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Release calc version 2.10.2t30
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help/nextcand
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help/nextcand
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NAME
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nextcand - next candidate for primeness
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SYNOPSIS
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nextcand(n [,count [, skip [, residue [,modulus]]]])
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TYPES
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n integer
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count integer with absolute value less than 2^24, defaults to 1
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skip integer. defaults to 1
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residue integer, defaults to 0
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modulus integer, defaults to 1
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return integer
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DESCRIPTION
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If modulus is nonzero, nextcand(n, count, skip, residue, modulus)
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returns the least integer i greater than abs(n) expressible as
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residue + k * modulus, where k is an integer, for which
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ptest(i,count,skip) == 1, or if there is no such integer, zero.
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If abs(n) < 2^32, count >= 0, and the returned value i is not zero, then
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i is definitely prime. If count is not zero and the returned
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value i is greater than 2^32, then i is probably prime, particularly
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if abs(count) > 1.
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If skip == 0, and abs(n) >= 2^32 or count < 0, the probabilistic test with
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random bases is used so that if n is composite the
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probability that it passes ptest() is less than 4^-abs(count).
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In any case, if the integer returned by nextcand() is not zero,
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all integers between abs(n) and that integer are composite.
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If skip == 1 (the default value), the bases used in the probabilistic
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test are the first abs(count) primes 2, 3, 5, ...
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For other values of skip, the bases used in the probabilistic tests
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are the abs(count) consecutive integers, skip, skip + 1, skip + 2, ...
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If modulus is zero, the value returned is that of residue if this is
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greater than abs(n) and ptest(residue,count,skip) = 1. Otherwise
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zero is returned.
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RUNTIME
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The runtime for v = nextcand(n, ...) will depend strongly on the
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number and nature of the integers between n and v. If this number
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is reasonably large the size of count is largely irrelevant as the
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compositeness of the numbers betweeen n and v will usually be
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determined by the test for small prime factors or one pseudoprime
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test with some base b. If N > 1, count should be positive so that
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candidates divisible by small primes will be passed over quickly.
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On the average for random n with large word-count N, the runtime seems
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to be roughly K/N^3 some constant K.
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EXAMPLE
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> print nextcand(50), nextcand(112140,-2), nextcand(112140,-3)
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53 112141 112153
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> print nextcand(100,1,1,1,6), nextcand(100,1,1,-1,6)
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103 101
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> print nextcand(100,1,1,2,6), nextcand(100,1,1,303,202)
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1 101
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> print nextcand(2e60, 1, 1, 31, 1e30)
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2000000000000000000000000000053000000000000000000000000000031
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LIMITS
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none
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LIBRARY
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int znextcand(ZVALUE n, long count, long skip, ZVALUE res, ZVALUE mod,
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ZVALUE *cand)
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SEE ALSO
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prevcand, ptest
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