cp's OEIS Frontend

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 1-6 of 6 results.

A256449 a(n) = A256447(n) - A256448(n).

Original entry on oeis.org

3, 2, 1, 0, 2, -3, 2, 0, 0, 3, 0, 1, 5, 3, -1, -3, 3, -1, -1, 3, 1, 0, -7, -10, 4, 5, 0, -1, -1, -31, -2, 0, -2, -14, -3, 1, -5, 5, 9, 0, 7, 7, 6, 5, 4, -6, -22, 5, 9, 7, -9, -1, -3, -6, 9, -15, 2, 5, 14, -4, 11, -24, 13, 0, 4, -9, -8, -10, 6, -2, 0, -2, 16, 11, -7, -13, 7, -11, 0
Offset: 1

Views

Author

Antti Karttunen, Mar 29 2015

Keywords

Comments

Positions of zeros: 4, 8, 9, 11, 22, 27, 32, 40, 64, 71, 79, 104, 113, 126, 140, 201, 225, 332, 333, 394, 451, ...
Corresponding primes: 7, 19, 23, 31, 79, 103, 131, 173, 311, 353, 401, 569, 617, 701, 809, 1229, 1427, 2237, 2239, 2707, 3187, ...

Crossrefs

Programs

Formula

a(n) = A256447(n) - A256448(n).
a(n) = 2*A250477(n) - A250474(n) - A250474(n+1).
a(n) = 3 - A256470(n).

A250477 Number of times prime(n) (the n-th prime) occurs as the least prime factor among numbers 1 .. (prime(n)^2 * prime(n+1)): a(n) = A078898(A251720(n)).

Original entry on oeis.org

6, 8, 12, 21, 33, 45, 63, 80, 116, 148, 182, 232, 265, 296, 356, 433, 490, 548, 625, 674, 740, 829, 919, 1055, 1187, 1252, 1313, 1376, 1446, 1657, 1897, 2029, 2134, 2301, 2484, 2605, 2785, 2946, 3110, 3301, 3439, 3654, 3869, 3978, 4086, 4349, 4811, 5147, 5273, 5395, 5604, 5787, 6049, 6403, 6684, 6954, 7153
Offset: 1

Views

Author

Antti Karttunen, Dec 14 2014

Keywords

Comments

a(n) = Position of 6 on row n of array A249821. This is always larger than A250474(n), the position of 4 on row n, as 4 is guaranteed to be the first composite term on each row of A249821.
From Antti Karttunen, Mar 29 2015: (Start)
a(n) = 1 + number of positive integers <= (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n).
That a(n) > A250474(n) can also be seen by realizing that prime(n) must occur at least as many times as the smallest prime factor for the numbers in range 1 .. (prime(n)^2 * prime(n+1)) than for numbers in (smaller) range 1 .. (prime(n)^3), and also by realizing that a(n) cannot be equal to A250474(n) because each row of A249822 is a permutation of natural numbers.
Or more simply, by considering the comment given in A256447 which follows from the new interpretation given above.
(End)

Crossrefs

Column 6 of A249822. Cf. also A250474 (column 4), A250478 (column 8).
First differences: A256446. Cf. also A256447, A256448.

Programs

  • Mathematica
    f[n_] := Count[Range[Prime[n]^2*Prime[n + 1]], x_ /; Min[First /@ FactorInteger[x]] == Prime@ n]; Array[f, 20] (* Michael De Vlieger, Mar 30 2015 *)
  • PARI
    allocatemem(234567890);
    A002110(n) = prod(i=1, n, prime(i));
    A250477(n) = { my(m); m = (prime(n) * prime(n+1)); sumdiv(A002110(n-1), d, (moebius(d)*(m\d))); };
    for(n=1, 23, print1(A250477(n),", "));
    \\ A more practical program:
    
  • PARI
    allocatemem(234567890);
    vecsize = (2^24)-4;
    v020639 = vector(vecsize);
    v020639[1] = 1; for(n=2,vecsize, v020639[n] = vecmin(factor(n)[, 1]));
    A020639(n) = v020639[n];
    A250477(n) = { my(p=prime(n),q=prime(n+1),u=p*q,k=1,s=1); while(k <= u, if(A020639(k) >= p, s++); k++); s; };
    for(n=1, 564, write("b250477.txt", n, " ", A250477(n)));
    \\ Antti Karttunen, Mar 29 2015

