Andy Martin - cp's OEIS frontend

User: Andy Martin

Andy Martin has authored 6 sequences.

A275939 Consider the prime race mod q (where q >= 2) between qk+1 and qk-1. Terms are numbers k where qk+1 first takes lead over qk-1.

3, 608981813029, 26861, 11, 608981813017, 71, 192252423729713, 37, 11, 23

Offset: 2

Andy Martin, Aug 12 2016

Values are available for all 2 <= q <= 999 except for 12 and 24. If q is odd and > 3 then 2*q will have the same value in the sequence as q.
Additional terms starting with q = 12 are:
unknown, 53, 71, 331, 17, 239, 37, 213827, 1381, 673, 23, 47, unknown, 101, 53, 379, 29, 59, 331
The longest q*k+1 versus q*k-1 races up to q = 999 are for q = 3,6,8,12,24 and 168. When q = 168 the race ends at prime 273084304417.
The mod 12 and 24 races were checked by computer to 1.1 * 10^14 without q*k+1 ever leading.
Kevin Ford (private communication) provides the following information on these races: "My paper with Richard Hudson contains a lot of information about the location of sign changes for pi(x,q,a)-pi(x,q,b).  Corollary 4 has rigorous upper bounds, but these will likely not be useful to you.  The information in Tables 2 and 3 will be more helpful, as these provide the most likely places to look for the first sign change.  In the case of the mod 12 race, it is probably around exp(187.536), or about 2.79 x 10^{81}.  For the mod 24 race, it's about exp(43.453)=7.437... x 10^{18}".

			For the fourth term q is 5. For primes 2,3,5 and 7 the mod 5 values are 2,3,0 and 2 respectively, so there is no change in the race. For the next prime 11, mod 5 gives 1, q*k+1 now leads 1 to 0, and the race is over.

Ford, Kevin; Konyagin, Sergei; Chebyshev's conjecture and the prime number race. IV International Conference "Modern Problems of Number Theory and its Applications": Current Problems, Part II (Russian) (Tula, 2001), 67-91.
Paulo Ribenboim, The Little Book of Big Primes, Springer 1991

Kevin Ford and Richard H. Hudson, Sign changes in pi q,a(x) - pi q,b(x), Acta Arithmetica 100 (2001), 297-314.
Kevin Ford and Sergei Konyagin, Chebyshev's conjecture and the prime number race
Andrew Granville and Greg Martin, Prime Number Races, Amer. Math. Monthly, 113 (No. 1, 2006), 1-33.
Andy Martin, Values for 2 to 999 as far as they are known

Cf. A007350, A007352.

/*
C language program used to investigate prime number races.
Computes the first lead of qn+1 over qn-1 for q from 2 to 999.
By Andy Martin oldadit@gmail.com 8/12/2016.
Requires Kim Walisch's primesieve library from http://primesieve.org
Iteration based on the primesieve_iterator.c example.
*/
#include 
#include 
#include 
#define UPDATE_COUNT 10000000000ull
void race(uint64_t q)
{
  uint64_t prime  = 0;
  uint64_t m1     = 0;
  uint64_t m_1    = 0;
  uint64_t rem    = 0;
  uint64_t update = UPDATE_COUNT;
  primesieve_iterator pi;
  primesieve_init(&pi);
  while (prime = primesieve_next_prime(&pi)) {
    if ((rem = prime % q) == 1){
      m1 += 1;
    } else if (rem == q-1) {
      m_1 += 1;
    }
    if (m1 > m_1){
      printf("Race mod %3llu ends at %12llu with %11llu pi(x;%llu,1) and %11llu pi(x;%llu,%llu)\n",
             q, prime, m1, q, m_1, q, q-1);
      break;
    }
    /* Enable for update on long races where q = 3,6,8,12,24,168 */
    if (prime > update) {
      printf("  Race mod %llu ongoing at prime %llu with m1 %llu and m_1 %llu diff: %llu\n",
             q, prime, m1, m_1, m_1 - m1);
      update += UPDATE_COUNT;
    }
  }
  primesieve_free_iterator(&pi);
}
int main()
{
  uint64_t i;
  for(i=2; i<1000; i++){ race(i); }
  return(0);
}

a(8)-a(11) from Andy Martin, Aug 15 2016

A272043 a(n) is the shyest prime in base n.

