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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) such that its 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(7n-5)/24,

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

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

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

V_0=n/2,

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

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

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

V_4=(n-8)(n-6)(n-4)(n-2)n(12683n^4-29372n^3+16228n^2+1040n-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

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

  c_n(k)=(1/2)*((n-2)/2)!*prod{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)*prod{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 conjecture of V. Shevelev for P_n(0) and P. Moses for P_n(1). - Vladimir Shevelev, Oct 11 2011

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 A210875 A238373

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 24 18:54 EDT 2017. Contains 292433 sequences.