January 2018 - mathfun@mailman.xmission.com

[math-fun] Reminder: FactorInteger@1 is {{1, 0}}, or better yet: {}!
by Bill Gosper 02 Jan '18

02 Jan '18

Either that, or 1 is a prime number: In[375]:= Cases[FactorInteger@Range@22, {{p_, 1}} -> p] Out[375]= {1, 2, 3, 5, 7, 11, 13, 17, 19} In[376]:= FindSequenceFunction@% Out[376]= FindSequenceFunction[{1, 2, 3, 5, 7, 11, 13, 17, 19}] In[377]:= FindSequenceFunction@Rest@%% Out[377]= Prime You should be embarrassed for all the damage to young students. Who might well ask: Why isn't it {{1,69}}? Can Maple be this stupid? Does it matter? €2,700.00 + €540.00 regular yearly upgrade ?? Macsyma: (c1) MAKELIST(FACTOR_NUMBER(N),N,1,9); 1 1 2 1 1 1 1 3 2 (d1) [[], [2 ], [3 ], [2 ], [5 ], [2 , 3 ], [7 ], [2 ], [3 ]] (c2) BLOCK([FANCY_DISPLAY : FALSE],PLAYBACK([1,2])); --Bill

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[math-fun] Possible linear recurrence
by Kimberling, Clark 02 Jan '18

02 Jan '18

Here's a question about possible linear recurrence for A296220, which gives the solution of the complementary equation a(n) = a(0)*b(n-1) + a(1)*b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences. In the Formula section, it is conjectured that a(n) = a(n-1) + a(n-7) - a(n-8) for n > 20. However, that sequence (using the same a(0),...a(20) as A296220) includes 244, but A296220 doesn't. So, the question is, does A296220 satisfy some other recurrence? Best wishes for the new year! Clark Kimberling

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[math-fun] a remarkable prime number to bring in the new year
by James Buddenhagen 01 Jan '18

01 Jan '18

This 350 digit prime number (broken into groups of 7 digits): 3756421 7246513 3261547 7452163 3612547 7152643 7126543 7324651 7432651 7561423 1246537 7562431 1564237 3546271 7451623 3456127 3462751 3156427 1265347 3456217 7256341 1476253 1726453 7642513 1342657 7356421 1672453 3152467 3241657 7624531 1354267 7362541 1246573 1562347 7435621 1572643 1452637 7542163 3612457 3526741 7164253 1674523 3542761 3462517 1436527 7216543 3254761 3524617 3456721 1276543 is reversible, i.e. it is prime whether you read it front to back or back to front. In addition each of the groups of 7 digits is a reversible prime 7-digits long, which is a different permutation of 1,2,3,4,5,6,7.

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[math-fun] Properties of 2018
by Allan Wechsler 01 Jan '18

01 Jan '18

2018's complete prime factorization is 2*1009. 2018 is the sum of two squares exactly one way: 2018 = 13^2 + 43^2. It is not the sum of two cubes.

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[math-fun] Russian math
by James Propp 01 Jan '18

01 Jan '18

Here's a (somewhat late) Diophantine puzzle in honor of the 365th day of the year: It is not hard to show that 365 can be written as both 10^2 + 11^2 + 12^2 and 13^2 + 14^2. (This fact came to my attention about twenty years ago through Nikolai Petrovich Bogdanov-Belsky's painting "Mental Calculation in the Public School Of S. A. Rachinsky", which you can view at https://goo.gl/images/yo88LY. Puzzle: Show that there are infinitely many integers that can be written both as a sum of two consecutive squares and as a sum of three consecutive squares. For extra credit, find the proof mentally. :-) Jim Propp

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[math-fun] today Jan. 1 2018
by Simon Plouffe 01 Jan '18

01 Jan '18

Hello, one of my friend remarked this particular fact about this day, Paul Simon (but not the singer). It is the first time in human history where all adults are born with a year beginning by 1 and all minors (less than 18 years of age) will be born in a year beginning with 2. Happy New Year. Best regards, Simon Plouffe

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Re: [math-fun] Russian math
by Dan Asimov 31 Dec '17

31 Dec '17

(My failed attempt to solve this is below.) ----- Here's a (somewhat late) Diophantine puzzle in honor of the 365th day of the year: It is not hard to show that 365 can be written as both 10^2 + 11^2 + 12^2 and 13^2 + 14^2. (This fact came to my attention about twenty years ago through Nikolai Petrovich Bogdanov-Belsky's painting "Mental Calculation in the Public School Of S. A. Rachinsky", which you can view at https://goo.gl/images/yo88LY. Puzzle: Show that there are infinitely many integers that can be written both as a sum of two consecutive squares and as a sum of three consecutive squares. For extra credit, find the proof mentally. :-) ----- Let's see: The sum of 3 consecutive squares is (x-1)^2 + x^2 + (x+1)^2 = 3x^2 + 2, x in Z while for 2 consecutive squares, y^2 + (y+1)^2 = 2y^2 + 2y + 1, y in Z. Equality means 2y^2 + 2y + 1 = 3x^2 + 2 or 2y^2 + 2y = 3x^2 + 1 and the LHS is even, hence the RHS is even, so x is odd: say x = (2K + 1). So 2y(y+1) = 3(2K+1)^2 + 1 = 12K^2 + 12K + 2 or y(y+1) = 6K^2 + 6K + 1. Now the LHS must be even but the RHS must be odd, so this is impossible. (Arggh, probably made some dumb mistake.) —Dan

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