

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
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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 PascalTriangle jjump sums and 2^(jm)/j. To clarify this, define the jjump 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^(n1). 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. 255267.
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 jth root of unity. Let L(k)=Sum_{p=0..j1} c(p)w^(kp) with c(0)=2 and c(i)=C(j,i) if i>0. Then p(j,X)=(XL(1))(XL(2))...(XL([(n1)/2])). For example: If j=3, then w is a primitive cube root of unity and [(31)/2]=1. We have L(1)=2+3w+3w^2=1 and XL(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(m1)=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(m1)=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^(n1), we have 1S(n,1)2^n=0 and 2S(n,2)2^(n1)=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[i1, j]+m[i1, j1]));
/*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, o1). 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, j1)>. Then n*G(m) = G(m+1). This gives a matrix representation for the sequence G. The jth 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, o1, for(i=1, oj, n[i, i+j]=m[o, o+1j]));
for(i=2, o, for(j=1, i1, n[i, j]=m[o, o+1i+j]));
/* Construct minimal polynomial by multiplying all factors
without multiplicity and avoiding factors X2^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



