Problems & Puzzles: Puzzles

 Puzzle 951. Mersenne primes as self-numbers. Carlos Rivera asks: In Oct. 2006 Luke Pebody demonstrated that the 41th Mersenne Prime 2^24036583-1 was a Self-number (see here what a Self number is). Pebody’s method was described in our Puzzle 373 and is also summarized in the Wikipedia. Nowadays we know 10 more Mersenne Primes 2^p-1 for the following p prime values: 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57,885,161, 74,207,281, 77,232,917 and 82,589,933 Q. Is any of these last 10 Mersenne primes a Self-number too? Contributions came from Emmanuel Vantieghem, Oscar Volpatti and Paul Cleary

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Emmanuel wrote on April 29, 2019:

Only the last of the ten primes gives a self prime.

Thus, the only primes  p  that make  2^p - 1  self are : 2, 3, 5, 44497, 13466917, 24036583  and  82589933.

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Oscar wrote on April 29, 2019:

The only self number within the given list is the Mersenne prime M(82589933).

Considering the prime p = 82589933, the number N1 = 2^p-1 has length L1 = 24862048.
Using Pebody's reduction method, N1 is partitioned into two blocks:
its 10 rightmost digits R1 = 5217902591 (a number larger than 9*L1 = 223758432);
its L1-10 leftmost digits, with sum S1 = 111879872.
Then N1 is self iff N2 = R1 - S1 = 5106022719 is self.
The reduction method can be used once again:
L2 = 10, R2 = 719, S2 = 16, N3 = 703.
But N3 is 73th self number (OEIS A003052), so N2 and N1 are self too.

In order to verify that
703 is actually self, consider the equation 703 = x + digitsum(x).
After the change of variable y = digitsum(x), we obtain the equation y = digitsum(703 - y).
The right hand is the sum of at most three digits, so its value belongs to the range [0,27]. As the function digitsum(x) is piecewise linear,
we only need to solve four linear equations:
[-6, 3]  ->  y = 7 + 0 + (3-y)   ->  y = 5    (out of range)
[ 4,13]  ->  y = 6 + 9 + (13-y)  ->  y = 14   (out of range)
[14,23]  ->  y = 6 + 8 + (23-y)  ->  y = 18.5 (within range, not integer)
[24,33]  ->  y = 6 + 7 + (33-y)  ->  y = 23   (out of range)

Similarly, I verified that the remaining Mersenne primes within the given list are not self.

p = 25964951:
L1 = 7816230, R1 = 77077247, S1 = 35170343, N2 = 41906904;
L2 = 8, R2 = 904, S2 = 20, N3 = 884;
L3 = 3, R3 = 84, S3 = 8, N4 = 76;
76 = 65 + (6 + 5).

p = 30402457:
L1 = 9152052, R1 = 652943871, S1 = 41201902, N2 = 611741969;
L2 = 9, R2 = 969; S2 = 20; N3 = 949;

L3 = 3, R3 = 49; S3 = 9; N4 = 40;
40 = 29 + (2 + 9).

p = 32582657:
L1 = 9808358, R1 = 4053967871, S1 = 44142101; N2 = 4009825770;
L2 = 10, R2 = 770, S2 = 28, N3 = 742;
L3 = 3, R3 = 42, S3 = 7, N4 = 35;
35 = 31 + (3 + 1).

p = 37156667:
L1 = 11185272, R1 = 308220927, S1 = 50349661, N2 = 257871266;

L2 = 9, R2 = 266, S2 = 30, N3 = 236;
L3 = 3, R3 = 36, S3 = 2, N4 = 34;
34 = 26 + (2 + 6).

p = 42643801:
L1 = 12837064, R1 = 562314751, S1 = 57766305, N2 = 504548446;
L2 = 9, R2 = 446, S2 = 26, N3 = 420;

420 = 399 + (3 + 9 + 9) = 408 + (4 + 0 + 8).

p = 43112609:
L1 = 12978189, R1 = 697152511, S1 = 58416600, N2 = 638735911;

L2 = 9, R2 = 911, S2 = 32, N3 = 879;
L3 = 3, R3 = 79, S3 = 8, N4 = 71;
71 = 58 + (5 + 8).

p = 57885161:
L1 = 17425170, R1 = 724285951, S1 = 78434718, N2 = 645851233;
L2 = 9, R2 = 233, S2 = 29, N3 = 204;
204 = 192 + (1 + 9 + 2) = 201 + (2 + 0 + 1).

p = 74207281:
L1 = 22338618, R1 = 1086436351, S1 = 100537506, N2 = 985898845;
L2 = 9, R2 = 845, S2 = 47, N3 = 798;
L3 = 3, R3 = 98, S3 = 7, N4 = 91;
91 = 77 + (7 + 7).

p = 77232917:
L1 = 23249425, R1 = 762179071, S1 = 104621220, N2 = 657557851;
L2 = 9, R2 = 851, S2 = 35, N3 = 816;
816 = 795 + (7 + 9 + 5) = 804 + (8 + 0 + 4).

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Paul wrote on April 29, 2019:

Ok so I have re-written my code and I hope its all good now, I now get a self number for the last one of the new ones, I.e. 2^82589933-1 Is a self number.

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