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 A174531 Coefficients of polynomials described below. 4

%I

%S 1,1,3,4,2,4,5,25,32,3,19,32,7,77,294,384,4,52,240,384,9,174,1323,

%T 4614,6144,5,110,967,3934,6144,11,330,4169,27258,90992,122880,6,200,

%U 2842,21040,79832,122880,13,559,10569,110513,664898,2161848,2949120,7,329,6867,79687,533630,1935048

%N Coefficients of polynomials described below.

%C These polynomials are defined by the recursion: P_1=1, P_2=1, and for n >= 2, P_{n+1}(k) = (k+n)*P_n(k) +(1/2)*(2*k+n+1)*P_n(k+1) + (1/4)*(4*k+n)*(n/2-1)!*binomial(k+n/2-1,n/2-1), if n is even;

%C P_{n+1}(k) = ((k+n)/(4*k+2*n))*P_n(k) + (1/4)*P_n(k+1) +((4*k+n)/(8*k+4*n))*((n-1)/2)!*binomial(k+(n-1)/2,(n-1)/2), if n is odd.

%C Conjectures: (1) all terms of the sequence are integers; (2) P_n(0) = (floor((n-1)/2))!*4^(floor((n-1)/2)); (3) if n is even, then the coefficients of P_n do not exceed the corresponding coefficients of P_(n-1) and the equality holds only for the last ones; (4) all coefficients of P_n, except of the last one, are multiples of n iff n is prime.

%C More conjectures: For even x, P[x, 1] = (2^x - 1) (x/2)! / (x+1), and for odd x, P[x, 1] = (2^x - 1) ((x-1) / 2)!. - _Peter J. C. Moses_, Sep 28 2011

%C Theorem: The polynomials take integer values for integer k. - _Vladimir Shevelev_ and _Peter J. C. Moses_, Oct 27 2011

%C Theorem: (a) P_n(x) is a polynomial of degree m=floor((n-1)/2) whose coefficients multiplied by m! are integers; (b) if P_n(x) = a_0(n){x}^{m} + a_1(n){x}^{m-1} + ... + a_{m-1}(n){x} + a_m(n), then a_i(n)=U_i(n), if n is odd, and a_i(n)=V_i(n), if n is even, where U_i, V_i are polynomials in n of degree 2i+1. The first such polynomials are

%C U_0 = n,

%C U_1 = (n-1)*n*(7*n-5)/24,

%C U_2 = (n-3)*(n-1)*n*(29*n^2 - 44*n + 7)/640,

%C U_3 = (n-5)*(n-3)*(n-1)*n*(1581*n^3 - 3775*n^2 + 1587*n + 223)/322560,

%C U_4 = (n-7)*(n-5)*(n-3)*(n-1)*n*(12683*n^4 - 42160*n^3 + 31378*n^2 + 3568*n - 2013)/30965760;

%C V_0 = n/2,

%C V_1 = (n-2)*n*(7*n-4)/48,

%C V_2 = (n-4)*(n-2)*n*(87*n^2 - 98*n + 16)/3840,

%C V_3 = (n-6)*(n-4)*(n-2)*n*(1581*n^3 - 2686*n^2 + 936*n + 64)/645120,

%C V_4 = (n-8)*(n-6)*(n-4)*(n-2)*n*(12683*n^4 - 29372*n^3 + 16228*n^2 + 1040*n - 768)/61931520. - _Vladimir Shevelev_ and _Peter J. C. Moses_, Nov 26 2011

%C Theorem: All terms of the sequence are integers. - _Vladimir Shevelev_ and _Peter J. C. Moses_, Dec 03 2011

%H Vincenzo Librandi, <a href="/A174531/b174531.txt">Table of n, a(n) for n = 1..2550</a>

%H V. Shevelev and P. Moses, <a href="http://arxiv.org/abs/1112.5715">On a sequence of polynomials with hypothetically integer coefficients</a>, arXiv:1112.5715 [math.NT], 2011.

%H V. Shevelev and P. Moses, <a href="http://www.emis.de/journals/INTEGERS/papers/n9/n9.Abstract.html">On a sequence of polynomials with hypothetically integer coefficients</a>, INTEGERS, V.13 (2013) Article #A9, 1-29.

%F From _Vladimir Shevelev_, Oct 11 2011: (Start)

%F P_n(k) = c_n(k)*(2^(n+k-1) - R_k(n)/(2k-2)!!), where c_n(k) = ((n-1)/2)!*Product_{i=1..k-1} (n+i)/(n+2i), if n is odd,

%F c_n(k) = (1/2)*((n-2)/2)!*Product_{i=0..k-1} (n+i)/(n+2i+1), if n is even, and R_k(n), k=0,1,..., are polynomials in n of degree k-1 (for k >= 1) with integer coefficients, defined by the recursion R_0(n)=0, R_1(n)=1, and for k >= 1, R_(k+1)(n) = 4*k*(R_k(n+1) - R_k(n)) + (n+4k)*Product_{i=1..k-1} (n+k+i).

