### Table 1: Summary of polyomino inversion relations

"... In PAGE 13: ... We shall generalize these results to polygons of any width. Table1 summarizes the self- reciprocity properties we have established. Most of them can be proved in various ways.... In PAGE 15: ... This very general result will be derived in Section 4. Likewise, the inversion relations for three-choice polygons and staircase polygons with a staircase hole, given in Table1 , are also special cases of more general formulae which will be derived in Section 5.... In PAGE 22: ... Theorem 4.1 then gives Hm(1=y; 1=q) = ? 1 yqm Hm(y; q); which implies G(x; y; q) + yG(xq; 1=y; 1=q) = 0: (33) Note that the rst two self-reciprocity relations of Table1 depend quadratically on the variable q: for this reason, they only yield an inversion relation for q = 1. 5 Self-reciprocity via Stanley apos;s general results 5.... ..."

### Table 1: Irreducible Polynomials, (I(x)), for use in N-D PRNS GF(pm) General Multipliers

"... In PAGE 6: ... Finally, (30) specifies a lower limit on p for which the method of this paper is feasible, as, for p = 2 or 3, the mi cannot be gt; 1, rendering a reduction in m impossible. Table1 shows examples where N-D PRNS is necessary (dimension gt; 1). 5 GF(pm) Multiplication Using GF(pm2) Operations, m2|m This section notes a subset of the N-Dimensional PRNS of Section 4.... ..."

### Table 7. Values of I(n), the number of irreducible binary polynomials of degree n.

1985

"... In PAGE 71: ...d) is the Mo. .bius m -function. The formula (A.1) provides an efficient method for computing I(k), the first few values of which are shown in Table7 . In addition, (A.... ..."

Cited by 64

### Table 1: Minimum value of for some power polynomials on F2n.

1997

"... In PAGE 10: ... 4.3 Some APN power polynomials Table1 lists all known exponents t (up to equivalence) such that x 7! xt is APN. But the only APN power polynomials Xt amongst these 4 families which can be used as a round permutation of a Feistel cipher are those corresponding to t 2 Ki with i 6.... In PAGE 11: ... The mapping x 7! xt is not APN over F2n, n 5, if t 2n4 + 4:5 Janwa, McGuire and Wilson [JMW95] proved that gt(X; Y ) is absolutely irreducible for any t 3 mod 4, t gt; 3 and for some values such that t 1 mod 4. They actually conjectured that this curve is absolutely irreducible for all values of t except those lying in the cyclotomic cosets Qi and Ki(see Table1 ). This statement also holds for any t lt; 100.... ..."

Cited by 2

### Table 4. Unresolved conjectured harmonic optima.

2006

"... In PAGE 6: ... However, including reducible configurations would substantially lengthen Table 3 without adding much more geometrical content. Table4 lists two more unresolved cases. They both appear to be harmonic optima and are balanced, and we have not been able to prove or disprove universal optimality.... In PAGE 7: ... On the other hand, each is also analogous to a configuration we know or believe is universally optimal (240 points in R8 and 64 points in R14, respectively). We have not been able to disprove universal optimality in the cases in Table4 , but they are sufficiently large that our failure provides little evidence in favor of universal optimality. Note that the data presented in Tables 3 and 4 may not specify the configurations uniquely.... ..."

Cited by 2

### Table 1: Criteria and Automated Reasoning Tools.

1999

"... In PAGE 3: ... It can also be used to proveunderivabilityof a conjecture,by attempting to generate a model of the premises in which the conjecture is false. Table1 summarizes how to test for the criteria.2... ..."

Cited by 3

### Table 2: Generators of irreducibles of type in A (q), for (321).

"... In PAGE 9: ...3. These bases are too large to present here; instead in Table2 we give one vector in each irreducible Sn-representation. Note that this example is somewhat trivial, since there are only two incomparable partitions, and thus the corresponding lattice is automatically distribu- tive; in addition, the multiplicity of representations in the graded components of the atoms never exceeds 1.... ..."

### Table 2: Generators of irreducibles of type in A (q), for (321).

1997

"... In PAGE 9: ...3. These bases are too large to present here; instead in Table2 we give one vector in each irreducible Sn-representation. Note that this example is somewhat trivial, since there are only two incomparable partitions, and thus the corresponding lattice is automatically distribu- tive; in addition, the multiplicity of representations in the graded components of the atoms never exceeds 1.... ..."

### Table 3. Irreducible trinomials in F3[x]withdegreen 1500:

"... In PAGE 8: ... Table3 counts the irreducible trinomials. A table entry is indexed by (10) e =(n0;k0;s0;a;b) with row index 0 n0 lt; 12, column index 0 k0 lt; 6, s0 2f01; 10; 11g and a; b 2 f 1g.... In PAGE 9: ... For each of these three entries, we also have 161 irreducible trinomials. In the bottom table of Table3 , the rst entry (1;0; ) means that we have 161 irreducible trinomials for e =(1; 0; 10; 1; 1), just as for (1; 0; 10; 1; 1) obtained by negating x.Theentryr(2;0; ) in the row n = 2 points to the (84; 76) irreducible trinomials xn xk +1forn 2, k 0, and s =(10; 11).... In PAGE 9: ... For n 1500 they say that (11) n 0mod4;k 2mod6;xn + axk + b irreducible =) s = 01 (that is, 2(n) gt; 2(k)): We have no explanation for this phenomenon. We also observe several repeated values in Table3 ; namely, (12) n 0mod4=) the entries for (n; k; s; a; b) and (n; (n k)rem6;s;a;b)areequal: Again, we do not know whether this is a coincidence. Open Question 13.... In PAGE 10: ... Table 4 gives the relevant values of T1500(e). The missing values are given by the pointers in Table3 , or else are zero. For every n0 lt; 12, there is some e as in (10) so that the proportion of irreducible polynomials in Tn(e)isatleast fteen times as high as in the set of all polynomials, within our experimental range.... ..."