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A223545 a(n+1) = Sum_{i=1..a(n)} floor(a(n)/i); a(1) = 2. 0
2, 3, 5, 10, 27, 95, 447, 2795, 22616, 230244, 2878355, 43253538, 767188892, 15813815440, 373816159742, 10018819334375, 301465449259275, 10097316273301640, 373656009129456297, 15176615488012528682, 672638507261177844871 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Equivalently, a(n+1) = Sum_{i=1..infinity} floor(a(n)/i).

The corresponding sequence with a(1) = 0 or 1 is a constant sequence. This is also true with a(1) = -1 for a(n+1) = Sum_{i=1..|a(n)|} floor(a(n)/i), but when a(1) = -2 that sequence begins -2, -3, -6, -16, -61, -322, -2227, .... The PARI program below uses the fact mentioned in the example; it calculates the first 15 terms over 16000 times faster than a brute force program simply using s = sum(i = 1, s, s\i). In particular, for any positive integer j, j is a summand exactly floor(a(n)/j) - floor(a(n)/(j+1)) times when calculating a(n+1).

LINKS

Table of n, a(n) for n=1..21.

EXAMPLE

a(5) = 10 + 5 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 27 since a(4) = 10. In general, this preponderance of relatively large numbers of each of the smallest summands simplifies these calculations. (This pattern is somewhat different when the terms are negative.)

PROG

(PARI)

{a(n, s = 2) =

local(j, ss, t);

if(s < 0,

  print("This function only accepts nonnegative starting");

  print("terms, for which it is (partially) optimized.");

  return());

for(k = 1, n, print1(s, ", ");

  if(k == n, return(s));

  ss = s; j = 1;

  while((t = s\j - s\(j+1)) > 1,

    ss += t*j;

    j++);

  s = ss + sum(i = 2, s\j, s\i))}

a(21)

CROSSREFS

Sequence in context: A133662 A204518 A336991 * A088938 A000617 A132183

Adjacent sequences:  A223542 A223543 A223544 * A223546 A223547 A223548

KEYWORD

nonn

AUTHOR

Rick L. Shepherd, Jul 19 2013

STATUS

approved

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Last modified June 21 00:57 EDT 2021. Contains 345329 sequences. (Running on oeis4.)