A241807 Numerators of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)) as defined in A241269.

1, 1, 2, 7, 11, 2, 11, 29, 37, 23, 28, 67, 79, 23, 53, 121, 137, 77, 86, 191, 211, 29, 127, 277, 301, 163, 176, 379, 407, 109, 233, 497, 529, 281, 298, 631, 667, 88, 371, 781, 821, 431, 452, 947, 991, 259, 541, 1129, 1177, 613, 638

Offset: 0

Author

Jean-François Alcover and Paul Curtz, Apr 29 2014

Keywords

• nonn
• frac

Comments

The subsequence 1, 23, 77, 163, 281, 431, 613, 827, ..., with indices congruent to 1 mod 8, is 16n^2+6n+1, that is, A000124(8n+1)/2 or A014206(8n+1)/4. Its second differences are constant: (16n^2+6n+1)'' = 32.

The sequence A014206/A241807 is integral and consists of the 16-periodic sequence (2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2, ...).

Examples

			1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, 37/495, 23/330, ...
		

Crossrefs

Cf. A000124, A014206, A241269, A188135, A054552, A185438.

Programs

Mathematica

Table[(n^2+n+2)/((n+1)*(n+2)*(n+3)) // Numerator, {n, 0, 50}]

Formula

a(n) = A014206(n)/period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2 (conjectured).

a(4k) = 8*k^2 +2*k +1,

a(4k+2) = 4*k^2 +5*k +2,

a(4k+3) = 8*k^2 +14*k +7,

a(8k+1) = 16*k^2 +6*k +1,

a(16k+5) = 16*k^2 +11*k +2,

a(16k+13) = 32*k^2 + 54*k +23.