A100287 First occurrence of n in A100002; the least k such that A100002(k) = n.

1, 2, 5, 9, 15, 25, 31, 43, 61, 67, 87, 103, 123, 139, 169, 183, 219, 241, 259, 301, 331, 361, 391, 447, 463, 511, 553, 589, 643, 687, 723, 783, 819, 867, 931, 979, 1027, 1099, 1179, 1227, 1309, 1347, 1393, 1479, 1539, 1603, 1699, 1759, 1863, 1909, 2019, 2029

Offset

1,2

Comments

Also, the first number that is crossed off at stage n in the Flavius sieve (A000960). - N. J. A. Sloane, Nov 21 2004

The sequence appears to grow roughly like 0.7825*n^2. Note that for n>2, the second occurrence of n in A100002 is at a(n)+1.

Equals main diagonal of triangle A101224, which is defined by the process starting with column 1: A101224(n,1) = n^2-n+1 for n>=1 and continuing with: A101224(n,k) = (n-k+1)*floor( (A101224(n,k-1) - 1)/(n-k+1) ) for k>1 until k=n. I.e., a(n) = A101224(n,n). - Paul D. Hanna, Dec 01 2004

Links

Donovan Johnson, Table of n, a(n) for n = 1..1000

Index entries for sequences related to the Josephus Problem

Formula

a(n) ~ Pi/4 * n^2 (via A000960). - Bill McEachen, Aug 08 2024

Mathematica

a[n_] := Fold[#2*Ceiling[#1/#2 + 1] &, 1, Reverse@Range[n - 1]]; Array[a, 30] (* Birkas Gyorgy, Feb 16 2011 *)

Prog

(PARI) {a(n)=local(A); for(k=1, n, if(k==1, A=n^2-n+1, A=(n-k+1)*floor((A-1)/(n-k+1)))); A}

Crossrefs

Cf. A000960, A099259, A100002, A101224.

Column 1 of A278507, column 2 of A278505 (without the initial 1-term).

Sequence in context: A098169 A055610 A134342 * A007176 A267757 A176691

Adjacent sequences: A100284 A100285 A100286 * A100288 A100289 A100290

Keyword

nonn

Author

T. D. Noe, Nov 11 2004

Status

approved