Original entry on oeis.org

2, 3, 31, 13, 523, 31, 3833, 491, 5483, 523, 18149, 661, 44657, 3833, 18869, 7333, 165479, 5483, 153953, 20411, 129127, 18149, 538651, 7079, 932257, 44657, 417037, 52639, 2223773, 18869, 3124217, 175229, 1993763, 165479, 2794811, 50461, 8678963, 153953

Offset: 1

Andy Martin, Apr 18 2016

Terminology: consider pairs of final digits of consecutive primes (a,b). Then of all pairs (3,1) is found last in the prime sequence, corresponding to (523, 541). This is termed the shyest pair, with 523 the shyest prime.
Consider final digit pairs (a,b) of consecutive primes.
There are three unique pairs: (2, 3) (3, 5) (5, 7)
For the remaining 16 pairs, record the first observed primes corresponding to the pair:
Initial prime           -- Second prime (mod 10) ---
      (mod 10)          1        3        7        9
            1     181,191   11, 13   31, 37  401,409
            3     523,541  283,293   13, 17   23, 29
            7       7, 11   47, 53  337,347   17, 19
            9      29, 31   19, 23   89, 97  139,149
523,541 is the largest pair, thus the last to occur in the sequence of primes. The first member of this pair is the shyest prime, base 10. (Note that if we consider two digit pairs (ab, cd) then 40191937, 40192037 is the shyest pair for base 10.)
For base 3 the table is:
Initial prime      Second prime (mod 3)
      (mod  3)     0         1       2
            0      -         -     3,5
            1      -     31,37    7,11
            2      2,3     5,7   23,29
and 31 is the shyest prime base 3.

Giovanni Resta, Table of n, a(n) for n = 1..200
Erica Klarreich, Mathematicians Discover Prime Conspiracy, Quanta Magazine, March 13, 2016
Robert J. Lemke Oliver and Kannan Soundararajan, Unexpected biases in the distribution of consecutive primes, arXiv:1603.03720 [math.NT], 2016.
Andy Martin, Ruby code output of first 35 terms and additional data
Terence Tao, Biases between consecutive primes, blog entry March 14, 2016.

Cf. A269364, A270310.

a[n_] := Block[{g,p,m,q,k, e= First /@ Select[ Tally[ Mod[ Prime@ Range[n* 100], n]], #[[2]] > 50 &], A}, A = Association@ Table[{i,j} -> 0, {i,e}, {j,e}]; g = Length[e]^2; m=p=2; While[g > 0, q = NextPrime@p; k = Mod[{p, q}, n]; If[ Lookup[A, Key@k, 1] == 0, A[k] = 1; g--]; m=p; p=q]; m]; Array[a, 25] (* Giovanni Resta, Apr 19 2016 *)

require 'Prime'
# Ruby Code
# Generates Hash with first occurrences of all possible pairs (a,b)
# of final digits for consecutive primes in specified base.
def gen_hash(h, base)
  last_prime = 2
  iteration = last_found = 0
  Prime.each() do |prime|
    # This check could be improved & may be invalid for bases above 35.
    return if (iteration+=1) > 10000 && iteration > 2 * last_found
    next if prime == 2
    l =  last_prime.to_s(base)[-1]
    p =  prime.to_s(base)[-1]
    if h[[l,p]].nil?
      h[[l,p]] = [last_prime,prime]
      last_found = iteration
    end
    last_prime = prime
  end
end
puts "First Prime  Second Prime  Base  Difference  Different  Final Digits In"
puts "                                                 Pairs    Base Notation"
puts "          2             3     1           1          1              1 1"
# For bases above 35 additional programming needed.
2.upto(35){|base|
  gen_hash(h = Hash.new, base)
  p0 = h.values.sort.last[0]
  p1 = h.values.sort.last[1]
  printf("%11d  %12d  %4d  %10d %10d              %s %s\n",
  p0, p1, base, p1 - p0, h.length, p0.to_s(base)[-1], p1.to_s(base)[-1])
}

a(22)-a(38) from Giovanni Resta, Apr 19 2016

A178983 The smallest cube containing n as a substring.

Original entry on oeis.org

0, 1, 27, 343, 64, 125, 64, 27, 8, 729, 1000, 91125, 125, 1331, 140608, 15625, 216, 1728, 85184, 2197, 205379, 216, 226981, 103823, 13824, 125, 9261, 27, 1728, 729, 39304, 1331, 5832, 1331, 343, 35937, 97336, 3375, 13824, 39304, 4096, 531441, 42875, 343

Offset: 0

Andy Martin, Jan 02 2011

			a(3) = 7^3 = 343, because it contains 3 as a substring and no smaller cube contains 3.

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A029943, A029947.

subs[n_] := Module[{d = IntegerDigits[n], len}, len = Length[d]; Union[Flatten[Table[FromDigits[Take[d, {i, k}]], {k, len}, {i, k}]]]]; Table[k = 0; While[! MemberQ[subs[k^3], n], k++]; k^3, {n, 0, 100}] (* T. D. Noe, Nov 06 2013 *)
With[{cbs=Range[0,100]^3},Table[SelectFirst[cbs,SequenceCount[IntegerDigits[#],IntegerDigits[n]]>0&],{n,0,50}]] (* Harvey P. Dale, Nov 17 2024 *)

# For a given nonnegative integer n,
# find the smallest nonnegative cube that contains it as a substring.
NUM_TERMS = 30
(0...NUM_TERMS).each{ |i|
  (0..(1.0/0.0)).each{ |j|
    (print "#{j*j*j}" + ", "; break) if "#{j*j*j}".include?("#{i}")
  }
}

Edited by Alois P. Heinz, Jan 02 2011

A161170 Least integer k such that the n-almost prime count is equal to the prime count.

