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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A256470 a(n) = A256469(n) - A256468(n).

Original entry on oeis.org

0, 1, 2, 3, 1, 6, 1, 3, 3, 0, 3, 2, -2, 0, 4, 6, 0, 4, 4, 0, 2, 3, 10, 13, -1, -2, 3, 4, 4, 34, 5, 3, 5, 17, 6, 2, 8, -2, -6, 3, -4, -4, -3, -2, -1, 9, 25, -2, -6, -4, 12, 4, 6, 9, -6, 18, 1, -2, -11, 7, -8, 27, -10, 3, -1, 12, 11, 13, -3, 5, 3, 5, -13, -8, 10, 16, -4, 14, 3, 12, -3, 23, 5, 4, 6, -8, 19, -13, 1, 0
Offset: 1

Views

Author

Antti Karttunen, Mar 30 2015

Keywords

Comments

a(n) = Difference between the number of primes occurring in range [prime(n)*prime(n+1), prime(n+1)^2] and the number of primes occurring in range [prime(n)^2, prime(n)*prime(n+1)].
In other words, a(n) tells how many more primes there are in the latter part of the range prime(n)^2 .. prime(n+1)^2 (after the geometric mean of its limits), than in its first part (before the geometric mean of its limits).

Links

Crossrefs

Positions of zeros: A256471. Cf. also A256472, A256473.
Positions of nonnegative terms: A256474, negative terms: A256475.
Positions of strictly positive terms: A256476, terms less than or equal to zero: A256477.
Cf. A256449, A256468, A256469.

Programs

Formula

a(n) = A256469(n) - A256468(n).
a(n) = 3 - A256449(n).

A251723 First differences of A054272, A250473 and A250474: a(n) = A054272(n+1) - A054272(n).

Original entry on oeis.org

1, 4, 5, 14, 8, 21, 10, 26, 46, 15, 56, 43, 19, 45, 79, 77, 31, 89, 65, 29, 105, 74, 113, 162, 88, 41, 86, 41, 99, 353, 98, 164, 48, 298, 57, 181, 185, 127, 197, 194, 75, 355, 76, 143, 74, 462, 478, 167, 81, 165, 269, 89, 437, 274, 273, 291, 90, 291, 198, 98, 511, 734, 219, 106, 214, 783, 340, 578, 124, 240, 362, 488, 380, 379, 251, 393, 529, 261, 530, 669, 150, 708, 150
Offset: 1

Views

Author

Antti Karttunen, Dec 15 2014

Keywords

Links

Crossrefs

One less than A050216, the first differences of A000879.
Cf. A054272, A250473, A250474, A251719, A256446, A256447, A256448, A256449.

Formula

a(n) = A054272(n+1) - A054272(n).
a(n) = A256447(n) + A256448(n). [Cf. also A256449.]

A256447 Number of integers in range (prime(n)^2)+1 .. (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n): a(n) = A250477(n) - A250474(n).

Original entry on oeis.org

2, 3, 3, 7, 5, 9, 6, 13, 23, 9, 28, 22, 12, 24, 39, 37, 17, 44, 32, 16, 53, 37, 53, 76, 46, 23, 43, 20, 49, 161, 48, 82, 23, 142, 27, 91, 90, 66, 103, 97, 41, 181, 41, 74, 39, 228, 228, 86, 45, 86, 130, 44, 217, 134, 141, 138, 46, 148, 106, 47, 261, 355, 116, 53, 109, 387, 166, 284, 65, 119, 181, 243, 198, 195, 122, 190, 268, 125, 265, 330, 78
Offset: 1

Views

Author

Antti Karttunen, Mar 29 2015

Keywords

Comments

a(n) = number of integers in range [(prime(n)^2)+1, (prime(n) * prime(n+1))] whose smallest prime factor is at least prime(n).
All the terms are strictly positive, because at least for the last number in the range we have A020639(prime(n)*prime(n+1)) = prime(n).
See the conjectures in A256448.

