A341606 Square array A(n,k) = A017666(A246278(n,k)), read by falling antidiagonals; denominator of abundancy index as applied onto prime shift array A246278.

2, 4, 3, 1, 9, 5, 8, 5, 25, 7, 5, 27, 35, 49, 11, 3, 21, 125, 77, 121, 13, 7, 15, 55, 343, 143, 169, 17, 16, 11, 175, 13, 1331, 221, 289, 19, 6, 81, 65, 539, 187, 2197, 323, 361, 23, 10, 75, 625, 119, 1573, 247, 4913, 437, 529, 29, 11, 63, 245, 2401, 209, 2873, 391, 6859, 667, 841, 31

Offset

1,1

Comments

See also comments and examples in A341605.

Links

Antti Karttunen, Table of n, a(n) for n = 1..22155; the first 210 antidiagonals

Index entries for sequences computed from indices in prime factorization

Index entries for sequences related to sigma(n)

Formula

A(n, k) = A017666(A246278(n, k)).

A(n, k) = A246278(n, k) / A355925(n, k). - Antti Karttunen, Jul 22 2022

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,
                            A017665(n)/A017666(n)
---------------------------------------------------------------------------
    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.
Note that if A006530 does not grow, then certainly A001222 drops.

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

Cf. A017666, A246278.

Cf. A341605 (numerators), A341626 (numerators of the columnwise first quotients of A341605/A341606), A341627 (and their denominators), A355925, A355927.

Cf. A341607 (the largest prime factor in this array), A341608 (the number of prime factors, with multiplicity).

Cf. also A007691, A341523, A341524.

Sequence in context: A304337 A274329 A322398 * A109158 A307500 A049245

Adjacent sequences: A341603 A341604 A341605 * A341607 A341608 A341609

Keyword

nonn,frac,tabl,look

Author

Antti Karttunen, Feb 16 2021

Status

approved