Original entry on oeis.org

10, 125, 1809, 37820, 2722768, 1037849736, 4204496515890, 476649763963226416

Offset: 2

Andy Martin, Jun 04 2009

Related to sequence A125149, but we compare the prime count to the semiprime count, then the product-of-three-primes count, and so on.
We start with a the number two, and a prime count of 1.
Then the prime count and semiprime count are first identical when k = 10, the prime count is 4 ({2, 3, 5, 7}) and the semiprime count is also 4 ({4, 6, 9, 10}).
Next, when k = 125 the prime count of 30 and product-of-three-primes count of 30 are first identical.
Based on the write up for A125149 and examination of the factor counts as k increases, we believe this sequence is infinite, but have not proved this.

			a(2) = 10 since there are now 4 primes ({2, 3, 5, 7}) and 4 semiprimes ({4, 6, 9, 10}) <= 10.
a(3) = 125 with 30 primes and 30 products of 3 primes.
a(4) = 1809 with 279 primes and 279 products of 4 primes.
a(5) = 37820 with 4000 primes and 4000 products of 5 primes.
a(6) = 2722768 with 198183 primes and 198183 products of 6 primes.
a(7) = 1037849736 with 52672391 primes and 52672391 products of 7 primes.
a(8) = 4204496515890 with 150007470826 primes and 150007470826 products of 8 primes.
a(9) = 476649763963226416 with 12012658440940682 primes and 12012658440940682 products of 9 primes.

Cf. A125149.

use ntheory ":all"; my($k,@S)=(0,map{0}1..20); forfactored { $S[@]++; while ($S[1] == $S[$k]) { print "$k $\n" if $k>1; $k++; } } 3e6; # Dana Jacobsen, Jan 18 2019

# A slow program to generate sequence
# Faster C code is available by request
# Tallies number of primes, semiprimes, trieneprimes ...
tally = Hash.new { |h,k| h[k] = 0}
# Returns number of factors of num
def factors(num)
(2..(Math.sqrt(num).to_i)).each{ |i|
return factors(num/i) + 1 if num % i == 0
}
1
end
# Testing number of primes against composites with num_factors
num_factors = 2
2.upto( 1.0/0.0 ) { |i|
tally[factors(i)] +=1
if tally[1] == tally[num_factors]
puts "k: #{i} Primes: #{tally[1]} Composites with #{num_factors} factors: #{tally[num_factors]}"
num_factors += 1
end
}

Edited example and a(8) from Donovan Johnson, Mar 10 2010

a(9) from Henri Lifchitz, Mar 17 2025

A173515 Consider positive integer solutions to x^3+ y^3 = z^3 - n or 'Fermat near misses' of 1, 2, 3 ... Arrange known solutions by increasing values of n. Sequence gives value of lowest z for a given n.

Original entry on oeis.org

9, 7, 2, 812918, 18, 217, 4, 3, 9730705, 332, 14, 135, 3, 19, 156, 16, 15584139827, 3, 139643, 6, 1541, 4, 2220422932, 5, 14, 4, 445, 12205, 9, 8, 16234, 815, 31, 4

Offset: 1

Andy Martin, Feb 20 2010

nonn

The submitted values are for z when 0 < n < 51. There is no solution for any n congruent to 4 or 5 mod 9. This eliminates 4,5,13,14,22,23,31,32,40,41,49 and 50 in the 0-to-50 range.
Per the Elsenhans and Jahnel link there are no solutions found for 3, 33, 39 and 42 in the 0-to-50 range, with a search bound of 10^14.
If sequences could contain 'nil' for no solution, and '?' for cases where a solution is not known, but might exist, then a more concise definition is possible: Least positive integer such that a(n)^3 - n is the sum of two positive cubes. The sequence would then start: 9, 7, ?, nil, nil, 2.

			6^3 + 8^3 = 9^3 - 1: There are no solutions when n = 1 for z < 9, thus the first term is 9.
5^3 + 6^3 = 7^3 - 2: There are no solutions for z < 7, thus the second term is 7.
It is unknown if there is a solution when n = 3.
It is known there are no solutions when n = 4 and 5.
1^3 + 1^3 = 2^3 - 6, thus the third term is 2.

