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PlanetMath is a virtual community which aims to help make mathematical knowledge more accessible. PlanetMath's content is created collaboratively: the main feature is the mathematics encyclopedia with entries written and reviewed by members. The entries are contributed under the terms of the Creative Commons By/Share-Alike License in order to preserve the rights of authors, readers and other content creators in a sensible way. We use LaTeX, the lingua franca of the worldwide mathematical community. On February 13th 2013, PlanetMath.org was updated to use the new software system Planetary. Some release notes are here. Please report bugs in the Planetary Bugs Forum or on Github.

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[p] • gg by Rushike SLR Aug 18
\begin{itemize} \item \end{itemize}\begin{flushleft} \end{flushleft} \textbf{}gg

[p] pseudoprimes in k(i) (contd) by akdevaraj Aug 17
561 is a pseudoprime to any base of shape (11*k+I) where k belongs to N. This is in addition to the other bases indicated in the previous message.

[p] years divisible by 100 by bbmath Aug 14
There is an exception to "years divisible by 4 being leap years": If a year is divisible by 100 (such years are divisible by 4), such a year is a leap year ONLY IF it is also divisible by 400. For example, 2000 was a leap year, but 1900 was not. Shockley's "Introduction to Number Theory" contains a "day of the week" formula that includes the above fact.

[P] Carmichael numbers and Devaraj numbers by akdevaraj Aug 9
Carmichael numbers constitute a sub-set of Devaraj numbers. To understand more about them refer sequences A 104016, A 104017 and A 162290. Some interesting facts pertaining to them will follow.

[P] The entry "division" became invisible. by pahio Aug 8
The reason was a dollar-sign error =o) Corrected! BTW, the PM search engine has long been out of order. It's quite difficult to find entries on a wanted subject.

[P] Pl. see the older entry http: by pahio Aug 7
Pl. see the older entry http://planetmath.org/divisibilitytest

[p] Mangammal primes by akdevaraj Aug 7
Definition: These are the impossible prime factors of 3^n - 2 (n belongs to N). This is identical with the sequence A123239 (OEIS ).

[p] Nice solution by burgess Aug 6
Nice solution

[P] division of two numbers by pahio Aug 4
Please see e.g. the entry http://planetmath.org/division BTW, one cannot write "P(R)*0 = 0" since your P(R) is not a _number_

