|
Search: id:A096693
|
|
|
| A096693 |
|
Balance index of the n-th prime. |
|
+0 21
|
|
| 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 4, 0, 0, 5, 1, 0, 0, 1, 0, 0, 2, 0, 1, 1, 0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 4, 1, 0, 1, 0, 1, 0, 2, 0, 2, 1, 1, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0
(list; graph; listen)
|
|
|
OFFSET
|
1,16
|
|
|
COMMENT
|
The number of orders to which the n-th prime belongs, i.e. the n-th prime is at the center (mode) of the average of k primes on either side a(n) number of times.
The average number of balanced prime indices, p_n, seems to reach a maximum at the 85-th prime, 439, of 50 balanced prime indices.
|
|
EXAMPLE
|
a(3) = 1 because the fifth prime, 5, equals (3 + 7)/2.
a(16) = 3 because the sixteenth prime, 53, equals (47 + 59)/2 = (41 + 43 + 47 + 59 + 61 + 67)/6 = (31 + 37 + 41 + 43 + 47 + 59 + 61 + 67 + 71 + 73)/10.
|
|
MATHEMATICA
|
f[n_] := Block[{c = 0, k = 1, p = Prime[n], s = Plus @@ Table[ Prime[i], {i, n - 1, n + 1}]}, While[k != n - 1, If[s == (2k + 1)p, c++ ]; k++; s = s + Prime[n - k] + Prime[n + k]]; c]; Table[ f[n], {n, 105}]
|
|
CROSSREFS
|
Cf. A090403, A096695, A096705, A096706, A096707, A096708, A096709, A096711.
Adjacent sequences: A096690 A096691 A096692 this_sequence A096694 A096695 A096696
Sequence in context: A046775 A085393 A128980 this_sequence A083206 A069531 A035677
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Robert G. Wilson v (rgwv(AT)rgwv.com), Jun 26 2004
|
|
|
Search completed in 0.002 seconds
|