A371104 Starting from k=7, each subsequent term is the next larger k such that the ratio A276086(k)/A003415(k) is nearer to 1 than for the previous k in the sequence.

7, 8, 213, 214, 2325, 2532, 4625, 30282, 32358, 32384, 60098, 570816, 572884, 575190, 9732128, 243513275

Offset: 1

Author

Antti Karttunen, Mar 12 2024

Keywords

• nonn
• hard
• more

Comments

Note that A276086(6) / A003415(6) = 5/5 = 1. If there are any x > 6, for which the ratio is 1, then the least one of them will terminate this sequence. Question: Could this sequence actually be infinite?

If it exists, a(17) > 1207959552.

Examples

			          k   A049345(k)    A276086(k)/A003415(k)  A276086(k)-A003415(k)
  ----------------------------------------------------------------------
          7,        101,       10/1        = 10,                9
          8,        110,       15/12       = 1.25,              3
        213,      10011,       66/74       = 0.89189189,       -8
        214,      10020,       99/109      = 0.90825688,      -10
       2325,     100211,     1950/1780     = 1.0955056,       170
       2532,     110200,     3575/3388     = 1.0551948,       187
       4625,     200021,     3042/2900     = 1.0489655,       142
      30282,    1011200,    32725/34181    = 0.95740324,    -1456
      32358,    1100300,    27625/26971    = 1.0242483,       654
      32384,    1101210,   116025/117696   = 0.98580241,    -1671
      60098,    2001110,    30345/30749    = 0.98686136,     -404
     570816,   12011100,  2114035/2093568  = 1.0097761,     20467
     572884,   12100020,   642447/643056   = 0.99905296,     -609
     575190,   12200000,   927979/927483   = 1.0005348,       496
    9732128,  101103110, 26152035/26148912 = 1.0001194,      3123
  243513275, 1220000021, 99685818/99683810 = 1.0000201,      2008.
		

Links

• Index entries for sequences related to primorial base

Crossrefs

Cf. A003415, A049345, A276086, A351228.

Programs

PARI

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
print1(7, ", "); r = A276086(7)/A003415(7); for(n=7, oo, t=A276086(n)/A003415(n); if(abs(1-t) < abs(1-r), r=t; print1(n, ", ")))