OFFSET
1,1
COMMENTS
See also comments and examples in A341605.
LINKS
FORMULA
EXAMPLE
The top left corner of the array:
k= 1 2 3 4 5 6 7 8 9 10 11 12
2k= 2 4 6 8 10 12 14 16 18 20 22 24
|
----+--------------------------------------------------------------------------
1 | 2, 4, 1, 8, 5, 3, 7, 16, 6, 10, 11, 2,
2 | 3, 9, 5, 27, 21, 15, 11, 81, 75, 63, 39, 9,
3 | 5, 25, 35, 125, 55, 175, 65, 625, 245, 275, 85, 875,
4 | 7, 49, 77, 343, 13, 539, 119, 2401, 121, 91, 133, 3773,
5 | 11, 121, 143, 1331, 187, 1573, 209, 14641, 1859, 2057, 253, 17303,
6 | 13, 169, 221, 2197, 247, 2873, 299, 28561, 3757, 3211, 377, 2197,
7 | 17, 289, 323, 4913, 391, 5491, 493, 83521, 6137, 6647, 527, 93347,
8 | 19, 361, 437, 6859, 551, 8303, 589, 130321, 10051, 10469, 37, 157757,
9 | 23, 529, 667, 12167, 713, 15341, 851, 279841, 19343, 16399, 943, 352843,
etc.
Arrays A341607 and A341608 give the largest prime factor (A006530) and the number of prime factors with multiplicity (A001222) of these terms. There are nonmonotonicities in both, for example, in columns 11, 12 and 14. This is illustrated below:
For column 11, with successive prime shifts of 22, we obtain:
n sigma(n) sigma(n)/n in lowest terms,
---------------------------------------------------------------------------
22 36 = (2^2 * 3^2), 18/11 = (2 * 3^2)/11
39 56 = (2^3 * 7), 56/39 = (2^3 * 7)/(3 * 13)
85 108 = (2^2 * 3^3), 108/85 = (2^2 * 3^3)/(5 * 17)
133 160 = (2^5 * 5), 160/133 = (2^5 * 5)/(7 * 19)
253 288 = (2^5 * 3^2), 288/253 = (2^5 * 3^2)/(11 * 23)
377 420 = (2^2 * 3 * 5 * 7), 420/377 = (2^2 * 3 * 5 * 7)/(13 * 29)
527 576 = (2^6 * 3^2), 576/527 = (2^6 * 3^2)/(17 * 31)
703 760 = (2^3 * 5 * 19), 40/37 = (2^3 * 5)/37 <-- A001222 drops!
943 1008 = (2^4 * 3^2 * 7), 1008/943 = (2^4 * 3^2 * 7)/(23 * 41)
-
On the second last row, the denominator of 760/703 (= 40/37) has only one prime factor (instead of two), namely 37, because sigma(703) has 19 as its divisor, which otherwise would be present in the denominator.
-
For column 12, with successive prime shifts of 24, we obtain:
n sigma(n) sigma(n)/n
---------------------------------------------------------------------------
24 60 = (2^2 * 3 * 5), 5/2 = (5)/(2)
135 240 = (2^4 * 3 * 5), 16/9 = (2^4)/(3^2)
875 1248 = (2^5 * 3 * 13), 1248/875 = (2^5 * 3 * 13)/(5^3 * 7)
3773 4800 = (2^6 * 3 * 5^2), 4800/3773 = (2^6 * 3 * 5^2)/(7^3 * 11)
17303 20496 = (2^4 *3 *7 *61), 20496/17303 = (2^4 *3 *7 *61)/(11^3 * 13)
37349 42840 = (2^3 *3^2 *5 *7 *17), 2520/2197 = (2^3 * 3^2 *5 *7)/(13^3) !!
93347 104400 = (2^4 *3^2 *5^2 *29), 104400/93347 = (2^4 *3^2 *5^2 *29)/(17^3 *19)
-
On the second last row, the denominator of 42840/37349 (= 2520/2197) has no prime factor 17 (which would be otherwise present), because sigma(37349) has it as its divisor.
-
For column 14, with successive prime shifts of 28, we obtain:
n sigma(n) sigma(n)/n
---------------------------------------------------------------------------
28 56 = (2^3 * 7), 2/1,
99 156 = (2^2 * 3 * 13), 52/33 = (2^2 * 13)/(3 * 11)
325 434 = (2 * 7 * 31), 434/325 = (2 * 7 * 31)/(5^2 * 13)
833 1026 = (2 * 3^3 * 19), 1026/833 = (2 * 3^3 * 19)/(7^2 * 17)
2299 2660 = (2^2 * 5 * 7 * 19), 140/121 = (2^2 * 5 * 7)/(11^2) <-- !!
3887 4392 = (2^3 * 3^2 * 61), 4392/3887 = (2^3 * 3^2 * 61)/(13^2 * 23)
On the second last row, the denominator of 2660/2299 (= 140/121) has no prime factor 19 (which would be otherwise present), because sigma(2299) has it as its divisor.
PROG
(PARI)
up_to = 105;
A246278sq(row, col) = if(1==row, 2*col, my(f = factor(2*col)); for(i=1, #f~, f[i, 1] = prime(primepi(f[i, 1])+(row-1))); factorback(f));
A017666(n) = denominator(sigma(n)/n);
A341606sq(row, col) = A017666(A246278sq(row, col));
A341606list(up_to) = { my(v = vector(up_to), i=0); for(a=1, oo, for(col=1, a, i++; if(i > up_to, return(v)); v[i] = A341606sq(col, (a-(col-1))))); (v); };
v341606 = A341606list(up_to);
A341606(n) = v341606[n];
CROSSREFS
AUTHOR
Antti Karttunen, Feb 16 2021
STATUS
approved