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A174531 Coefficients of polynomials described below. 4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

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

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.

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

EXAMPLE

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,

...

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 *)

PROG

(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

CROSSREFS

Sequence in context: A084511 A084521 A214923 * A021296 A323100 A210875

Adjacent sequences:  A174528 A174529 A174530 * A174532 A174533 A174534

KEYWORD

nonn

AUTHOR

Vladimir Shevelev, Mar 21 2010

EXTENSIONS

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

STATUS

approved

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Last modified September 19 08:48 EDT 2019. Contains 327189 sequences. (Running on oeis4.)