A000057 -id:A000057 - cp's OEIS frontend

A370526 Square array read by descending antidiagonals: Define function b(i,n,k) where b(0,n,k) = n, b(1,n,k) = k, b(i,n,k) = A038502(b(i-1,n,k) + b(i-2,n,k)). T(n,k) is the number of steps until reaching the cyclic part of {b(i,n,k)}, or -1 if no cycle exists.

0, 0, 0, 12, 0, 4, 3, 24, 13, 7, 6, 0, 14, 0, 1, 11, 12, 11, 5, 5, 18, 17, 12, 23, 0, 15, 4, 3, 2, 4, 19, 2, 8, 5, 1, 1, 17, 3, 4, 24, 0, 13, 12, 7, 5, 4, 11, 14, 10, 27, 23, 10, 19, 14, 5, 4, 6, 10, 0, 11, 14, 9, 0, 12, 1, 4, 14, 13, 11, 10, 22, 10, 29, 15

Offset: 1

Yifan Xie, Feb 21 2024

b(i,n,k) = (b(i-1,n,k) + b(i-2,n,k))/p^valuation(b(i-1,n,k) + b(i-2,n,k), p), i.e., b(i,n,k) is b(i-1,n,k) + b(i-2,n,k) with all factors of p removed, where p = 3 in this sequence. Therefore, b(i,n,k) is not divisible by 3 for i >= 3.
At least one of b(i,n,k) < b(i-1,n,k) + b(i-2,n,k) and b(i+1,n,k) < b(i-1,n,k) + b(i,n,k) is true for i >= 5.
It appears that all repetends have the form of (1, 1, 2) (the position of 2 possibly changed), multiplied by G = A038502(gcd(n,k)).
Conjecture: T(n,k) >= 0.
This conjecture can be supported by a heuristic argument: Using dynamic programming, we can compute that for p's in A000057, the probability that b(i-1,n,k) + b(i-2,n,k) is not divisible by p is (p-2)/p, and the probability that valuation(b(i-1,n,k) + b(i-2,n,k), p) = x (x >= 1) is 2*(p-1)/p^(x+1). Therefore, the mathematical expectation of a(i) is (a(i-1,n,k) + a(i-2,n,k))*(p-1)/(p+1), which is exactly the average of the earlier two terms when p = 3, and larger when p >= 5.

			Array begins:
  n\k|  1   2   3   4   5   6   7
  ---+--------------------------------
  1  |  0,  0, 12,  3,  6, 11, 17, ...
  2  |  0,  0, 24,  0, 12, 12,  4, ...
  3  |  4, 13, 14, 11, 23, 19,  4, ...
  4  |  7,  0,  5,  0,  2, 24, 10, ...
  5  |  1,  5, 15,  8,  0, 27, 11, ...
  6  | 18,  4,  5, 13, 23, 14, 10, ...
  7  |  3,  1, 12, 10,  9,  7, 29, ...
  ...
T(1,4) = 3 because its sequence b begins with b(0) = 1, b(1) = 4, b(2) = A038502(1+4) = 5, b(3) = A038502(4+5) = 1, b(4) = 2, b(5) = 1, b(6) = 1, which has reached the cyclic part of (1, 2, 1) at i=3.

Yifan Xie, Rows n = 1..100 of triangle, flattened

Cf. A000057, A038502, A369876.

T(n, k)={my(i=-1, z=0); while((z != 2*n || z != 2*k) && (n != 2*z || n != 2*k) && (k != 2*n || k != 2*z), z=n; n=k; k=(z+n)/3^(valuation(z+n, 3)); i++); i; };

T(n,k) = 0 iff n = k, n = 2*k or k = 2*n and gcd(x,y) is not divisible by 3.

A161500 Primes dividing some member of A073833.

Original entry on oeis.org

2, 5, 29, 41, 89, 101, 109, 269, 421, 509, 521, 709, 929, 941, 1549, 1861, 2281, 2521, 2749, 2801, 2909, 3121, 3169, 3469, 5821, 5881, 7109, 8069, 8969, 9041, 9181, 10061, 10601, 11549, 15121, 16061, 16889, 16981, 21929, 30089, 30169, 32561, 41149

Offset: 1

Franklin T. Adams-Watters, Jun 11 2009

nonn

Primes that divide A073833(n) will divide A073834(m) for any m > n, and this is all the prime divisors of A073834(m).
Iterating f(x) = x + 1/x modulo p will eventually either produce a zero (in which case p is in this sequence), or it will loop to an earlier term (in which case it is not). Since f(-x) = -f(x), encountering the negation of an earlier term means that the iteration is looping.
Note that A073833(6) = 969581 = 521 * 1861 is the first composite member of that sequence.

Cf. A073833, A073834, A000057.

ina(p)=local(m,k,v);m=Mod(1,p);v=vector(p\2);while(m!=0,k=lift(m);if(2*k>p,k=p-k);if(v[k],return(0));v[k]=1;m+=1/m);1

A214033 Places n where A214030(n) = n or A214030(n) = n-2.

Original entry on oeis.org

13, 17, 19, 23, 37, 41, 47, 67, 89, 109, 137, 139, 157, 181, 191, 211, 229, 233, 239, 257, 277, 281, 283, 307, 311, 331, 349, 353, 359, 373, 379, 397, 479, 499, 503, 521, 523, 547, 571, 593, 599, 613, 617, 619, 641

Offset: 1

Art DuPre, Jul 12 2012

It always has been one of the great mysteries of mathematics, that the superdiagonal sequence of A001177 consists of prime numbers A000057.
Here, regarding A214031 and A214032,there is the further conjecture that these two disjoint sequences are primes and roughly comparable in density. It isn't clear that these two sequences have a density, without appealing to the Riemann Hypothesis, but they are certainly close to one another in growing size.
Since these two sequences are disjoint, it is natural to take their union.

Cf. A000057, A001177, A214030, A214031, A213032.

{b23(n)=local(t,m=1,s=[n]); if (n<2,0,while(1,
if(m%2,s=concat(s,2),s=concat(s,3));
t=contfracpnqn(concat(s,n));
t=contfrac(n*t[1,1]/t[2,1]);
if(t[1]

cp's OEIS Frontend

A370526 Square array read by descending antidiagonals: Define function b(i,n,k) where b(0,n,k) = n, b(1,n,k) = k, b(i,n,k) = A038502(b(i-1,n,k) + b(i-2,n,k)). T(n,k) is the number of steps until reaching the cyclic part of {b(i,n,k)}, or -1 if no cycle exists.

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A161500 Primes dividing some member of A073833.

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A214033 Places n where A214030(n) = n or A214030(n) = n-2.

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