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Given a sequence of length n , an approximation to h can be reconstructed using Max[MapIndexed[#1/First[#2] &, FoldList[Plus, First[list], Rest[list]]]] The fractional part of the result obtained is always an element of the Farey sequence Union[Flatten[Table[a/b, {b, n}, {a, 0, b}]]] (See also pages 892 , 932 and 1084 .)
[Turing] machine 596440 For any list of initial colors init , it turns out that successive rows in the first t steps of the compressed evolution pattern turn out to be given by NestList[Join[{0}, Mod[1 + Rest[FoldList[Plus, 0, #]], 2], {{0}, {1, 1, 0}} 〚 Mod[Apply[Plus, #], 2] + 1] 〛 &, init, t] Inside the right-hand part of this pattern the cell values can then be obtained from an upside-down version of the rule 60 additive cellular automaton, and starting from a sequence of 1 's the picture below shows that a typical rule 60 nested pattern can be produced, at least in a limited region.
But for multiway systems where each rule p  q is accompanied by its reverse q  p , and such pairs are represented say by "AAB" ↔ "BBAA" , an equivalent operator system can immediately be obtained either from Apply[Equal, Map[Fold[#2[#1] &, x, Characters[#]] &, rules, {2}], {1}] or from (compare page 1172 ) Append[Apply[Equal, Map[(Fold[f, First[#], Rest[#]] &)[Characters[#]] &, rules, {2}], {1}], f[f[a, b], c]  f[a, f[b, c]]] where now objects like "A" and "B" are treated as constants—essentially functions with zero arguments.
Based on this LE[list_] := Module[{n = Length[list], i = Max[MapIndexed[ #1 - #2 &, PrimePi[list]]] + 1}, CRT[PadRight[ list, n + i], Join[Array[Prime[i + #] &, n], Array[Prime, i]]]] will yield a number x that can be decoded into a list of length n using essentially the so-called Gödel β function Mod[x, Prime[Rest[NestList[NestWhile[# + 1 &, # + 1, Mod[x, Prime[#]]  0 &] &, 0, n]]]]
Starting with a list of the initial conditions for s steps, the configurations for the next s steps are given by Append[Rest[list], Map[Mod[Apply[Plus, Flatten[c #]], 2]&, Transpose[ Table[RotateLeft[list, {0, i}], {i, -r, r}], {3, 2, 1}]]] where r = (Length[First[c]] - 1)/2 .
But an important difference is that it would occur even if one could in effect perfectly isolate a system from the rest of the universe.
The first m rules (which yield far more than m elements of the original sequence) are obtained for any h that is not a rational number from the continued fraction form (see page 914 ) of h by Map[(({0  Join[#, {1}], 1  Join[#, {1, 0}]} &)[Table[0, {# - 1}]]) &, Reverse[Rest[ContinuedFraction[h, m]]]] Given these rules, the original sequence is given by Floor[h] + Fold[Flatten[#1 /. #2] &, {0}, rules] If h is the solution to a quadratic equation, then the continued fraction form is repetitive, and so there are a limited number of different substitution rules.
There is no known simple formula for the number of ways of expressing an integer as a sum of three squares, although part of the condition in the main text for integers to be expressible in this way was established by René Descartes in 1638 and the rest by Adrien Legendre in 1798. … Hardy and John Littlewood in 1922 to be proportional to 2n Apply[Times, Map[(# - 1)/(# - 2)&, Map[First, Rest[FactorInteger[n]]]]]/Log[n] 2 It was proved in 1937 by Ivan Vinogradov that any large odd integer can be expressed as a sum of three primes.
In the first 200 billion digits, the frequencies of 0 through 9 differ from 20 billion by {30841, -85289, 136978, 69393, -78309, -82947, -118485, -32406, 291044, -130820} An early approximation to π was 4 Sum[(-1) k /(2k + 1), {k, 0, m}] 30 digits were obtained with 2 Apply[Times, 2/Rest[NestList[Sqrt[2 + #]&, 0, m]]] An efficient way to compute π to n digits of precision is (# 〚 2 〛 2 /# 〚 3 〛 )& [NestWhile[Apply[Function[{a, b, c, d}, {(a + b)/2, Sqrt[a b], c - d (a - b) 2 , 2 d}], #]&, {1, 1/Sqrt[N[2, n]], 1/4, 1/4}, # 〚 2 〛 ≠ # 〚 2 〛 &]] This requires about Log[2, n] steps, or a total of roughly n Log[n] 2 operations (see page 1134 ).
If all possible sequences of colors were allowed, then there would be k possibilities for what block could follow a given block, given by Map[Rest, Table[Append[list, i], {i, 0, k - 1}]] .
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