[P] Egyptian and Greek square by milogardner Aug 2
under repair ... For now consider a \PMlinkexternal{generalized scribal square root method}{http://planetmath.org/squarerootof3567and29} was published in Dec. 2012. I INTRODUCTION: An ancient square root method was decoded in 2012. Scholars for 100 years failed to fully decode Archimedes’ three steps that estimated unit fraction series answers to four to six places (modern standards), a method used by Fibonacci and Galileo. Unresolved aspects of ancient ESTIMATION OF PI problem were reported by Kevin Brown and E.B. Davis with upper and lower limits; (1351/780)^2 is greater than PI is greater than (265/153)^2 A. Archimedes’ actual square root of pi^2 and n^2 method, decoded in Dec 2012 and Jan 2013, calculated the higher PI limit(1351/780)^2 by: 1. step 1. guess (1 + 2/3)^2 = 1 + 4/3 + 4/9 = 2 + 3/9 + 4/9, meant 2/9 = error1 2. step 2 reduced error1 2/9 by dividing 2/9 by 2(1 + 2/3) steps that meant 2/9 x (3/10) = 1/15 such that (1 + 2/3 + 1/15)^2, error2 (1/15)^2 = 1/225 = error2 knowing (1 + 11/15) = 26/15 3. step 3 reduced error2 = 1/225 by dividing by 2 x (26/15) = 52/15 1/225 x (15/52) = 1/15 x (1/52) = (1/780)^2 = error 3 reached (26/15 - 1/780)^2 = (1351/780)^2 in modern fractions recorded a unit fraction series that began with step 2 data and subtracted 1/780 (1 + 2/3 + 1/15 - 1/780)^2 as Archimedes would have written (1 + 2/3 + 1/30 + (13 + 6 + 4 + 2)/780)^2 = (1 + 2/3 + 1/30 + 1/60 + 1/+ 1/195 + 1/390)^2 B . The lower limit 265/153 modified step 2, used 1/17 rather than 1/15, (1+ 2/3 + 1/17) = (1 + 37/51) such that (1 + 111/153)changed to (1 + 112/153) = 265/153 II. PROOF: Modern translations of scribal square roots of five (5), six (6), seven (7), (29) and any rational numberPlanetmathPlanetmathPlanetmath are demonstrated below (A, B, C, D). Greek and Egyptian algebraic steps were finite. Decoding algorithmic looking finite arithmetic steps 2, 3 and 4 have been demystified. A. Computed the square root of five(5) that estimated (Q + R)^2, R= 1/(1/2Q) step 1: estimated Q = 2. R,= (1/4) such that (2 + 1/4)^2 = 5 + (1/4)^2; error1 = 1/16 step 2, reduced error1 that divided 1/16 by 2(2 + 1/4)= 18/4 such that 1/16 x (4/18) = 1/72 = error2 = (1/72)^2 2 + 1/4 - (1/72)^2)^2 = (2 + 1285/5184)^2; as Archimedes and Ahmes would have re-written (2 + 1285/5184) as a unit fraction series: a [2 + (864 + 398 + 51 + 1/17 + 6)/5184) = [2 + 1/6 + 1/13 + 1/39 + 1/864]^2 b [2 + (864 + 370 + 64 + 6 +1 )/5184] = [2 + 1/6 + 1/14 + 1/81 + 1/864 + 1/5184]^2 Note that scribal shorthand notes suggested by academics prior to Dec. 2012 suggested incomplete square root steps even though major operational aspects of the same class of arithmetic and algebraic pesu steps have been found in the medieval era. In May 2013 the shorthand notes of Galileo reveal the same method was also used by Fibonacci and Archimedes. In the square root of five step 3 was not needed. B. square root of six (6) , step 1: estimated Q = 2, R = (6 -4)/4 = 1/2 such that (2 + 1/2)^2 = 6 + (1/2)^2, error1 = 1/4 step 2: reduced 1/4 error that divided by (2 + 1/2) such that 1/4 x (2/10) = 1/20. hence (2 + 1/4 - 1/400) =((2 + 99/400)*2, error2 = 1/400 Ahmes, Archimes and Fibonacci may have stopped at this point and recorded (2 + 99/400) as a unit fraction series [2 + (80 + 10 + 8 + 1)/400] = [2 + 1/5 + 1/40 + 1/50 + 1/400 ] step 3 (as included Archimedes square root of three method) was optionasl divided 1/400 by (400/1798) = 1/1798, hence (2 + 99/400 - (1/1798)^2 = accurate (1/1798)^2 Archimedes’ actual square root method would have recorded [2+ 1/5 + 1/40 + 1/50 + 1/400] with a note that a longer series, with an error of (1/1798)^2 was easily found. C. square root of seven (7) step 1: estimates Q = 2, R = 3/4 and (2 + 3/4)^2 = 7 + (3/4)^2 , error1 = 9/16 step 2: divides 9/16 by twice (2 + 3/4) = (9/16)(4/22) = 9/88 = (1/11 + 1/88) (2 + 3/4 -9/88) = [2 + 1/2 + 13/88]^2 = [2 + 1/2 + (8 + 4 + 1)/88]^2 = [2 + 1/2 + 1/11 + 1/22 + 1/88]^2 step 3 may have been required divide 9/88 by twice (2 + 1/2 + 13/88) = (9/88)(88/466) = 9/466 = (1/155 + 1/155 + 1/466) hence [2 + 1/2 + 1/11 + 1/22 + 1/88 - (2/155 +1/466)]^2 would have been recorded as a unit fraction series Note that Archimedes and Ahmes paired (1/22 - 2/155) = and (1/88 - 1/466) = readers may choose the most likely final unit fraction series, D. ESTIMATE the square root of 29. step 1 found R = (29-25)/2Q = 4/10 = 2/5 such that (5 + 2/5)^2 = 29 + 4/25 = error1 STEP 2 reduce error1 4/25 by dividing by 2(5 + 2/5) 4/25 x 5/54 = 2/27 hence a final unit fraction series converted (5 + 2/5 - 2/27)^2 by considering (5 + 1/5 + (27-10)/135)^2 = (5 + 1/5 + 1/9 + 2/135)^2 was accurate to (2/27)^2 NOTE: the conversion of 2/135 to a unit fraction series followed Ahmes 2/n table rules (2/5)(1/27) = (1/3 + 1/15)(1/27) = 1/51 + 1/405 MEANT THE SQUARE ROOT OF 29 WAS ESTIMATED IN TWO STEPS BY (5 + 1/5 + 1/9 + 1/51 + 1/405)^2 footnote: Fibonacci’s square root of 17 method was appropriately cited as used by Galileo though not properly analyzed in every detail. Fibonacci guessed (4 + 1/8)^2 = (17 + 1/64) , and Fibonacci reduced the estimated 1/64 error foumd an inversePlanetmathPlanetmathPlanetmath proportion:1/64 x 8/66 = 1/528 which meant (4 + 1/8 - 1/528)^2 = (2177/528)^2 = 17.000003 is accurate to (1/528)^2 perArchimedes and not by Newton, as suggested by scholars prior to 2012. III CONCLUSION Unit fractionPlanetmathPlanetmath square root was formalized by 2050 BCE and used by Egyptians, Greeks, Arabs, medieval scribes and as late as Galileo. The method estimated irrational square roots of N by 1-step, 2-step, 3-step and 4-steps methods. Step 1 guessed quotients (Q) and remainders (R) = n/(2Q) with n = (N - Q^2). Step 2, 3, and 4 reduced error 1, 2 and 3 associated with the previous step by dividing by 2(Q + R). References 1 A.B. Chace, Bull, L, Manning, H.P., Archibald, R.C., The Rhind Mathematical Papyrus, Mathematical 2 Marshall Clagett Ancient Egyptian Science, Volume III, American Philosophical Society, Philadelphia, 1999. 3 Richard Gillings, Mathematics in the Time of the Pharaohs, Dover Books, 1992, PAGE 214-217. 4 H. Schack-Schackenburg, ”Der Berliner Papyreys 6619”, Zeitscrift fur Agypyische Sprache , Vol 38 (1900), pp. 135-140 and Vol. 40 (1902), p. 65f. 5 L. E. Sigler, Fibonacci’s Liber Abaci, Leonardo’s Book of Calculation ,Springer, NY, 2002, page 491.