%F The sequence of polynomials R's for k >= 0 begins: 0, 1, n+4, n^2 + 11*n + 32, n^3 + 21*n^2 + 152*n + 384, n^4 + 34*n^3 + 443*n^2 + 2642*n + 6144, ... This formula proves the conjecture of V. Shevelev for P_n(0) and P. Moses for P_n(1). (End)

%F Formula for positive argument k: P_n(k) = ((n-1)/2)! * binomial((n-1)/2+k-1,k-1) / binomial(n+2*k-2,k-1) * T_n(k), if n is odd; P_n(k) = (n/2-1)! * binomial(n/2+k-1,k) / binomial(n+2*k-1,k) * T_n(k), if n is even, where T_n(k) = Sum_{i=1..n} * 2^(i-1) * binomial(n+2*k-i-1,k-1). - _Vladimir Shevelev_, Oct 23 2011

%F For even n >= 4, P_n(x) = (n+x-1)*P_(n-2)(x+1) + (x - 1 + n/2)*(x - 2 + n/2)...(x+1); for odd n >= 3, P_n(x) = 2(n + x - 1)*P_(n-1)(x) + (x - 1 + (n-1)/2)*(x - 2 + (n-1)/2)...x. These formulas yield the integrality of all coefficients of P_n(x). - _Vladimir Shevelev_ and _Peter J. C. Moses_, Dec 03 2011

%F Formula for negative argument: P_n(-m) = 2^(n-2m-1) * binomial(n-m-1,m-1)/binomial((n-2)/2,m-1) * ((n-2)/2)!, if n is even and 0 < m <= (n-2)/2; P_n(-m) = 2^(n-2m-1) * (binomial(n-m-1,m)/binomial((n-1)/2,m)) * ((n-1)/2)!, if n is odd and 0 < m <= (n-1)/2. It is interesting that this formula is significantly simpler than the above formula for positive arguments. - _Vladimir Shevelev_, Feb 14 2013; edited by _Joerg Arndt_ and _M. F. Hasler_, Feb 23 2013

%e Sequence of the polynomials begins:

%e P_1 = 1,

%e P_2 = 1,

%e P_3 = 3*k + 4,

%e P_4 = 2*k + 4,

%e P_5 = 5*k^2 + 25*k + 32,

%e P_6 = 3*k^2 + 19*k + 32,

%e P_7 = 7*k^3 + 77*k^2 + 294*k + 384,

%e P_8 = 4*k^3 + 52*k^2 + 240*k + 384,

%e P_9 = 9*k^4 + 174*k^3 + 1323*k^2 + 4614*k + 6144,

%e P_10 = 5*k^4 + 110*k^3 + 967*k^2 + 3934*k + 6144,

%e P_11 = 11*k^5 + 330*k^4 + 4169*k^3 + 27258*k^2 + 90992*k + 122880,

%e P_12 = 6*k^5 + 200*k^4 + 2842*k^3 + 22040*k^2 + 79832*k + 122880,

%e ...

%p B := proc(x,n)

%p mul((x-i+1)/i,i=1..n) ;

%p expand(%) ;

%p end proc:

%p A174531p := proc(n,x)

%p if n = 1 or n =2 then

%p 1;

%p elif type(n,'odd') then

%p thalf := (n-1)/2 ;

%p (x+n-1)*procname(n-1,x)+(2*x+n)*procname(n-1,x+1)/2 +

%p (4*x+n-1)*(thalf-1)!*B(x+thalf-1,thalf-1)/4 ;

%p else

%p thalf := (n-2)/2 ;

%p (x+n-1)*procname(n-1,x)/(4*x+2*n-2)

%p +procname(n-1,x+1)/4 +

%p (4*x+n-1)*thalf!*B(x+thalf,thalf)/(8*x+4*n-4) ;

%p fi ;

%p expand(%) ;

%p factor(%) ;

%p expand(%) ;

%p end proc:

%p for n from 1 to 14 do

%p for k from floor((n-1)/2) to 0 by -1 do

%p printf("%d,",coeftayl(A174531p(n,x),x=0,k)) ;

%p end do:

%p end do: # _R. J. Mathar_, Sep 27 2011

%t p[1,k_]:=1; p[2,k_]:=1; p[t_?OddQ,k_]:=p[t,k]=Expand[Factor[FunctionExpand[(k+t-1) p[t-1,k]+1/2 (2 k+t) p[t-1,k+1]+1/4 (4 k+t-1) (1/2 (t-3))! Binomial[1/2 (2 k+t-3),1/2 (t-3)]]]]; p[t_?EvenQ,k_]:=p[t,k]=Expand[Factor[FunctionExpand[((k+t-1) p[t-1,k])/(4 k+2 (t-1))+1/4 p[t-1,k+1]+((4 k+t-1) (1/2 (t-2))! Binomial[k+t/2-1,1/2 (t-2)])/(8 k+4 (t-1))]]]; A174531=Flatten[Table[Reverse[CoefficientList[p[n,k],k]],{n,1,14}]] (* _Peter J. C. Moses_, Sep 28 2011 *)

%o (PARI)

%o P(n)={ n<3 & return(k->1); if( n%2, k-> (k+n-1)*P(n-1)(k)+(1/2)*(2*k+n)*P(n-1)(k+1)+(1/4)*(4*k+n-1)*((n-3)/2)!*binomial(k+(n-3)/2,(n-3)/2), k-> ((k+n-1)/(4*k+2*n-2))*P(n-1)(k)+(1/4)*P(n-1)(k+1)+((4*k+n-1)/(8*k+4*n-4))*(n/2-1)!* binomial(k+n/2-1,n/2-1))}

%o for(n=1,19,print(P(n)(k))) /* or, to list coefficients: */

%o for(n=1,19,apply(t->print1(t","),Vec(P(n)(k)));print) \\ _M. F. Hasler_, Sep 27 2011

%K nonn

%O 1,3

%A _Vladimir Shevelev_, Mar 21 2010

%E Coefficients of P_12 corrected by _D. S. McNeil_, Sep 27 2011

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Last modified April 7 11:07 EDT 2020. Contains 333301 sequences. (Running on oeis4.)