A244608 - OEIS

%I #58 Jul 17 2020 07:19:49

%S 0,0,1,1,1,4,1,11,-1,1,26,-27,1,57,-289,-1,1,120,-2160,-256,1,247,

%T -13359,-13604,1,1,502,-73749,-383750,3125,1,1013,-378283,-7682623,

%U 1006734,1,1,2036,-1845522,-124221692,126018521,46656

%N Jump Sum Recursion Triangle.

%C The Jump Sum Recursion Triangle is defined as a row by row reading of the coefficients of the recursions satisfied by the error term between Pascal-Triangle j-jump sums and 2^(jm)/j. To clarify this, define the j-jump sum, S(j,n)=Sum_{m= -oo..oo} C(n,jm), with C(x,y) the binomial coefficient if 0<=y<=x and 0 otherwise. It is known that S(1,n)=2^n and S(2,n)=2^(n-1). This suggests the heuristic that S(j,n) is approximately 2^n/j and motivates looking at the integral error terms, jS(j,n)-2^n. So, for fixed j, define the sequence, G(j,m)=G(m)=jS(j,mj)-2^(mj), and let p(j,X) be the characteristic polynomials of the sequence. The Jump Sum Recursion Triangle is defined as a row by row reading of the coefficients of p(1,X), p(2,X), p(3, X), p(4, X)... .

%D Charles Cook and Rebecca Hillman, Some Jump Sum Patterns For the Rows of Pascal's and Related Triangles, Congressus Numerantium, 2010, pp. 255-267.

%D Russell Jay Hendel, Jump Sum Recursions, 16th Fibonacci Conference, Rochester NY, 2014.

%H Michel Marcus, <a href="/A244608/b244608.txt">Table of n, a(n) for n = 1..650</a>

%H John B. Dobson, <a href="http://arxiv.org/abs/1610.09361">A matrix variation on Ramus's identity for lacunary sums of binomial coefficients</a>, arXiv preprint arXiv:1610.09361 [math.NT], 2016.

%F The following theorem can be proved: For j>=3, let w be a primitive j-th root of unity. Let L(k)=Sum_{p=0..j-1} c(p)w^(kp) with c(0)=2 and c(i)=C(j,i) if i>0. Then p(j,X)=(X-L(1))(X-L(2))...(X-L([(n-1)/2])). For example: If j=3, then w is a primitive cube root of unity and [(3-1)/2]=1. We have L(1)=2+3w+3w^2=-1 and X-L(1)=X+1 which is p(3,X) and corresponds to the 3rd row (terms 3 and 4) in the Jump Sum Recursion Triangle.

%e Suppose j=3. Then using our definition, S(3,3)=C(3,0)+C(3,3)=2, S(3,6)=C(6,0)+C(6,3)+C(6,6) = 1+20+1=22 (To explain the name "jump sum," we note, that we don't sum the Pascal Row across, but jump in steps of 3 when taking the sum). G(3,1)=G(1)=3S(3,3)-2^3=-2 and G(2)=3S(3,6)-2^6=2. Clearly, G(2)=-G(1). In fact, the sequence G(m)= G(3,m) - -2,2,-2,2,... - satisfies the recursion G(m)+G(m-1)=0 with characteristic equation X+1. Hence, the 3rd row in the Jump Sum Recursion Triangle is 1,1. Similarly, we can calculate that the sequence G(m)=G(m,4) which is, -8,32,-256,1024,..., which satisfies the recursions G(m)+4G(m-1)=0 with characteristic equation X+4. Hence, the 4th row of the Jump Sum Recursion Triangle is 1,4.  Since S(n,1)=2^n and S(n,2)=2^(n-1), we have 1S(n,1)-2^n=0 and 2S(n,2)-2^(n-1)=0 showing that the sequences G(m,1) and G(m,2) satisfy the zero recursion, G(m)=0. Hence the first two rows of the Jump Sum triangle are identically 0.

%e The Jump Sum Triangle starts:

%e 0;

%e 0;

%e 1, 1;

%e 1, 4;

%e 1, 11, -1;

%e 1, 26, -27;

%e 1, 57, -289, -1;

%e 1, 120, -2160, -256;

%e 1, 247, -13359, -13604, 1;

%e 1, 502, -73749, -383750, 3125;

%e ...

%o (PARI)

%o /* Function CreateTriangle produces first r rows of the Jump Sum Recursion Triangle*/

%o CreateTriangle(r) ={

%o /* First two rows created manually */

%o print(1, "    ", [0]);

%o print(2, "    ", [0]);

%o /* Create r rows of Pascal's triangle: m[i,j]= C(i,j)*/

%o m=matrix(r,r,i,j,0);

%o m[1,1]=1;

%o m[1,2]=1;

%o for(i=2,r,m[i,1]=1);

%o for(i=2,r,for(j=2,r,m[i,j]=m[i-1,j]+m[i-1,j-1]));

%o /*Loop to create rows 3,4,5,...,r of Jump Sum Recursion Triangle */

%o for(o=3,r,

%o /* We create a modified Binomial Coefficient Circulant Matrix,n. The first row of n is 2,C(o,2),C(o,3),..., C(o,o-1).  To motivate this we extend our definitions above as follows:Define S(j,n,k)=Sum_{m=-oo..oo} C(n,k+jm), define G(m,k)=jS(j,mj,k)-2^(mj), and define G(m)= <G(m,0), G(m,1),...,G(m,j-1)>. Then n*G(m) = G(m+1). This gives a matrix representation for the sequence G. The j-th row of the Jump Sum Recursion Triangle which are the coefficients of the recursion satisfied by G is equal to the minimal polynomial satisfied by n. Hence we can print out the triangle by printing out the coefficients of the minimal polynomials*/

%o n=matrix(o,o,i,j,0);

%o for(i=1,o,n[i,i]=2);

%o for(j=1,o-1,for(i=1,o-j,n[i,i+j]=m[o,o+1-j]));

%o for(i=2,o,for(j=1,i-1,n[i,j]=m[o,o+1-i+j]));

%o /* Construct minimal polynomial by multiplying all factors

%o without multiplicity and avoiding factors X-2^n and X */

%o   f=factor(charpoly(n));

%o g=length(f[,1]);

%o h=1;

%o for(i=2,g,if(f[i,1]==x,h,h=h*f[i,1]));

%o print(o,"   ",Vec(h));

%o )};

%o /*The first 11 rows of the Jump Sum Recursion Triangle, listed in the DATA step, can be obtained by calling CreateTriangle with argument r=11.*/

%o CreateTriangle(11);

%K sign,eigen,easy,tabf

%O 1,6

%A _Russell Jay Hendel_, Jul 01 2014