Problems & Puzzles:
Puzzles
Puzzle 951.
Mersenne primes as selfnumbers.
Carlos
Rivera asks:
In Oct. 2006 Luke Pebody
demonstrated that the 41th Mersenne Prime 2^240365831 was a
Selfnumber (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^p1 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 Selfnumber too?


Contributions came from Emmanuel Vantieghem, Oscar Volpatti and Paul
Cleary
***
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.
***
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^p1
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 L110 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 +
(3y) > y = 5 (out of range)
[ 4,13] >
y = 6 + 9 + (13y) >
y = 14 (out of range)
[14,23] > y = 6 +
8 + (23y)
> y = 18.5 (within range, not integer)
[24,33] > y = 6 +
7 + (33y) >
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).
***
Paul wrote on April 29, 2019:
Ok so I have rewritten 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. 2^825899331 Is a self number.
***
