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User: Peter Rowlett

Peter Rowlett has authored 2 sequences.

A371896 a(n) is the length of the uninterrupted sequence of primes generated by the polynomial f(x) = x^2 + x + p for x=0,1,..., where p=A001359(n).

2, 4, 10, 16, 2, 40, 2, 2, 4, 3, 2, 2, 2, 3, 2, 4, 2, 2, 2, 3, 5, 2, 2, 3, 2, 2, 2, 2, 5, 2, 2, 3, 2, 3, 3, 2, 2, 2, 2, 4, 2, 2, 7, 2, 3, 2, 5, 2, 4, 4, 6, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 5, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2

Offset: 1

Peter Rowlett, Apr 11 2024

nonn

p=A001359(n) is the smaller prime of a twin prime pair so that f(0) = p and f(1) = p+2 are both primes so a(n) >= 2 and this sequence is the terms >= 2 in A208936.

			For n=6, p = A001359(n) = 41 and f(x) = x^2 + x + 41 is Euler's polynomial which generates primes f(x) for x=0,1,2,...,39, which is 40 terms so a(6) = 40 (cf. A202018).

L. Euler, Nouveaux Mémoires de l'Académie royale des Sciences, 1772, p. 36.

Peter Rowlett, Table of n, a(n) for n = 1..10000

Cf. A001359, A202018, A208936.

A001359[1] = 3;
A001359[n_] := A001359[n] = (p = NextPrime[A001359[n - 1]];
  While[!PrimeQ[p + 2], p = NextPrime[p]]; p)
A371896[n_] := With[{p = A001359@n}, k = 1;
  While[PrimeQ[k^2 + k + p], k++]; k]
(* Leo C. Stein, May 09 2024 *)

A363166 Bouton numbers: a(n) is the number of P positions in games of Nim with three nonzero heaps each containing at most n sticks.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 7, 7, 8, 10, 13, 17, 22, 28, 35, 35, 36, 38, 41, 45, 50, 56, 63, 71, 80, 90, 101, 113, 126, 140, 155, 155, 156, 158, 161, 165, 170, 176, 183, 191, 200, 210, 221, 233, 246, 260, 275, 291, 308, 326, 345, 365, 386, 408, 431, 455, 480, 506, 533, 561, 590, 620, 651, 651, 652

Offset: 1

Peter Rowlett, Jul 06 2023

In combinatorial game theory, if the next player cannot win (assuming perfect play) we call this a P position.
Bouton (1901) called P positions in Nim "safe combinations" and provided a list of 35 such positions in games with less than 16 sticks in each heap. Bouton excluded positions where any heap has been reduced to zero sticks. This seems reasonable since if a pile is removed the game is reduced to a two-heap Nim game, so we follow Bouton here.
These positions have Nim sum zero. Nim sum is calculated by converting the heap sizes to binary and applying bitwise XOR addition.
The paper cited named the game Nim and provided the first known mathematical analysis.

			With n=1, there is one stick in each pile. This is an N position (the next player can force a win) since the next player can take any stick and 0 1 1 is a P position, so a(1)=0.
With n=3, the only P position is 1 2 3, so a(3)=1. This is the simplest P position in three-heap Nim in that it uses the least number of sticks overall.
Bouton's list is a(15)=35: 1 2 3, 1 4 5, 1 6 7, 1 8 9, 1 10 11, 1 12 13, 1 14 15, 2 4 6, 2 5 7, 2 8 10, 2 9 11, 2 12 14, 2 13 15, 3 4 7, 3 5 6, 3 8 11, 3 9 10, 3 12 15, 3 13 14, 4 8 12, 4 9 13, 4 10 14, 4 11 15, 5 8 13, 5 9 12, 5 10 15, 5 11 14, 6 8 14, 6 9 15, 6 10 12, 6 11 13, 7 8 15, 7 9 14, 7 10 13, 7 11 12.

Peter Rowlett, Table of n, a(n) for n = 1..256
C. L. Bouton, Nim, A Game with a Complete Mathematical Theory, Annals of Mathematics, 3 (1901), 35-39.

from itertools import combinations_with_replacement
for n in range(1,16):
    num_digits = len('{0:b}'.format(n))
    count=0
    comb = combinations_with_replacement(['{0:b}'.format(i).zfill(num_digits) for i in range(1,n+1)], 3)
    for heaps in comb:
        heapsums = [0 for i in range(0,num_digits)]
        for heap in heaps:
            for d in range(0,num_digits):
                heapsums[d] += int(heap[d])
        if not True in (heapsum % 2 != 0 for heapsum in heapsums):
            count +=1
    print(n,count)

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User: Peter Rowlett

A371896 a(n) is the length of the uninterrupted sequence of primes generated by the polynomial f(x) = x^2 + x + p for x=0,1,..., where p=A001359(n).

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