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Consider coefficients U(m,l,k) defined by the identity Sum_{k=1..l} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,l,k) * T^k that holds for all positive integers l,m,T. This sequence gives 2-column table read by rows, where n-th row lists coefficients U(1,n,k) for k = 0, 1 and n >= 1.
5

%I #45 Feb 09 2024 12:38:19

%S 5,6,28,18,81,36,176,60,325,90,540,126,833,168,1216,216,1701,270,2300,

%T 330,3025,396,3888,468,4901,546,6076,630,7425,720,8960,816,10693,918,

%U 12636,1026,14801,1140,17200,1260,19845,1386,22748,1518,25921,1656

%N Consider coefficients U(m,l,k) defined by the identity Sum_{k=1..l} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,l,k) * T^k that holds for all positive integers l,m,T. This sequence gives 2-column table read by rows, where n-th row lists coefficients U(1,n,k) for k = 0, 1 and n >= 1.

%C For l=T, the identity takes the form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

%H Max Alekseyev, <a href="https://mathoverflow.net/q/309470">Derivation of the general formula for U(m,n,k)</a>, MathOverflow, 2018.

%H Petro Kolosov, <a href="https://arxiv.org/abs/1603.02468">Another Power Identity involving Binomial Theorem and Faulhaber's formula</a>, arXiv:1603.02468 [math.NT], 2016-2018.

%H Petro Kolosov, <a href="https://kolosovpetro.github.io/pdf/OEIS_Um(n,k)_coefficients.pdf">More details on derivation of present sequence</a>.

%H Petro Kolosov, <a href="https://kolosovpetro.github.io/arxiv_1603_02468/identity_1_1_r_h_s.txt">Mathematica program</a>, verifies the identity T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k for m=0,1,...,12.

%H Petro Kolosov, <a href="https://kolosovpetro.github.io/pdf/HistoryAndOverviewOfPolynomialP.pdf">History and overview of the polynomial P_b^m(x)</a>, 2024.

%F U(m,l,t) = (-1)^m * Sum_{k=1..l} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, l >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after _Max Alekseyev_, see links.

%F Conjectures from _Colin Barker_, Aug 03 2019: (Start)

%F G.f.: x*(5 + 6*x + 8*x^2 - 6*x^3 - x^4) / ((1 - x)^4*(1 + x)^4).

%F a(n) = (4 - 4*(-1)^n - 3*(-5+(-1)^n)*n - 3*(-3+(-1)^n)*n^2 + (1+(-1)^(1+n))*n^3) / 8.

%F a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n>8.

%F (End)

%e column column

%e l k=0 k=1

%e --- ------ ------

%e 1 5 6

%e 2 28 18

%e 3 81 36

%e 4 176 60

%e 5 325 90

%e 6 540 126

%e 7 833 168

%e 8 1216 216

%e 9 1701 270

%e 10 2300 330

%e 11 3025 396

%e 12 3888 468

%e ...

%t (* Define the R[n,k] := A302971(n,k)/A304042(n,k) *)

%t R[n_, k_] := 0

%t R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*

%t Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*

%t BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n

%t R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;

%t (* Define the U(m,l,t) coefficients *)

%t U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];

%t (* Define the value of the variable 'm' to be m = 1 for A320047 *)

%t m = 1;

%t (* Print first 10 rows of U(m,l,t) coefficients for 'm' defined above *)

%t Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

%Y The case m=2 is A316349.

%Y The case m=3 is A316387.

%Y Column k=0 is A275709.

%Y Column k=1 is A028896.

%Y Cf. A302971, A304042, A287326, A300656, A300785.

%K nonn,tabf

%O 1,1

%A _Kolosov Petro_, Oct 04 2018