125 2 - 1,was factored on Tuesday, January 5, 1971, in 371 seconds run time as follows:

125 2 - 1 = 31 * 601 * 1801 * 269,089,806,001 * 4,710,883,168,879,506,001.John Brillhart at the University of Arizona had already done this. M137 was factored on Friday, July 9, 1971 in about 50 hours of computer time:

137 2 - 1 = 32,032,215,596,496,435,569 * 5,439,042,183,600,204,290,159.[Current prime records --

- greater than sqrt(X)=X^(1/2) is ln 2.
- less than X^(1/3) is about 4.86%.

RELEVANT DATA: ([] denote the expected value of adjacent entries.)

where:RANGE COUNT CUMULATIVE SUM OF COUNT 10^12 to 10^6 7198 [6944] 10018 10^6 to 10^4 2466 2820 10^4 to 10^3 354 402 [487] 10^3 to 252 40 48 ;252 = 10^2.4 252 to 100 7 8 100 to 52 1 1 ;52 = 10^1.7 51 to 1 0 0

"COUNT" is the number of numbers between 10^12 + 1 and 10^12 + 10018 whose largest prime factor is in "RANGE". The number of primes in 10^12 + 1 to 10^12 + 10018 is 335; the prime number theorem predicts 363 in this range. This is relevant to Knuth's discussion of Legendre's factoring method, vol. 2, p. 351-354.

- 166,666,666,667 = (10^12 + 2)/6
- 166,666,666,669

The primes which bracket 10^12 are 10^12 + 39 and 10^12 - 11.

The primes which bracket 10^15 are 10^15 + 37 and 10^15 - 11.

The number 23,333,333,333 is prime.

Various primes, using T = 10^12, are:

40 T + 1, 62.5 T + 1, 200 T - 3, 500 T - 1, 500 T - 7.

N 2 2 - 7 = Xwas searched to about N = 10^40; only his solutions (N = 3, 4, 5, 7, 15) were found. It has recently been proven that these are the only ones. Another Ramanujan problem: Find all solutions of n! + 1 = x^2.

*phi*= (1 + sqrt(5))/2- all - 1 < X < 1
- sqrt(2) (half are integers, other half are probably uniformly distributed)
- 1 + sqrt(2) -- Proof:
N N (1 + sqrt(2)) + (1 - sqrt(2)) = integer (by induction); N the (1 - sqrt(2)) goes to zero.

- 2 + sqrt(2) -- similar to 1 + sqrt(2)
- any algebraic number whose conjugates are all inside the unit circle

(ln 2)/(ln 3) N .

- XX.XX is a known solution for N = 5
- XX.XX....XX.XX is a known solution for N = 14

0 1 16 81 256 625 1 15 65 175 369 14 50 110 194 36 60 84 24 24 0

1 4 16 64 256 3 12 48 192 9 36 144 27 108 81

- N = 2 -> finite maximum string
- N = 10 -> known infinite

SUB-PROBLEM: How many sequence exist for any particular length?

N ==== \ L N + 1 N + 1 L > BINOMIAL(N + L, L) ((1 - X) X + (1 - X) X ) = 1 / ==== L = 0set N to 20 and observe that it is the probability that one or the other player wins at pingpong. (X = probability of first player gaining one point, L = loser's score, deuce rule irrelevant.) If this seems silly, try more conventional methods.

PROBLEM: If somehow you determine A should spot B 6 points for their probabilities of winning to be equal, and B should spot C 9 points, how much should A give C?

(A + B + C....)! ---------------- A! B! C! ...This is equal modulo the prime p to

(A0, B0, C0...) (A1, B1, C1...) (A2, B2, C2...)...where AJ is the Jth from the right digit of A base p.

Thus BINOMIAL(A+B, A) mod 2 is 0 iff (AND A B) is not.

The exponent of the largest power of p which divides (A, B, C,...) is equal to the sum of all the carries when the base p expressions for A, B, C, ... are added up.

(A,B,C,...) = (A+B,C,...)(A,B) = (A+B+C,...)(A,B,C) = ...

- Take a unit step at some heading (angle).
- Double the angle, step again.
- Redouble, step, etc.

PARTIAL ANSWER (Schroeppel, Gosper): When the initial angle is a
rational multiple of *pi,* it seems that your locus is bounded (in fact,
eventually periodic) iff the denominator contains as a factor the
square of an odd prime other than 1093 and 3511, which must occur at
least cubed. (This is related to the fact that 1093 and 3511 are the
only known primes satisfying

P 2 2 = 2 mod P ).But a denominator of 171 = 9 * 19 never loops, probably because 9 divides

The length of powers of primes seems to be

power-1 L = (length of prime) * prime .The length of products of powers of primes seems to be

L = least common multiple of lengths of powers of primes which are factors.There can be only 1, 2, or 4 sub-cycles in the cycle of a prime.

