A003067 -id:A003067 - cp's OEIS frontend

A003068 Problimes (third definition).

2, 4, 7, 11, 15, 19, 23, 28, 33, 38, 43, 48, 53, 58, 63, 68, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127, 133, 139, 145, 151, 157, 163, 169, 175, 181, 187, 193, 199, 205, 211, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329

Offset: 1

N. J. A. Sloane

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=1..1000
M. D. Hirschhorn, How unexpected is the prime number theorem?, Amer. Math. Monthly, 80 (1973), 675-677.
M. D. Hirschhorn, How unexpected is the prime number theorem?, Amer. Math. Monthly, 80 (1973), 675-677. [Annotated scanned copy]
R. C. Vaughan, The problime number theorem, Bull. London Math. Soc., 6 (1974), 337-340.

Cf. A003066, A003067.

a[1] := 2: for i from 1 to 150 do a[i+1] := ceil(a[i]+1/product((1-1/a[j]), j=1..i)): od: # James Sellers, Mar 07 2000

a[1] = 2; a[n_] := a[n] = Ceiling[a[n-1] + 1/Product[1 - 1/a[j], {j, 1, n-1}]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Nov 18 2013 *)

More terms from James Sellers, Mar 07 2000

A003066 Problimes (first definition).

Original entry on oeis.org

2, 4, 6, 9, 12, 15, 19, 23, 27, 31, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 86, 92, 98, 104, 110, 116, 122, 128, 134, 140, 146, 152, 158, 164, 170, 176, 182, 188, 194, 200, 206, 213, 220, 227, 234, 241, 248, 255, 262, 269, 276, 283, 290, 297, 304, 311, 318, 325

Offset: 1

N. J. A. Sloane

From Dean Hickerson, Jan 13 2003: (Start)
Suppose you have a list of the first n prime numbers p_1, ..., p_n and you want to estimate the next one. The probability that a random integer is not divisible by any of p_1, ..., p_n is (1-1/p_1) * ... * (1-1/p_n). In other words, 1 out of every 1/((1-1/p_1) * ... * (1-1/p_n)) integers is relatively prime to p_1, ..., p_n.
So we might expect the next prime to be roughly this much larger than p_n; i.e. p_(n+1) may be about p_n + 1/((1-1/p_1) * ... * (1-1/p_n)). This sequence and A003067, A003068 are obtained by replacing this approximation by an exact equation, using 3 different ways of making the results integers. (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..1000
M. D. Hirschhorn, How unexpected is the prime number theorem?, Amer. Math. Monthly, 80 (1973), 675-677.
M. D. Hirschhorn, How unexpected is the prime number theorem?, Amer. Math. Monthly, 80 (1973), 675-677. [Annotated scanned copy]
R. C. Vaughan, The problime number theorem, Bull. London Math. Soc., 6 (1974), 337-340.

Cf. A003067, A003068.

a[1] := 2: for i from 1 to 150 do a[i+1] := floor(a[i]+1/product((1-1/a[j]), j=1..i)): od: # James Sellers, Mar 07 2000

a[1] = 2; a[n_] := a[n] = Floor[a[n-1] + 1/Product[1-1/a[j], {j, 1, n-1}]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Mar 09 2012, after James Sellers *)

More terms from James Sellers, Mar 07 2000

A368784 a(0) = 1. For n > 0, a(n) is the smallest integer k > n such that (Sum_{i = 1..n} i)/(Sum_{i = n + 1..k} i) < 1/n.

Original entry on oeis.org

1, 2, 4, 7, 10, 13, 17, 21, 25, 30, 35, 40, 45, 50, 56, 62, 68, 74, 81, 87, 94, 101, 108, 115, 122, 130, 138, 145, 153, 162, 170, 178, 187, 195, 204, 213, 222, 231, 240, 250, 259, 269, 279, 289, 298, 309, 319, 329, 339, 350, 361, 371, 382, 393, 404, 415, 427, 438

Offset: 0

Felix Huber, Feb 15 2024

(Sum_{i = 1..n} i)/(Sum_{i = n + 1..k} i) = n*(n + 1)/((k - n)*(n + 1 + k)) < 1/n. It follows that k > -1/2 + sqrt(4*n^3 + 8*n^2 + 4*n + 1)/2.

			a(3) = 7, because (1 + 2 + 3)/(4 + 5 + 6 + 7) = 3/11 < 1/3 and (1 + 2 + 3)/(4 + 5 + 6) = 2/5 > 1/3.

Felix Huber, Table of n, a(n) for n = 0..10000

Cf. A003067, A061465, A126022, A130243.

A368784 := n -> floor(-1/2 + 1/2*sqrt(4*n^3 + 8*n^2 + 4*n + 1)) + 1;
seq(A368784(n), n = 0 .. 57);

a[n_]:= Floor[-1/2 + Sqrt[4*n^3 + 8*n^2 + 4*n + 1]/2] + 1; Array[a,58,0] (* Stefano Spezia, Feb 17 2024 *)

a(n) = floor(-1/2 + sqrt(4*n^3 + 8*n^2 + 4*n + 1)/2) + 1.

a(n) = round(sqrt(n*(n+1)^2 + 1/4)). - Chai Wah Wu, Mar 11 2024

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A003068 Problimes (third definition).

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A003066 Problimes (first definition).

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A368784 a(0) = 1. For n > 0, a(n) is the smallest integer k > n such that (Sum_{i = 1..n} i)/(Sum_{i = n + 1..k} i) < 1/n.

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