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.

A080569 a(n) is the first number in the first run of at least n successive numbers, all having exactly 3 distinct prime factors.

Original entry on oeis.org

30, 230, 644, 1308, 2664, 6850, 10280, 39693, 44360, 48919, 218972, 526095, 526095, 526095, 17233173, 127890362, 29138958036, 146216247221, 23671413563491, 36966736685739
Offset: 1

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Author

Randy L. Ekl, Feb 21 2003

Keywords

Comments

The 19th term, if it exists, is at least 1.1 * 10^12. - Fred Schneider, Jan 05 2008
There can be at most 209 terms in this sequence. Any list of 210 consecutive numbers must contain a number n which is multiple of 2*3*5*7 = 210. So omega(n) would be >3. - Fred Schneider, Jan 05 2008
Eggleton and MacDougall show that there are no more than 59 terms in this sequence. [From T. D. Noe, Oct 13 2008]
a(19) > 10^13. - Donovan Johnson, Jun 11 2013
a(19) <= 7523987244435061. - Donovan Johnson, Jul 08 2013
a(21) > 2 * 10^15, if it exists. - Toshitaka Suzuki, Jun 23 2025

Examples

			a(3) = 644 because 644 = 2^2 * 7 * 23, so omega(644) = 3, 645 = 3*5*43, so omega(645) = 3 and 646 = 2*17*19, so omega(646) = 3 and no other number n < 644 has omega(n)=omega(n+1)=omega(n+2)=3.
		

Crossrefs

Programs

  • Mathematica
    k = 1; Do[ While[ Union[ Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]] != {3}, k++ ]; Print[k], {n, 1, 16}]
  • PARI
    k=1; for(i=1,600000,s=1; for(j=1,k,if(omega(i+j-1)!=3,s=0,)); if(s==1,print1(i,", "); k++; i--,) )

Extensions

Edited and extended by Robert G. Wilson v, Feb 22 2003
More terms from Don Reble, Mar 02 2003
a(19)-a(20) from Toshitaka Suzuki, Apr 01 2025

A048971 Let d(n) = number of distinct primes dividing n (A001221); sequence gives t such that d(t)=d(t+1)=...=d(t+n-1) is a run of record length.

Original entry on oeis.org

1, 14, 54, 91, 323, 44360, 48919, 218972, 534078, 2699915, 3355906, 17233173
Offset: 1

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Author

Keywords

Examples

			54, 55, 56, 57, 58 is a run of length 5 in A001221 all with d = 2.
		

Crossrefs

Monotonic subsequence of A048932. Cf. A048972.

Extensions

More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Sep 04 2006

A048972 Length of record run mentioned in A048971.

Original entry on oeis.org

1, 2, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16
Offset: 1

Views

Author

Keywords

Examples

			54, 55, 56, 57, 58 is a run of length 5 in A001221 all with d = 2.
		

Crossrefs

Extensions

More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Sep 04 2006

A305235 Smallest positive number k such that there are exactly n successive equal values of A001221 starting at k, i.e., such that A305234(k) = n.

Original entry on oeis.org

1, 4, 3, 2, 54, 91, 142, 141, 44360, 48919, 218972, 526097, 526096, 526095, 17233173, 127890362, 29138958036, 118968284929, 118968284928, 585927201065, 585927201064, 585927201063, 585927201062, 313978488186061, 453918847597185, 453918847597184, 455626105596320
Offset: 0

Views

Author

Felix Fröhlich, May 28 2018

Keywords

Comments

a(27) > 2 * 10^15. - Toshitaka Suzuki, Jun 22 2025

Examples

			For n = 5: A001221(91+k) = 2 for k = 0..5 and 91 is the smallest number x with exactly 5 successors that have the same value of A001221 as x, so a(5) = 91.
		

Crossrefs

Programs

  • PARI
    a305234(n) = my(k=n+1, i=0); while(omega(k)==omega(n), i++; k++); i
    a(n) = my(k=1); while(1, if(a305234(k)==n, return(k)); k++)

Extensions

a(16)-a(22) from Toshitaka Suzuki, Apr 01 2025
a(23)-a(26) from Toshitaka Suzuki, Jun 22 2025

A349262 a(n) is the start of the least run of exactly n consecutive numbers with the same value of A349258.

Original entry on oeis.org

1, 14, 20, 2, 91, 6850, 2302, 141, 56014, 184171, 2800171, 27805034, 35297611, 8313366182, 1791416073, 3618621410
Offset: 1

Views

Author

Amiram Eldar, Nov 12 2021

Keywords

Comments

a(17) > 10^11, if it exists.

Examples

			a(2) = 14 since A349258(14) = A349258(15) = 2, but A349258(13) != 2 and A349258(16) != 2.
		

Crossrefs

Cf. A349258.
Similar sequences: A006558, A045983, A048932, A067813, A077657, A318166.

Programs

  • Mathematica
    f[p_, e_] := 2^DigitCount[e, 2, 1] - 1; d[1] = 0; d[n_] := Plus @@ f @@@ FactorInteger[n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], dprev = 0, n = 2, c = 1, k = 1}, s[[1]] = 1; While[k < len && n < nmax, d1 = d[n]; If[d1 == dprev, c++, If[c > 0 && c <= len && s[[c]] == 0, k++; s[[c]] = n - c]; c = 1]; n++; dprev = d1]; TakeWhile[s, # > 0 &]]; seq[8, 10^4]

A349305 a(n) is the start of the least run of exactly n consecutive numbers with the same number of nonunitary divisors.

Original entry on oeis.org

4, 10, 1, 19940, 54584, 204323, 2789143044, 27092041443
Offset: 1

Views

Author

Amiram Eldar, Nov 14 2021

Keywords

Comments

a(9) > 10^11, if it exists.

Examples

			a(2) = 10 since A048105(10) = A048105(11) = 0, and A048105(9) != 0 and A048105(12) != 0.
		

Crossrefs

Similar sequences: A006558, A045983, A048932, A067813, A077657, A318166.

Programs

  • Mathematica
    d[n_] := DivisorSigma[0, n] - 2^PrimeNu[n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], dprev = -1, n = 1, c = 0, k = 0}, While[k < len && n < nmax, d1 = d[n]; If[d1 == dprev, c++, If[c > 0 && c <= len && s[[c]] == 0, k++; s[[c]] = n - c]; c = 1]; n++; dprev = d1]; TakeWhile[s, # > 0 &]]; seq[6, 10^6]
Showing 1-6 of 6 results.