A174531 Coefficients of polynomials described below.

1, 1, 3, 4, 2, 4, 5, 25, 32, 3, 19, 32, 7, 77, 294, 384, 4, 52, 240, 384, 9, 174, 1323, 4614, 6144, 5, 110, 967, 3934, 6144, 11, 330, 4169, 27258, 90992, 122880, 6, 200, 2842, 21040, 79832, 122880, 13, 559, 10569, 110513, 664898, 2161848, 2949120, 7, 329, 6867, 79687, 533630, 1935048

Offset: 1

Author

Vladimir Shevelev, Mar 21 2010

Keywords

• nonn

Comments

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;

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.

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.

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

Theorem: The polynomials take integer values for integer k. - Vladimir Shevelev and Peter J. C. Moses, Oct 27 2011

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

U_0 = n,

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

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

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

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

V_0 = n/2,

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

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

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

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

Theorem: All terms of the sequence are integers. - Vladimir Shevelev and Peter J. C. Moses, Dec 03 2011

Examples

			Sequence of the polynomials begins:
P_1  =  1,
P_2  =  1,
P_3  =  3*k + 4,
P_4  =  2*k + 4,
P_5  =  5*k^2 +  25*k + 32,
P_6  =  3*k^2 +  19*k + 32,
P_7  =  7*k^3 +  77*k^2 +  294*k + 384,
P_8  =  4*k^3 +  52*k^2 +  240*k + 384,
P_9  =  9*k^4 + 174*k^3 + 1323*k^2 + 4614*k + 6144,
P_10 =  5*k^4 + 110*k^3 +  967*k^2 + 3934*k + 6144,
P_11 = 11*k^5 + 330*k^4 + 4169*k^3 + 27258*k^2 + 90992*k + 122880,
P_12 =  6*k^5 + 200*k^4 + 2842*k^3 + 22040*k^2 + 79832*k + 122880,
...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..2550
• V. Shevelev and P. Moses, On a sequence of polynomials with hypothetically integer coefficients, arXiv:1112.5715 [math.NT], 2011.
• V. Shevelev and P. Moses, On a sequence of polynomials with hypothetically integer coefficients, INTEGERS, V.13 (2013) Article #A9, 1-29.

Programs

Maple

B := proc(x,n)
        mul((x-i+1)/i,i=1..n) ;
        expand(%) ;
end proc:
A174531p := proc(n,x)
        if n = 1 or n =2 then
                1;
        elif type(n,'odd') then
                thalf := (n-1)/2 ;
                (x+n-1)*procname(n-1,x)+(2*x+n)*procname(n-1,x+1)/2 +
                (4*x+n-1)*(thalf-1)!*B(x+thalf-1,thalf-1)/4 ;
        else
                thalf := (n-2)/2 ;
                (x+n-1)*procname(n-1,x)/(4*x+2*n-2)
                +procname(n-1,x+1)/4 +
                (4*x+n-1)*thalf!*B(x+thalf,thalf)/(8*x+4*n-4) ;
        fi ;
        expand(%) ;
        factor(%) ;
        expand(%) ;
end proc:
for n from 1 to 14 do
        for k from floor((n-1)/2) to 0 by -1 do
                printf("%d,",coeftayl(A174531p(n,x),x=0,k)) ;
        end do:
end do: # R. J. Mathar, Sep 27 2011

Mathematica

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 *)

PARI

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))}
for(n=1,19,print(P(n)(k))) /* or, to list coefficients: */
for(n=1,19,apply(t->print1(t","),Vec(P(n)(k)));print)  \\ M. F. Hasler, Sep 27 2011

Formula

From Vladimir Shevelev, Oct 11 2011: (Start)

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,

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).

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)

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

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

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

Extensions

Coefficients of P_12 corrected by D. S. McNeil, Sep 27 2011