A302334 A weighted smoothing applied to the primes as a data set: a(n) = floor(A007443(2n-1)/2^(2n-2)), where A007443 is binomial transform of primes.

2, 3, 5, 7, 10, 13, 16, 20, 24, 28, 32, 36, 40, 44, 48, 53, 57, 61, 66, 70, 75, 79, 84, 89, 94, 98, 103, 108, 113, 119, 124, 129, 135, 140, 146, 151, 156, 162, 167, 172, 178, 183, 189, 194, 200, 205, 211, 216, 222, 228, 233, 239, 244, 250, 255, 261, 267, 273, 278, 284

Offset: 1

Author

Peter Munn, Apr 05 2018

Keywords

• nonn

Comments

a(n) is the weighted average of the first 2n - 1 primes, using row 2n - 2 of Pascal's triangle as weights, with the result rounded down. a(n) is thus based on the longest ordered list of consecutive primes that has prime(n) in the central position, while giving substantially greater weight to the primes near prime(n).

A guiding aim when framing the definition was having the arithmetic mean of the first k terms close to the arithmetic mean of the first k primes. In this respect, a simplified analysis suggested the binomial weighting might perform equally well for large k as small k, and empirical results were encouraging. For all k <= 500 the difference between the means is < 0.541, with 0.5 being exceeded only for 394 <= k <= 401. (These figures become not quite as good if floor rounding is replaced by nearest-integer, though a rounding midway between the two does better than either.)

The early terms (playing the role of primes) correspond closely to A053620 (in the role of primepi function), but the correspondence gets better if nearest-integer rounding is used instead of the floor rounding used here. - Peter Munn, Feb 26 2024

Conjecture: the second differences are in [-2,2].

Examples

			For n=3, we calculate a weighted average of the first 2n - 1 = 5 primes. Row 2n - 2 = 4 of Pascal's triangle, (1,4,6,4,1), provides the weights, and its row sum is 2^4 = 16.
Specifically, using the first formula, a(3) = floor( Sum_{k=0..4}(binomial(4,k)*prime(k+1)) / 2^4 ).
The sum in the formula = 1*prime(1) + 4*prime(2) + 6*prime(3) + 4*prime(4) + 1*prime(5) = 1*2 + 4*3 + 6*5 + 4*7 + 1*11 = 2 + 12 + 30 + 28 + 11 = 83.
So a(3) = floor(83/2^4) = floor(83/16) = 5.
Comparison with the primes: (Start)
Analysis table showing the difference between the start of this sequence and the start of the list of primes. a(n) is subtracted from prime(n) to give a sense of how prime(n) is lower or higher than it might be if the primes were more smoothly distributed. The column headed "cumulative" gives the partial sums of the previous column, which are then divided by n and rounded to 3 decimal places to give the final column. The final column therefore shows the difference between the arithmetic means of the first n primes and the first n terms of this sequence.
  n        prime(n)      a(n)    difference  cumulative average
   1           2           2          0          0       0.000
   2           3           3          0          0       0.000
   3           5           5          0          0       0.000
   4           7           7          0          0       0.000
   5          11          10          1          1       0.200
   6          13          13          0          1       0.167
   7          17          16          1          2       0.286
   8          19          20         -1          1       0.125
   9          23          24         -1          0       0.000
  10          29          28          1          1       0.100
  11          31          32         -1          0       0.000
  12          37          36          1          1       0.083
  13          41          40          1          2       0.154
  14          43          44         -1          1       0.071
  15          47          48         -1          0       0.000
  16          53          53          0          0       0.000
  17          59          57          2          2       0.118
  18          61          61          0          2       0.111
  19          67          66          1          3       0.158
  20          71          70          1          4       0.200
  21          73          75         -2          2       0.095
  22          79          79          0          2       0.091
  23          83          84         -1          1       0.043
  24          89          89          0          1       0.042
  25          97          94          3          4       0.160
  26         101          98          3          7       0.269
  27         103         103          0          7       0.259
  28         107         108         -1          6       0.214
  29         109         113         -4          2       0.069
  30         113         119         -6         -4      -0.133
  31         127         124          3         -1      -0.032
  32         131         129          2          1       0.031
(End)
		

Links

• Peter Munn, Table of n, a(n) for n = 1..500
• P. Marchand and L. Marmet, Binomial smoothing filter: A way to avoid some pitfalls of least square polynomial smoothing, Review of Scientific Instruments, 54, 1034-41, 1983.
• Wikipedia, Smoothing

Crossrefs

Cf. A000040, A000079, A007443, A053620, A060620.

Programs

Mathematica

a[n_] := Floor[ Sum[ Binomial[2n -2, k]*Prime[k +1]/2^(2n -2), {k, 0, 2n -2}]]; Array[a, 60] (* Robert G. Wilson v, Jun 10 2018 *)

PARI

a(n) = floor(sum(k=0, 2*n-2, (binomial(2*n-2,k) * prime(k+1))/2^(2*n-2))); \\ Michel Marcus, Aug 21 2018

Formula

a(n) = floor(Sum_{k=0..2n-2} (binomial(2n-2,k) * prime(k+1))/2^(2n-2)).

a(n) = floor(A007443(2n-1)/2^(2n-2)).

Extensions

a(51)-a(60) from Robert G. Wilson v, Jun 10 2018