Examples

			For n=1, we have in range [(prime(1)^2)+1, (prime(1) * prime(2))], that is, in range [5,6], two numbers, 5 and 6, whose smallest prime factor (A020639) is at least 2, thus a(1) = 2.
For n=2, we have in range [10, 15] three numbers, {11, 13, 15}, whose smallest prime factor is at least 3, thus a(2) = 3.
For n=3, we have in range [26, 35] three numbers, {29, 31, 35}, whose smallest prime factor is at least prime(3) = 5, thus a(3) = 3.

Links

Crossrefs

One more than A256468.
Cf. A020639, A250474, A250477, A251723, A256446, A256448, A256449.

Programs

  • Mathematica
    f[n_] := Count[Range[Prime[n]^2 + 1, Prime[n] Prime[n + 1]],
  x_ /; Min[First /@ FactorInteger[x]] >=
Prime@n]; Array[f, 81] (* Michael De Vlieger, Mar 30 2015 *)
  • Scheme
    (define (A256447 n) (- (A250477 n) (A250474 n)))

Formula

a(n) = A250477(n) - A250474(n).
a(n) = A251723(n) - A256448(n).
a(n) = A256448(n) + A256449(n).
a(n) = A256468(n) + 1.
Other identities. For all n >= 1:
a(n+1) = A256446(n) - A256448(n).

A256448 a(n) = A250474(n+1) - A250477(n).

Original entry on oeis.org

-1, 1, 2, 7, 3, 12, 4, 13, 23, 6, 28, 21, 7, 21, 40, 40, 14, 45, 33, 13, 52, 37, 60, 86, 42, 18, 43, 21, 50, 192, 50, 82, 25, 156, 30, 90, 95, 61, 94, 97, 34, 174, 35, 69, 35, 234, 250, 81, 36, 79, 139, 45, 220, 140, 132, 153, 44, 143, 92, 51, 250, 379, 103, 53, 105, 396, 174, 294, 59, 121, 181, 245, 182, 184, 129, 203, 261, 136, 265, 339, 72
Offset: 1

Views

Author

Antti Karttunen, Mar 29 2015

Keywords

Comments

a(n) tells how many more positive integers there are <= prime(n+1)^2 whose smallest prime factor is at least prime(n+1), as compared to how many positive integers there are <= (prime(n) * prime(n+1)) whose smallest prime factor is at least prime(n).
Conjecture 1: for n >= 2, a(n) > 0.
Conjecture 2: ratio a(n)/A256447 converges towards 1. See the associated plots in A256447 and A256449 and comments in A050216.
As what comes to the second conjecture, it's not necessarily true. See the plots linked into A256468. - Antti Karttunen, Mar 30 2015

Examples

			For n=1, the respective primes are prime(1) = 2 and prime(2) = 3, and the ranges in question are [1, 9] and [1, 6]. The former range contains 4 such numbers whose lpf (A020639) is at least 3, namely {3, 5, 7, 9}, while the latter range contains 5 such numbers whose lpf is at least 2, namely {2, 3, 4, 5, 6}, thus a(1) = 4 - 5 = -1.
For n=2, the respective primes are prime(2) = 3 and prime(3) = 5, and the ranges in question are [1, 25] and [1, 15]. The former range contains 8 such numbers whose lpf is at least 5, namely {5, 7, 11, 13, 17, 19, 23, 25}, while the latter range contains 7 such numbers whose lpf is at least 3, namely {3, 5, 7, 9, 11, 13, 15}, thus a(2) = 8 - 7 = 1.
For n=3, the respective primes are prime(3) = 5 and prime(4) = 7, and the ranges in question are [1, 49] and [1, 35]. The former range contains 13 such numbers whose lpf is at least 7, namely {7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49}, while the latter range contains 11 such numbers whose lpf is at least 5, namely {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35}, thus a(3) = 13 - 11 = 2.

Links

Crossrefs

Two less than A256469.
Cf. A050216, A250474, A250477, A251723, A256446, A256447, A256449.

Programs

Formula

a(n) = A256469(n) - 2.
a(n) = A250474(n+1) - A250477(n).
a(n) = A251723(n) - A256447(n).
a(n) = A256446(n) - A256447(n+1).
a(n) = A256447(n) - A256449(n).
Showing 1-4 of 4 results.

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