[seqfan] Re: distribution of totatives of a Primorial

[Jamie Morken]

Hi Seqfans,
Just one update to the last message, I checked to see if this patternworks for larger Primorials.  For 2310 and 30030, the equidistant
partitions don't give equal number of totatives in each partition,but the errors are interesting, and creates this irregular triangle, wherethe 7 rows are for the Primorial numbers 1, 2, 6, 30, 210, 2310, 30030.There are no errors until Primorial number 2310.  Where the error valuerefers to non equal distribution of totatives in each equidistant partition of a Primorial.For row 6, first value error value of -1, this refers to the fraction 480/2310=16/77,
using the denominator 77, and the numerator 16 for the 16th totative of 2310, whichis 71.  Since 73 is also a totative of 2310 but there are no other totatives between 73 and 77,this gives an error of -1.
row1: 0.

row2: 0.

row3: 0,0.

row4: 0,0.

row5: 0,0,0,0,0,0.

row6: -1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, -1, 0, -1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 1, 0.

row7: 2, 0, 0, 0, 0, 0, -2, 0, 0, 0, -2, 0, -2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, -2, 0.

length of rows = 1,1,2,2,6,30,30... (after first term this is A058250).
sum of values on a row = 0.Non-zero value positions on row 6 and row 7 are: 1,7,11,13,17,19,23,29.I think this could be used to count totatives in a known range since thesequence has a symmetry, also if the row sum is always zero.I have asked Michael De Vlieger to coauthor this sequence.  Thanks.cheers,Jamie
----- Original Message -----
From: Jamie Morken <jmorken at shaw.ca>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Wed, 10 Apr 2019 20:51:52 -0600 (MDT)
Subject: [seqfan] distribution of totatives of a Primorial

Hi Seqfans,
An algorithm for calculating the number of totatives in an specified range,also showing that totatives of a Primorial can be partitioned into sectionswhich contain equal numbers of totatives.
For Primorial number 2*3*5*7=210, which has 48 totatives, The specified ranges are given by 48/210. 48/210 as a reduced fraction = 8/35. Creating a set of GCD(210,48) = 6 fractions by adding 8 and 35 respectively to the numerator and denominator of the reduced fraction 8/35 gives:
8/35, 16/70, 24/105, 32/140, 40/175, 48/210. Finding the totatives of 210 which are smaller and nearest to each ofthe denominators: 35,70,105,140,175,210. gives the totatives: 31,67,103,139,173,209. 
Checking the remaining 48-6 = 42 totatives,there are 7 totatives in 6 ranges separated by thetotatives 31,67,103,139,173,209.(1,11,13,17,19,23,29), 
(37,41,43,47,53,59,61), 
(71,73,79,83,89,97,101), 
(107,109,113,121,127,131,137), (143,149,151,157,163,167,169), (179,181,187,191,193,197,199).
The formula to calculate 7 is (48-6)/6=7.where 48 is the number of totatives of Primorial number 210 = 48. and the GCD of 210 and 48 = 6.
Related OEIS sequences:https://oeis.org/A058250https://oeis.org/A038110https://oeis.org/A038111
cheers,Jamie

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