), for n >= 0.
This triangle is the e.g.f. companion of A060187 which gives the coefficients of the row polynomials for the o.g.f.s of {Sodd(n, m)}_{m>=0}, which are
G(n, x) = Sum_{k=0..n} A060187(n+1,k+1) * x^k / (1 - x)^(n+2), for n >= 0.
The inverse Laplace transform L^[-1] is used to obtain the present triangle from A060187. To accomplish this the following reordering identity is used:
(Sum_{j=0..n} a(n, j)*x^j)/(1-x)^(n+1) = Sum_{k=0..n} (b(n, k)*x^k/(1-x)^(k+1)), with b(n, k) = Sum_{p=0..min(k,n)} binomial(n-p, k-p)*a(n, p), for n >= 0. This can be proved by multiplying with (1-x)^(n+1) and using the binomial theorem to find first a(n, j) = Sum_{i=0..min(n,j)} (-1)^(j-i)*binomial(n-i, j-i)*b(n,i). This is then inverted by using the binomial identity (5.24) of Graham et al., p. 169, to find b in terms of a.
This leads finally to the inverse Laplace transform formula L^[-1]{(Sum_{j=0..n} a(n, j) * x^j) / (1-x)^(n+1)} = exp(t)*Sum_{k=0..n} b(n, k)*t^k/k!, for n >= 0, with the given expression for b(n, k). This is then applied on the o.g.f. G(n, x) given above.
On can obtain Sodd(n, m) in two ways from S(n,m) = Sum_{j=1..n} j^n by bisection of the j sum: Sodd(n, m) = S(n, 2*(m+1)) - 2^n*S(n, m+1) = S(n, 2*m+1) - 2^n*S(n, m).
The diagonal for n >= 0 is A000165. This is compatible with the second formula for T(n, k) given below.
(with different offsets).
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REFERENCES
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Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2nd ed.; Addison-Wesley, 1994, p. 169, eq. (5.24).
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LINKS
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FORMULA
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T(n, k) = 0 for k > n+1, T(-1, 0) = 1, and
T(n ,k) = Sum_{j=0..min(n+1,k)} binomial(n+1-j,k-j)*A060187(n+1,j+1), for n >= 0 and k = 0..n+1.
T(n, k) = Sum_{j=0..k} (-1)^(k-j) * binomial(k-1, j-1) * (2*j+1)^n, for n >= 0 and k = 0..n+1 (if one puts here binomial(-1, -1) = 1).
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EXAMPLE
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The triangle T(n, k) begins (row n=-1 is not used):
n\k 0 1 2 3 4 5 6 7 8
-1: 1
+0: 1 1
+1: 1 3 2
+2: 1 9 16 8
+3: 1 27 98 120 48
+4: 1 81 544 1232 1152 384
+5: 1 243 2882 10800 17760 13440 3840
+6: 1 729 14896 87128 224640 289920 184320 46080
+7: 1 2187 75938 669480 2544528 4986240 5295360 2903040 645120
...
row n=8: 1 6561 384064 4990112 26917632 75204864 118702080 107089920 51609600 10321920,
row n=9: 1 19683 1933442 36467040 272199360 1042594560 2295175680 3030773760 2376622080 1021870080 185794560. ...
n = 0: Eodd(0, t) = R(0, t)*exp(t) = (1 + 1*t)*exp(t). G(0, x) = 1/(1 - x)^2.
n = 2: Eodd(3, t) = (1 + 9*t + 16*t^2/2! + 8*t^3/3!)*exp(t), G(2, x) = (1 + 6*x + x^2)/(1 - x)^4.
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MATHEMATICA
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Table[Sum[(-1)^(k - j) Binomial[k - 1, j - 1] (2 j + 1)^n, {j, 0, k}], {n, -1, 8}, {k, 0, n + 1}] // Flatten (* Michael De Vlieger, Mar 17 2017 *)
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PROG
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(PARI) {for(n=-1, 8, for(k=0, n + 1, print1(if(k==0, 1, sum(j=0, k, (-1)^(k - j) * binomial(k - 1, j - 1) * (2*j + 1)^n)), ", "); ); print(); ); } \\ Indranil Ghosh, Mar 18 2017
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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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