Michael C. Case - cp's OEIS frontend

User: Michael C. Case

Michael C. Case has authored 2 sequences.

A331032 Number of iterations of n -> n + gpf(n) needed for the trajectory of n to join the trajectory of A076271, where gpf(n) is the greatest prime factor of n.

0, 0, 1, 0, 2, 0, 4, 2, 0, 1, 6, 0, 10, 3, 0, 4, 12, 3, 16, 0, 2, 5, 18, 2, 0, 9, 1, 1, 22, 0, 28, 12, 4, 11, 0, 9, 30, 15, 8, 5, 36, 0, 40, 3, 4, 17, 42, 11, 0, 3, 10, 7, 46, 15, 2, 0, 14, 21, 52, 7, 58, 27, 0, 2, 6, 1, 60, 9, 16, 0, 66, 11, 70, 29, 10, 13

Offset: 1

Michael C. Case, Jan 08 2020

nonn

Record values occur at prime values of n, and equal one less than the next lowest prime number (see Formula). Because of this, a(n) is always less than n, so for any positive integer starting value n, iterations of n -> n + gpf(n) will eventually join A076271.

			a(8)=2 because the trajectory for 1 (sequence A076271) starts 1->2->4->6->9->12->15->20... and the trajectory for 8 starts 8->10->15->20... so the sequence beginning with 8 joins A076271 after 2 steps.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A006530, A076271.

gpf(n) = if (n==1, 1, my (f=factor(n)); f[#f~, 1])
a(n) = { my (o=1); for (k=0, oo, while (oRémy Sigrist, Apr 05 2020

a(k*p) = prevprime(p) - k for all k <= prevprime(p).

a(p) = prevprime(p) - 1 for p > 2.

A331394 Number of ways of 4-coloring the Fibonacci square spiral tiling of n squares with colors introduced in order.

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 6, 7, 11, 16, 19, 25, 38, 51, 63, 88, 127, 165, 214, 303, 419, 544, 731, 1025, 1382, 1819, 2487, 3432, 4583, 6125, 8406, 11447, 15291, 20656, 28259, 38185, 51238, 69571, 94703, 127608, 172047, 233845

Offset: 1

Michael C. Case, Jan 15 2020

nonn

The Fibonacci square spiral tiling is the pattern formed by tiling the plane using squares with side-lengths of successive Fibonacci numbers (so the k-th square is of size F(k)), in a spiral pattern.
The Fibonacci square spiral tiling for 6 squares:
   ___ ___ _____________
  |     |   |               |
  |     | |               |
  |___|_|_|               |
  |         |               |
  |         |               |
  |         |               |
  |         |               |
  |_______|_______________|
In a 4-coloring of the Fibonacci square spiral tiling, the square k cannot be the same color as squares k-4, k-3, or k-1. When k-1 is the same color as k-3, k can be colored in 2 different ways.
The first 3 squares must be colored ABC but for k>3 square k can be the same color as square k-2.

			There are 3 ways to 4-color a Fibonacci square spiral tiling of 5 squares:
   _____ ___   _____ ___   _____ ___
  |     | C | |     | C | |     | C |
  |  B  |_ _| |  B  |_ _| |  D  |_ _|
  |_____|A|B| |_____|A|B| |_____|A|B|
  |         | |         | |         |
  |         | |         | |         |
  |    C    | |    D    | |    C    |
  |         | |         | |         |
  |_________| |_________| |_________| so a(5)=3.
There are 7 ways to 4-color a Fibonacci square spiral tiling of 8 squares (ABCBCADA, ABCBCDAD, ABCBDADA, ABCBDADC, ABCDCABA, ABCDCDAB, ABCDCDBA), so a(7) = 8.

Michael C. Case, Table of n, a(n) for n = 1..200 [a(192) corrected by _Georg Fischer_, Apr 15 2020]
Wikipedia, Fibonacci number
Index entries for linear recurrences with constant coefficients, signature (1, -1, 2).

Cf. A000045, A216790, A077951.

Rest@ CoefficientList[Series[x (1 + x) (1 - x + 2 x^2 - 2 x^3 + 2 x^4)/(1 - x + x^2 - 2 x^3), {x, 0, 42}], x] (* Michael De Vlieger, Jan 31 2020 *)

a(n) = a(n-1) - a(n-2) + 2*a(n-3) for n >= 7.
G.f.: x*(1 + x)*(1 - x + 2*x^2 - 2*x^3 + 2*x^4)/(1 - x + x^2 - 2*x^3).
a(n)/a(n-1) approaches the only real solution of x^3 - x^2 + x - 2 = 0, x = (1 - 2*(2/(47 + 3*sqrt(249)))^(1/3) + ((47 + 3*sqrt(249))/2)^(1/3))/3 = 1.35320996419932... .

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User: Michael C. Case

A331032 Number of iterations of n -> n + gpf(n) needed for the trajectory of n to join the trajectory of A076271, where gpf(n) is the greatest prime factor of n.

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