|
%I #36 Sep 14 2017 03:56:38
%S 1,1,1,1,3,2,1,9,16,8,1,27,98,120,48,1,81,544,1232,1152,384,1,243,
%T 2882,10800,17760,13440,3840,1,729,14896,87128,224640,289920,184320,
%U 46080,1,2187,75938,669480,2544528,4986240,5295360,2903040,645120,1,6561,384064,4990112,26917632,75204864,118702080,107089920,51609600,10321920
%N 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}.
%C The n = -1 row with T(-1, 0) = 1 has been added in order to have a triangle, and is not used.
%C For n >= 0 the exponential row polynomials are R(n, t) = Sum_{k=0..n+1} T(n, k)*t^k/k!.
%C 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.
%C 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
%C G(n, x) = Sum_{k=0..n} A060187(n+1,k+1) * x^k / (1 - x)^(n+2), for n >= 0.
%C The inverse Laplace transform L^[-1] is used to obtain the present triangle from A060187. To accomplish this the following reordering identity is used:
%C (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.
%C 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.
%C 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).
%C The first columns of the triangle are A000012, A000244, 2*A005059, 8*A017389, 48*A028060, ...
%C The diagonal for n >= 0 is A000165. This is compatible with the second formula for T(n, k) given below.
%C For the generated sequences for n = 0..4 see A000027, A000290, A000447, A002593, A002309.
%C (with different offsets).
%D Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2nd ed.; Addison-Wesley, 1994, p. 169, eq. (5.24).
%F T(n, k) = 0 for k > n+1, T(-1, 0) = 1, and
%F 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.
%F 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).
%e The triangle T(n, k) begins (row n=-1 is not used):
%e n\k 0 1 2 3 4 5 6 7 8
%e -1: 1
%e +0: 1 1
%e +1: 1 3 2
%e +2: 1 9 16 8
%e +3: 1 27 98 120 48
%e +4: 1 81 544 1232 1152 384
%e +5: 1 243 2882 10800 17760 13440 3840
%e +6: 1 729 14896 87128 224640 289920 184320 46080
%e +7: 1 2187 75938 669480 2544528 4986240 5295360 2903040 645120
%e ...
%e row n=8: 1 6561 384064 4990112 26917632 75204864 118702080 107089920 51609600 10321920,
%e row n=9: 1 19683 1933442 36467040 272199360 1042594560 2295175680 3030773760 2376622080 1021870080 185794560. ...
%e n = 0: Eodd(0, t) = R(0, t)*exp(t) = (1 + 1*t)*exp(t). G(0, x) = 1/(1 - x)^2.
%e 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.
%t 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 *)
%o (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
%Y Cf. A060187, A000165 (diagonal).
%Y Columns: A000012, A000244, 2*A005059, 8*A017389, 48*A028060.
%Y Cf. Generated sequences (with offset differing): A000027, A000290, A000447, A002593, A002309.
%K nonn,easy,tabl
%O -1,5
%A _Wolfdieter Lang_, Mar 14 2017
|