Formula

a(n) = A078898(A251720(n)).
a(1) = 1, a(n) = Sum_{d | A002110(n-1)} moebius(d) * floor(A006094(n) / d). [Follows when A251720, (p_n)^2 * p_{n+1} is substituted to the similar formula given for A078898. Here p_n is the n-th prime (A000040(n)), A006094(n) gives the product p_n * p{n+1} and A002110(n) gives the product of the first n primes. Because the latter is always squarefree, one could use here also Liouville's lambda (A008836) instead of Moebius mu (A008683)].
a(n) = A250474(n) + A256447(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

Crossrefs

One less than A050216, the first differences of A000879.

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.
		

Crossrefs

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).

A256446 First differences of A250477: a(n) = A250477(n+1) - A250477(n).

Original entry on oeis.org

2, 4, 9, 12, 12, 18, 17, 36, 32, 34, 50, 33, 31, 60, 77, 57, 58, 77, 49, 66, 89, 90, 136, 132, 65, 61, 63, 70, 211, 240, 132, 105, 167, 183, 121, 180, 161, 164, 191, 138, 215, 215, 109, 108, 263, 462, 336, 126, 122, 209, 183, 262, 354, 281, 270, 199, 192, 249, 139, 312, 605, 495, 156, 162, 492, 562, 458, 359, 178, 302, 424, 443, 377, 306
Offset: 1

Views

Author

Antti Karttunen, Mar 29 2015

Keywords

Crossrefs

Programs

Formula

a(n) = A250477(n+1) - A250477(n).
a(n) = A256447(n+1) + A256448(n).

A256469 Number of primes between prime(n)*prime(n+1) and prime(n+1)^2.

Original entry on oeis.org

1, 3, 4, 9, 5, 14, 6, 15, 25, 8, 30, 23, 9, 23, 42, 42, 16, 47, 35, 15, 54, 39, 62, 88, 44, 20, 45, 23, 52, 194, 52, 84, 27, 158, 32, 92, 97, 63, 96, 99, 36, 176, 37, 71, 37, 236, 252, 83, 38, 81, 141, 47, 222, 142, 134, 155, 46, 145, 94, 53, 252, 381, 105, 55, 107, 398, 176, 296, 61
Offset: 1

Views

Author

Antti Karttunen, Mar 30 2015

Keywords

Examples

			For n=1, there is only one prime in range prime(1)*prime(2) .. prime(2)^2, [6 .. 9], namely 7, thus a(1) = 1.
For n=2, the primes in range prime(2)*prime(3) .. prime(3)^2, [15 .. 25] are {17, 19, 23}, thus a(2) = 3.
		

Crossrefs

Programs

  • Mathematica
    Table[Count[Range[Prime[n] Prime[n + 1], Prime[n + 1]^2], ?PrimeQ], {n, 69}] (* _Michael De Vlieger, Mar 30 2015 *)
    Table[PrimePi[Prime[n+1]^2]-PrimePi[Prime[n]Prime[n+1]],{n,70}] (* Harvey P. Dale, Jul 31 2021 *)
  • PARI
    allocatemem(234567890);
    default(primelimit,4294965247);
    A256469(n) = (primepi(prime(n+1)^2) - primepi(prime(n)*prime(n+1)));
    for(n=1, 6541, write("b256469.txt", n, " ", A256469(n)));
    
  • Scheme
    (define (A256469 n) (let* ((p (A000040 n)) (q (A000040 (+ 1 n))) (q2 (* q q))) (let loop ((s 0) (k (* p q))) (cond ((= k q2) s) (else (loop (+ s (if (prime? k) 1 0)) (+ k 1)))))))

Formula

a(n) = A256448(n)+2.
a(n) = A050216(n) - A256468(n).
a(n) = A256468(n) + A256470(n).
Showing 1-6 of 6 results.