Primes with 1 sub-cycle seem to have lengths

prime - 1 L = --------- , NN covering all integers. Primes with 2 sub-cycles seem to have lengths

M prime - (-1) L = ------------- , MM covering all integers except of form 10 K + 5.

Primes with 4 sub-cycles seem to always be of form 4 K + 1, and seem to have lengths

prime + 1 prime - 1 L = --------- or --------- , R SR covering all integers of form 10 K + 1, 3, 7 or 9; S covering all integers.

At Schroeppel's suggestion, the primes have been separated mod 40, which
usually determines their number of sub-cycles:

Attention was directed to primes which are 1 or 9 mod 40 but have 1 or 4 subcycles.PRIME mod 40 SUB-CYCLES 1, 9 usually 2, occasionally 1 or 4 (about equally) 3, 7, 23, 27 2 11, 19, 31, 39 1 13, 17, 33, 37 4 21, 29 1 or 4 (about equally) 2 (only 2) 1 5 (only 5) 4

2 2 25 X + 16 Yseems to express those which are 9 mod 40;

2 2 (10 X +- 1) + 400 Yseems to express those which are 1 mod 40.

PROBLEM: Can some of the "seems" above be proved? Also, can a general test be made which will predict exact length for any number?

3 17 18 19 7 1 11 16 2 5 6 9 12 4 8 14 10 13 15

Proof: Let K (= 130) be the sum of a row.

Lemma 1: In a magic square of order four, the sum of the corners is K.

Proof: Add together each edge of the square and the two diagonals. This covers the square entirely, and each corner twice again. This adds to 6 K, so twice the corner sum is 2 K.

Lemma 2: In a magic cube of order 4, the sum of any two corners connected by an edge of the cube is K/2.

Proof: Call the corners a and b. Let c, d and e, f be the corners of any two edges of the cube parallel to ab. Then abcd, abef, and cdef are all the corners of magic squares. So a+b+c+d + a+b+e+f + c+d+e+f = 3K; a+b+c+d+e = 3K/2; a+b = K/2.

Proof of magic cube impossibility: Consider a corner x. There are three corners connected by an edge to x.

Each must have value K/2 - x. QED

COROLLARY: There is no magic tesseract of order 5.

PSEUDO-PROOF: Let X be the probability. Let S be the set of points in the
integer lattice whose coordinates are relatively prime, so that S occupies a
fraction X of the lattice points. Let S(D) be the set of points whose
coordinates have a GCD of D. S(D) is S expanded by a factor of D from the
origin. So S(D) occupies a fraction X/D^2 of the lattice, or the probability
that two random integers have a GCD of D is X/D^2. If D unequals D', then S(D)
intersect S(D') is empty, and union of all S(D) is the entire lattice.
Therefore X*(1/1^2+1/2^2+1/3^2+...) = 1, so X = 6/*pi*^2. This argument is not
rigorous, but can be made so.

Figure 1(a). This diagram is to
substantiate the claim that every Gaussian integer has a unique bit
combination. Running through bit combinations 0, 1, 10, 11, ..., the
diagram is a map of values, radix *i*-1. The origin is circled;
the dot is at the 127th combination (1111111 = 2 + 5*i*), which
is merely the last point drawn.

Figure 1(b). As 1(a), but radix
*i*+1. Large circle is origin. Dashes indicate continuity of
curve at confusing places. Dotted curve is with an infinity of ones
to the left (big dot = ...1111 = i). The solid and dotted curves are
symmetrical about the point marked with a small circle.

Figure 2. Similar to 1(a), but showing
fraction parts as well. Reprinted by special permission from Knuth,
*The Art of Computer Programming, Volume 2, Seminumerical
Algorithms,* 1969, Addison-Wesley, Reading, Mass.

N MAX L DISTINCT 2 4 54 3 5 219 4 6 714 5 7 2,001 6 7 5,004 7 8 11,439 8 9 24,309 9 9 48,619 10 10 92,377 11 10 167.959 12 10 293,929

86 30739014 2 = 77,371,252,455,336,267,181,195,264 and 2 ,where the program was stopped. If digits of such powers were random, the probability that there is another zeroless power would be about 1/10^411816. Assuming there aren't any raises the question:

How many final nonzero digits can a power of two have?

