Marcus Hedbring - cp's OEIS frontend

User: Marcus Hedbring

Marcus Hedbring has authored 3 sequences.

A225539 Numbers n where 2^n and n have the same digital root.

5, 16, 23, 34, 41, 52, 59, 70, 77, 88, 95, 106, 113, 124, 131, 142, 149, 160, 167, 178, 185, 196, 203, 214, 221, 232, 239, 250, 257, 268, 275, 286, 293, 304, 311, 322, 329, 340, 347, 358, 365, 376, 383, 394, 401, 412, 419, 430, 437, 448

Offset: 1

Marcus Hedbring, May 17 2013

The digital roots of n have a cycle length of 9 (A010888) and the digital roots of 2^n have a cycle length of 6 (A153130). Therefore, if n is a term so is n+18.

The only values of the digital roots of a(n) are 5 and 7 (A010718).

			For n=23, the digital root of n is 5. 2^n equals 8388608 so the digital root of 2^n is 5 as well.

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A010888, A153130, A010718.

digitalRoot[n_] :=  Module[{r = n}, While[r > 9, r = Total[IntegerDigits[ r]]]; r]; Select[Range[448], digitalRoot[2^#] == digitalRoot[#] &] (* T. D. Noe, May 19 2013 *)
LinearRecurrence[{1,1,-1},{5,16,23},60] (* Harvey P. Dale, Dec 29 2018 *)

forstep(n=16,500,[7,11],print1(n", ")) \\ Charles R Greathouse IV, May 19 2013

a(n) = 9*n - 3 + (-1)^n.
a(n) = a(n-1) + 7 (odd n), a(n) = a(n-1) + 11 (even n) with a(1) = 5.
G.f. x*(5 + 11*x + 2*x^2) / ((1-x)^2 * (1+x)). - Joerg Arndt, May 17 2013

A225381 Elimination order of the first person in a Josephus problem.

Original entry on oeis.org

1, 2, 2, 4, 3, 5, 4, 8, 5, 8, 6, 11, 7, 11, 8, 16, 9, 14, 10, 18, 11, 17, 12, 23, 13, 20, 14, 25, 15, 23, 16, 32, 17, 26, 18, 32, 19, 29, 20, 38, 21, 32, 22, 39, 23, 35, 24, 47, 25, 38, 26, 46, 27, 41, 28, 53, 29, 44, 30, 53, 31, 47, 32, 64, 33, 50, 34, 60, 35

Offset: 1

Marcus Hedbring, May 17 2013

nonn

In a Josephus problem such as A006257, a(n) is the order in which the person originally first in line is eliminated.
The number of remaining survivors after the person originally first in line has been eliminated, i.e., n-a(n), gives the fractal sequence A025480.
For the linear version, see A225489.

			If there are 7 persons to begin with, they are eliminated in the following order: 2,4,6,1,5,3,7. So the first person (the person originally first in line) is eliminated as number 4. Therefore a(7) = 4.

Stefano Spezia, Table of n, a(n) for n = 1..10000
Cristina Ballantine and Mircea Merca, Plane Partitions and Divisors, Symmetry (2024), Vol. 16, Iss. 5. See page 9.
Mircea Merca, Plane Partitions and a Problem of Josephus, Mathematics (2023), Vol. 11, Iss. 4996. See page 2.
Index entries for sequences related to the Josephus Problem

Cf. A006257, A006519, A025480, A225489.

t = {1}; Do[AppendTo[t, If[OddQ[n], (n + 1)/2, t[[n/2]] + n/2]], {n, 2, 100}]; t (* T. D. Noe, May 17 2013 *)

a(n) = (n+1)/2 (odd n); a(n) = a(n/2) + n/2 (even n).
a(n) = n - A025480(n).
G.f.: Sum{n>=1} x^n/(1-x^A006519(n)). - Nicolas Nagel, Mar 19 2018

A225489 Elimination order for the first person in a linear Josephus problem.

Original entry on oeis.org

1, 2, 2, 3, 5, 6, 5, 6, 8, 9, 8, 9, 12, 13, 11, 12, 17, 18, 14, 15, 21, 22, 17, 18, 23, 24, 20, 21, 27, 28, 23, 24, 32, 33, 26, 27, 36, 37, 29, 30, 38, 39, 32, 33, 42, 43, 35, 36, 48, 49, 38, 39, 52, 53, 41, 42, 53, 54, 44, 45, 57, 58, 47, 48, 65, 66, 50, 51

Offset: 1

Marcus Hedbring, May 08 2013

nonn

The process is identical to that of A090569 where n persons are arranged on a line and every second person is eliminated. When we reach the end of the line the direction is reversed without double-counting the person at the end. a(n) is the order in which the person originally first in line is eliminated.

			If there are 7 persons to begin with, they are eliminated in the following order: 2,4,6,5,1,7,3. So the first person (the person originally first in line) is eliminated as number 5. Therefore a(7) = 5.

Chris Groër, The Mathematics of Survival: From Antiquity to the Playground, Amer. Math. Monthly, 110 (No. 9, 2003), 812-825.
Index entries for sequences related to the Josephus Problem

Cf. A090569, A088442.

t = {1}; Do[AppendTo[t, Switch[Mod[n,4], 0, 3*n/4, 1, t[[1 + (n-1)/4]] + 3*(n-1)/4, 2, t[[1 + (n-2)/4]] + 3*(n-2)/4 + 1, 3, 3*(n-3)/4 + 2, 4, Mod[n,4] + 1]], {n, 2, 100}]; t (* T. D. Noe, May 17 2013 *)

For n=4m then a(n) = 3*n/4;
for n=4m+1 then a(n) = a(1+(n-1)/4) + 3*(n-1)/4;
for n=4m+2 then a(n) = a(1+(n-2)/4) + 3*(n-2)/4 + 1;
for n=4m+3 then a(n) = 3*(n-3)/4 + 2.

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User: Marcus Hedbring

A225539 Numbers n where 2^n and n have the same digital root.

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A225381 Elimination order of the first person in a Josephus problem.

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A225489 Elimination order for the first person in a linear Josephus problem.

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