Andreas-Stephan Elsenhans and Joerg Jahnel, List of solutions of x^3 + y^3 + z^3 = n for n < 1000 neither a cube nor twice a cube
D.R. Heath-Brown, W.M. Lioen and H.J.J. te Riele, on Solving the Diophantine Equation x^3 + y^3 + z^3 = k on a Vector Computer
Kenji Koyama, Yukio Tsuruoka, and Hiroshi Sekigawa, On Searching For Solutions of the Diophantine Equation x^3 + y^3 + z^3 = n, Math. Comp. 66 (1997), 841-851.
Eric S. Rowland, Known Families of Integer Solutions of x^3 + y^3 + z^3 = n

Cf. A050788, A159935.

# x^3 + y^3 = z^3 - n
# Solve for all z less than z_limit, and
# n less than n_limit.
# When n = 7, z = 812918 and faster code and language are needed.
# However, by optimizing this code slightly and running for 2 days
# the author was able to search all z < 164000 and n < 100
#
n_limit = 7 # Configure as desired
z_limit = 20 # Configure as desired
h = {}
(2..z_limit).each{ |z|
  (1..(z-1)).each{ |y|
  (1..(y)).each{ |x|
  n = z*z*z - x*x*x - y*y*y
  if n > 0 && n < n_limit && h[n].nil?
  puts "Found z = #{z} when #{x}^^3 + #{y}^^3 = #{z}^^3 - #{n}"
  h[n] = z
  end
} } } print "\nPartial sequence generated when n < #{n_limit} and z is searched to #{z_limit} is:\n"
h.sort.each{|k,v| print "#{v}, " }
print "\b\b \n"

Author conjectures that no explicit formula or recurrence exists.

A137442 n^2 followed by smallest integer not yet listed.

Original entry on oeis.org

1, 2, 4, 3, 9, 5, 16, 6, 25, 7, 36, 8, 49, 10, 64, 11, 81, 12, 100, 13, 121, 14, 144, 15, 169, 17, 196, 18, 225, 19, 256, 20, 289, 21, 324, 22, 361, 23, 400, 24, 441, 26, 484, 27, 529, 28, 576, 29, 625, 30, 676, 31, 729, 32, 784, 33, 841, 34, 900, 35, 961, 37, 1024, 38

Offset: 1

Andy Martin, Apr 18 2008

Sequence is a permutation of the positive integers.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000463.

f[s_List] := Block[{k = 1}, While[ MemberQ[s, k], k++ ]; Flatten@ Append[s, {((2 + Length@s)/2)^2, k}]]; Nest[f, {1, 2}, 33] (* Robert G. Wilson v, May 31 2009 *)
Module[{nn=40,sq,int,len},sq=Range[nn]^2;int=Complement[Range[nn],sq];len=Min[Length[int],nn];Riffle[Take[sq,len],Take[int,len]]](* Harvey P. Dale, Nov 05 2013 *)

lista(nn) = {for (n=1, nn, print1(n^2, ", ", n+round(sqrt(n)), ", "););} \\ Michel Marcus, Nov 02 2014

a(n) = if (n % 2, ((n+1)/2)^2, (n/2)+round(sqrt(n/2))); \\ Michel Marcus, Nov 02 2014

# correct to any term:
sk_ct = 2
skip = 4
at = 1
(1..(1.0/0)).each{ |i|
if (at+=1) == skip
at+=1
sk_ct +=1
skip = sk_ct * sk_ct
end
print i*i, " ", at, " "
}

# Simpler Ruby code, correct until i is so large that floating point rounding causes errors. I estimate this will be before i reaches 10000000000000000
(1..(1.0/0)).each{ |i|
print i*i, " ", i + (Math.sqrt(i) + 0.5).to_i, " "
}

Formula, generating two terms for every m: m^2, m + round(sqrt(m)).

IFTE(n mod 2 ==1, ((n+1)/2)^2, (n/2)+round(sqrt(n/2),0)). - Gerald Hillier, Nov 15 2010

More terms from Robert G. Wilson v, May 31 2009

cp's OEIS Frontend

User: Andy Martin

A275939 Consider the prime race mod q (where q >= 2) between q*k+1 and q*k-1. Terms are numbers k where q*k+1 first takes lead over q*k-1.

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A272043 a(n) is the shyest prime in base n.

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A178983 The smallest cube containing n as a substring.

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A161170 Least integer k such that the n-almost prime count is equal to the prime count.

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A173515 Consider positive integer solutions to x^3+ y^3 = z^3 - n or 'Fermat near misses' of 1, 2, 3 ... Arrange known solutions by increasing values of n. Sequence gives value of lowest z for a given n.

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A137442 n^2 followed by smallest integer not yet listed.

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A275939 Consider the prime race mod q (where q >= 2) between qk+1 and qk-1. Terms are numbers k where qk+1 first takes lead over qk-1.