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 1-5 of 5 results.

A093772 a(n) is the smallest integer at which the value of the "truncated Mertens function" (= A088004) equals n.

Original entry on oeis.org

1, 6, 10, 14, 15, 21, 22, 26, 34, 35, 38, 39, 51, 55, 57, 58, 62, 65, 77, 85, 86, 87, 91, 93, 94, 95, 119, 122, 123, 129, 134, 142, 143, 145, 146, 158, 159, 161, 185, 202, 203, 205, 206, 209, 210, 213, 214, 215, 217, 218, 219, 221, 253, 254, 265, 278, 299, 301, 302
Offset: 1

Views

Author

Labos Elemer, Apr 28 2004

Keywords

Comments

Truncated Mertens function = summatory Moebius when argument runs through nonprimes.

Crossrefs

Cf. A002321, A008682, A059071, A088004, A093773.

Programs

  • Mathematica
    mer[x_] :=mer[x]=mer[x-1]+MoebiusMu[x]; mer[0]=0;$RecursionLimit=1000; t=Table[mer[w]+PrimePi[w], {w, 1, 1000}] Table[Min[Flatten[Position[t, j]]], {j, 1, 200}]

A093773 a(n) is the smallest integer at which the value of the "truncated Mertens function" (= A088004) equals the n-th prime number.

Original entry on oeis.org

6, 10, 15, 22, 38, 51, 62, 77, 91, 123, 134, 159, 203, 206, 214, 253, 302, 305, 330, 341, 365, 395, 454, 489, 526, 542, 545, 554, 566, 586, 723, 753, 781, 794, 866, 870, 914, 933, 966, 1059, 1138, 1141, 1198, 1202, 1214, 1219, 1293, 1351, 1383, 1387, 1403
Offset: 1

Views

Author

Labos Elemer, Apr 28 2004

Keywords

Comments

Truncated Mertens function = summatory Moebius when argument runs through nonprimes. See A088004(n) = A002321(n) + A000720(n).

Crossrefs

Cf. A000720, A002321, A008682, A059071, A088004, A093772.

Programs

  • Mathematica
    mer[x_] :=mer[x]=mer[x-1]+MoebiusMu[x]; mer[0]=0;$RecursionLimit=1000; t=Table[mer[w]+PrimePi[w], {w, 1, 1000}] Table[Min[Flatten[Position[t, Prime[j]]]], {j, 1, 200}]

Formula

a(n) = A093772(prime(n)) = A093772(A000040(n)). Solutions to min{x; A002321(x) + A000720(x) = A000040(n) = prime(n)} = a(n).

A093774 Smallest integers at which the value of truncated Mertens function equals n!.

Original entry on oeis.org

1, 6, 21, 93, 698, 5377, 48198, 483817, 5231438, 61160870, 774696437, 10552025533, 153901234789
Offset: 1

Views

Author

Labos Elemer, Apr 28 2004

Keywords

Comments

Observe that magnitude of a(n) is not far from that of (n-1)!.

Crossrefs

Cf. A002321, A000720, A088004, A093772, A093773.

Programs

  • Mathematica
    s = 0; k = 1; Do[ While[s = s + MoebiusMu[k]; s + PrimePi[k] < n!, k++ ]; Print[k]; k++, {n, 10}] (* Robert G. Wilson v, Apr 30 2004 *)

Formula

Solutions to Min(x : A088004(x) = n!), i.e., a(n) = Min(x : A002321(x) + A000720(x) = n!).

Extensions

a(9)-a(10) from Robert G. Wilson v, Apr 30 2004
a(11)-a(13) from Donovan Johnson, Jun 21 2012

A093775 Smallest integers at which the value of truncated Mertens function equals the n-th primorial, the product of first n prime numbers.

Original entry on oeis.org

6, 21, 129, 1290, 20209, 353018, 7537961, 173772587, 4735433401, 160157951005
Offset: 1

Views

Author

Labos Elemer, May 03 2004

Keywords

Crossrefs

Cf. A002321, A000720, A088004, A093772, A093773, A002110, A093774.

Programs

  • Mathematica
    pri[x_] :=pri[x-1]*Prime[x];pri[0]=1; s = 0; k = 1; Do[ While[s = s + MoebiusMu[k]; s + PrimePi[k] < pri[n], k++ ]; Print[k]; k++, {n, 10}]

Formula

Solutions to Min(x : A088004(x) = n!), i.e. a(n) = Min(x: A002321(x) + A000720(x) = A002110(n))

Extensions

a(8)-a(10) from Donovan Johnson, Jun 21 2012

A093776 Smallest integer at which the value of truncated Mertens function equals 2^n.

Original entry on oeis.org

6, 14, 26, 58, 142, 326, 734, 1713, 3713, 8057, 17869, 38985, 84046, 180010, 385846, 823687, 1737474, 3680099, 7755978, 16282918, 34142786, 71419857, 148960009, 310320958, 645191390, 1339363921, 2777235410, 5750237373, 11891042257, 24563702542, 50684981730
Offset: 1

Views

Author

Labos Elemer, May 03 2004

Keywords

Comments

It appears that the ratio of a(j+1)/a(j) is a bit larger than 2 and perhaps tends to 2. Why?

Crossrefs

Cf. A002321, A000720, A088004, A093772, A093773, A002110, A093774, A093775.

Programs

  • Mathematica
    s = 0; k = 1; Do[ While[s = s + MoebiusMu[k]; s + PrimePi[k] < 2^n, k++ ]; Print[k]; k++, {n, 20}]

Formula

Solutions to Min(x : A088004(x) = 2^n}, i.e. a(n) = Min(x: A002321(x) + A000720(x) = 2^n)

Extensions

a(21) - a(24) from Robert G. Wilson v, May 06 2004
a(25)-a(31) from Donovan Johnson, Jun 21 2012
Showing 1-5 of 5 results.

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