ANSWER (Schroeppel): Arbitrarily many. If we look at the last n digits of consecutive powers of 2, we see:

- None end in zero.
- After the nth, they are all multiples of 2^n.
- They get into a loop of length 4 * 5^(n-1). (Because 2 is a primitive root of powers of 5.)

3 3 3 3 3 + 4 + 5 = 6 .

2 91038 90995 89338 00226 07743 74008 17871 09376 = 82880 83126 51085 58711 66119 71699 91017 17324 91038 90995 89338 00226 07743 74008 17871 09376

3 120 = 2 * 3 * 5 = 1111000 base 2 5 672 = 2 * 3 * 7 = 1010100000 base 2 9 523,776 = 2 * 3 * 11 * 31 = 1111111111000000000 base 2

- sum of factors of 220 = 284
- sum of factors of 284 = 220

A program to search for loops of length > 2, all of whose members are <
6,600,000,000 found the known loops of length 5 (lowest member is 12496) and 28
(lowest member is 14316), but also 13 loops of 4 members (lowest member is
given):

1,264,460 = 2^2 * 5 * 17 * 3,719 2,115,324 = 2^2 * 3^2 * 67 * 877 2,784,580 = 2^2 * 5 * 29 * 4,801 4,938,136 = 2^3 * 7 * 109 * 809 7,169,104 = 2^4 * 17 * 26,357 18,048,976 = 2^4 * 11 * 102,551 18,656,380 = 2^2 * 5 * 932,819 46,722,700 = 2^2 * 5^2 * 47 * 9,941 81,128,632 = 2^3 * 13 * 19 * 41,057 174,277,820 = 2^2 * 5 * 29 * 487 * 617 209,524,210 = 2 * 5 * 7 * 19 * 263 * 599 330,003,580 = 2^2 * 5 * 16,500,179 498,215,416 = 2^3 * 19 * 47 * 69,739

Two-member loops (amicable pairs) are:

(Exhaustive to smaller member <= 196724 and larger member < 2^35.)220 <-> 284 1184 <-> 1210 2620 <-> 2924 5020 <-> 5564 6232 <-> 6368 10744 <-> 10856 12285 <-> 14595 17296 <-> 18416 63020 <-> 76084 66928 <-> 66992 67095 <-> 71145 69615 <-> 87633 79750 <-> 88730 100485 <-> 124155 122265 <-> 139815 122368 <-> 123152 141644 <-> 153176 142310 <-> 168730 171856 <-> 176336 176272 <-> 180848 185368 <-> 203432 196724 <-> 202444

A prime decade is where N+1, N+3, N+7 and N+9 are all prime. The first occurrence of two prime decades with the theoretical minimum separation is N = 1006300 and N = 1006330. The 335th prime decade is N = 2342770. There are 172400 primes < 2342770.

*pi*= 16 arctan (1/5) - 4 arctan(1/239),- which is related to the fact that 2 * 13^4 - 1 = 239^2,
- which is why 239/169 is an approximant (the 7th) of sqrt(2).
- arctan(1/239) = arctan(1/70) - arctan(1/99) = arctan(1/408) + arctan(1/577)
- 239 needs 4 squares (the maximum) to express it.
- 239 needs 9 cubes (the maximum, shared only with 23) to express it.
- 239 needs 19 fourth powers (the maximum) to express it.
- (Although 239 doesn't need the maximum number of fifth powers.)
- 1/239 = .00418410041841..., which is related to the fact that
- 1,111,111 = 239 * 4,649.
- The 239th Mersenne number, 2^239 - 1, is known composite, but no factors are known.#
- 239 = 11101111 base 2.
- 239 = 22212 base 3.
- 239 = 3233 base 4.
- There are 239 primes < 1500.
- K239 is Mozart's only work for 2 orchestras.
- Guess what memo this is.
- And 239 is prime, of course.

*# Actually, this statement was not correct, even in 1972. Many
thanks to these sharp-eyed Netizens of sci.math --
kfoster@rainbow.rmii.com (Kurt Foster) 4-Nov-1996, rawling@erols.com
(Mark Rawlings) 4-Nov-1996, Dan Shanks ~1972 [according to
rcs@cs.arizona.edu (Richard Schroeppel)] -- for pointing out that
*

M239 = 2^239 - 1 = 479 * 1913 * 5737 * 176383 * 134000609 * 7110008717824458123105014279253754096863768062879.

Laboratoire A2X, Universite Bordeaux I 351 Cours de la Liberation 33405 TALENCE Cedex, FRANCE pari@math.u-bordeaux.frftp://megrez.math.u-bordeaux.fr/pub/pari/