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A244608 Jump Sum Recursion Triangle. 1
0, 0, 1, 1, 1, 4, 1, 11, -1, 1, 26, -27, 1, 57, -289, -1, 1, 120, -2160, -256, 1, 247, -13359, -13604, 1, 1, 502, -73749, -383750, 3125, 1, 1013, -378283, -7682623, 1006734, 1, 1, 2036, -1845522, -124221692, 126018521, 46656 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,6

COMMENTS

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)... .

REFERENCES

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

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

LINKS

Michel Marcus, Table of n, a(n) for n = 1..650

John B. Dobson, A matrix variation on Ramus's identity for lacunary sums of binomial coefficients, arXiv preprint arXiv:1610.09361 [math.NT], 2016.

FORMULA

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.

EXAMPLE

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.

The Jump Sum Triangle starts:

0;

0;

1, 1;

1, 4;

1, 11, -1;

1, 26, -27;

1, 57, -289, -1;

1, 120, -2160, -256;

1, 247, -13359, -13604, 1;

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

...

PROG

(PARI)

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

CreateTriangle(r) ={

/* First two rows created manually */

print(1, "    ", [0]);

print(2, "    ", [0]);

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

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

m[1, 1]=1;

m[1, 2]=1;

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

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

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

for(o=3, r,

/* 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*/

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

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

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

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

/* Construct minimal polynomial by multiplying all factors

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

  f=factor(charpoly(n));

g=length(f[, 1]);

h=1;

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

print(o, "   ", Vec(h));

)};

/*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.*/

CreateTriangle(11);

CROSSREFS

Sequence in context: A301390 A283433 A121463 * A115022 A230534 A177822

Adjacent sequences:  A244605 A244606 A244607 * A244609 A244610 A244611

KEYWORD

sign,eigen,easy,tabf

AUTHOR

Russell Jay Hendel, Jul 01 2014

STATUS

approved

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Last modified February 26 11:15 EST 2021. Contains 341631 sequences. (Running on oeis4.)