[p] failure functions - another example by akdevaraj Aug 1
Let our definition of a failure be a non-Carmichael number. Note 561, a Carmichael number, can be represented by the quadratic polynomial x^2 + x +55 where x = 22. Hence x^2 + x + 55 is the parent function. The relevant failure function is x = 29 + 25k; here k belongs to N. When the values of x generated by the failure function are substituted in the parent function we get only failures i.e. non-Carmichael numbers.

[p] pseudoprimes in k(i) (contd) by akdevaraj Jul 31
Although 561 is a Carmichael number in k(1) it is only a pseudoprime in k(i). If we have a composite number consisting of two primes of form 4m+3 and if it happens to be a pseudo to a base, say 2, it is also pseudo to the base (number + i). Example: 341 = 11*31; this is pseudo to base 2. It is also pseudo to base (341 + i) and base ( 341 + 2i).

[p] pseudoprimes in k(i) by akdevaraj Jul 29
341 is a pseudoprime to base 2 (which is in k(1). It is also a pseudoprime to base (341 + i). Hence it is a pseudoprime in k(i) also. Similarly 561, a Carmichael number is a pseudoprime in k(i) as it is pseudo to the base (187 + i) as well as the base (561 + i). Interestingly it is also pseudo to the base (561 + 2i).

Some formulas of Arithmetic progression/series by burgess Jul 29
An arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. Example: 2,4,6,8,10….. Arithmetic Series : The sum of the numbers in a finite arithmetic progression is called as Arithmetic series. Example: 2+4+6+8+10….. nth term in the finite arithmetic series Suppose Arithmetic Series a1+a2+a3+…..an Then nth term an=a1+(n-1)d Where a1- First number of the series an- Nth Term of the series n- Total number of terms in the series d- Difference between two successive numbers Sum of the total numbers of the arithmetic series Sn=n/2*(2*a1+(n-1)*d) Where Sn – Sum of the total numbers of the series a1- First number of the series n- Total number of terms in the series d- Difference between two successive numbers Example: Find n and sum of the numbers in the following series 3 + 6 + 9 + 12 + x? Here a1=3, d=6-3=3, n=5 x= a1+(n-1)d = 3+(5-1)3 = 15 Sn=n/2*(2a1+(n-1)*d) Sn=5/2*(2*3+(5-1)3)=5/2*18 = 45 I hope the above formulae are helpful to solve your math problems>