A100298 -id:A100298 - cp's OEIS frontend

A112878 A100298(n) = 0.

14, 34, 35, 41, 42, 56, 92, 100, 124, 132, 133, 134, 135, 153, 167, 204, 216, 217, 218, 236, 261, 269, 283, 295, 330, 342, 369, 373, 392, 443, 464, 480, 488, 521, 587, 599, 634, 648, 688, 714, 723, 742, 786, 814, 815, 816, 817, 818, 819, 820, 821, 829, 833

Offset: 1

Klaus Brockhaus, Oct 24 2005

nonn

A100298 is similar to Cald's sequence A006509, where zeros are much less frequent (cf. A112877).

			A100298(14) = 0 and A100298(k) > 0 for k < 14, so a(1) = 14.

Cf. A100298, A006509, A112877.

A111118 a(1) = 1; skipping over integers occurring earlier in the sequence, count down c(n) (c(n) = n-th composite) from a(n) to get a(n+1). If this is <= 0, instead count up from a(n) c(n) positions (skipping already occurring integers) to get a(n+1).

Original entry on oeis.org

1, 5, 11, 2, 13, 23, 9, 26, 7, 28, 3, 31, 52, 29, 55, 25, 57, 22, 59, 19, 62, 17, 64, 15, 66, 10, 68, 6, 71, 115, 69, 117, 63, 119, 60, 121, 56, 124, 53, 126, 50, 128, 47, 131, 45, 133, 43, 135, 40, 137, 38, 140, 35, 142, 33, 144, 30, 147, 24, 149, 18, 151, 14, 153, 8, 156

Offset: 1

Leroy Quet, Oct 15 2005

nonn

If we did not skip earlier occurring integers when counting, we would instead have sequence A100298.

			The first 4 terms of the sequence can be plotted on the number line as:
1,2,*,*,5,*,*,*,*,*,11,*,*.
Now a(4) is 2. Counting c(4) = 9 down from 2 gets a negative integer. So we instead count up 9 positions, skipping the 5 and 11 as we count, to arrive at 13 (which is at the rightmost * of the number line above).

Cf. A100298, A110080, A002808.

More terms from Klaus Brockhaus, Oct 17 2005

cp's OEIS Frontend

A112878 A100298(n) = 0.

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