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 1-10 of 10 results.

A072875 Smallest start for a run of n consecutive numbers of which the i-th has exactly i prime factors.

Original entry on oeis.org

2, 3, 61, 193, 15121, 838561, 807905281, 19896463921, 3059220303001, 3931520917431241
Offset: 1

Views

Author

Rick L. Shepherd, Jun 30 2002 and Jens Kruse Andersen, Jul 28 2002

Keywords

Comments

By definition, each term of this sequence is prime.
a(11) <= 1452591346605212407096281241 (Frederick Schneider), see primepuzzles link. - sent by amd64(AT)vipmail.hu, Dec 21 2007
Prime factors are counted with multiplicity. - Harvey P. Dale, Mar 09 2021

Examples

			a(3)=61 because 61 (prime), 62 (=2*31), 63 (=3*3*7) have exactly 1, 2, 3 prime factors respectively, and this is the smallest solution;
a(6)=807905281: 807905281 is prime; 807905281+1=2*403952641;
807905281+2=3*15733*17117; 807905281+3=2*2*1871*107951;
807905281+4=5*11*43*211*1619; 807905281+5=2*3*3*3*37*404357;
807905281+6=7*7*7*7*29*41*283; 807905281 is the smallest number m such that m+k is product of k+1 primes for k=0,1,2,3,4,5,6.
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 61, p. 22, Ellipses, Paris 2008.

Crossrefs

a(1) = A000040(1), a(2) = A005383(1), a(3) = A112998(1), a(4) = A113000(1), a(5) = A113008(1), a(6) = A113150(1).

Programs

  • Mathematica
    (* This program is not suitable to compute a large number of terms. *) nmax = 6; kmax = 10^6; a[1] = 2; a[n_] := a[n] = For[k = a[n-1]+n-1, k <= kmax, k++, If[AllTrue[Range[0, n-1], PrimeOmega[k+#] == #+1&], Return[k] ] ]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 1, nmax}] (* Jean-François Alcover, Sep 06 2017 *)

Extensions

a(7) found by Mark W. Lewis
a(8) and a(9) found by Jens Kruse Andersen
a(10) found by Jens Kruse Andersen; probably a(11) > 10^20. - Aug 24 2002
Entry revised by N. J. A. Sloane, Jan 26 2007
Cross-references and editing by Charles R Greathouse IV, Apr 20 2010

A093552 a(n) is the smallest number m such that m-k is product of k+1 primes for k=0,1,2,...,n.

Original entry on oeis.org

2, 5, 47, 107, 71999, 392279, 305778479, 24405534719, 899133243623
Offset: 0

Views

Author

Farideh Firoozbakht, Apr 14 2004

Keywords

Comments

Each term of this sequence is prime.

Examples

			a(5)=392279 because:
392279 is prime;
392279-1 = 2*196139;
392279-2 = 3*229*571;
392279-3 = 2*2*281*349;
392279-4 = 5*5*13*17*71;
392279-5 = 2*3*3*19*31*37;
392279 is the smallest number m such that m-k is product of k+1 primes for k=0,1,2,3,4 & 5.
		

Crossrefs

Programs

  • Mathematica
    Table[m = 1; While[Table[Plus @@ Last /@ FactorInteger[Prime[m] - k], {k, 0, n}] != Range[1, n + 1], m++]; Prime[m], {n, 0, 5}] (* Robert Price, Jun 09 2019 *)

Extensions

a(6) corrected by Mike Oakes, Jun 20 2012
a(7) from Zak Seidov, Apr 23 2017
a(8) from Giovanni Resta, Apr 27 2017

A072664 End of smallest run of n consecutive integers with n, n-1, ..., 1 distinct prime factors in that order.

Original entry on oeis.org

2, 7, 107, 2187, 578309, 12239309, 45640671719, 29591195200067
Offset: 1

Views

Author

Rick L. Shepherd, Jun 30 2002

Keywords

Comments

Using a musical analogy, each run is a "crescendo" of primality where each subsequent member of the run is gradually "more prime" in the sense of having one fewer distinct prime factors (see A001221). These a(n) are the peaks of crescendos of increasing length. a(7) is greater than 60000000.
This sequence was inspired by A068069, where the members of the runs have n different numbers of distinct prime factors, 1 through n, but where the order is not specified.

Examples

			a(1)=2 because 2 is prime and therefore the smallest integer with exactly one distinct prime factor. a(2)=7 because 6=2*3 and 7 (prime) is the smallest run of consecutive integers with exactly 2 and 1 distinct prime factors in that order. a(3)=107 because 105=3*5*7, 106=2*53 and 107 (prime) is the smallest run with exactly 3, 2 and 1 distinct prime factors in that order. Note that a(1), a(2), a(3), a(5) and a(6) are prime but that a(4)=2187=3^7 is not.
		

Crossrefs

Cf. A086560 (smallest start with run pattern 1, 2, ..., n), A072665 (center with run pattern n+1, n, ..., 2, 1, 2, ..., n, n+1), A068069 (run order not specified), A001221 (omega(n)).

Extensions

a(7) from Donovan Johnson, Jan 24 2009
a(8) from Donovan Johnson, Jul 19 2011

A072665 Center of smallest run of 2n+1 consecutive numbers with exactly n+1,n,...,2,1,2,...,n,n+1 distinct prime factors, respectively.

Original entry on oeis.org

2, 11, 2917
Offset: 0

Views

Author

Rick L. Shepherd, Jul 30 2002

Keywords

Comments

Borrowing from musical terminology, these could be considered "swells" of primality - first a crescendo ("more prime"), then a decrescendo ("less prime"). a(3), if it exists, is greater than 70750000. The corresponding sequence but counting prime factors with multiplicity (A001222) has only two terms (2, 5) because either the number immediately before or after any odd center > 5 equals 4k for some k >= 2, and thus has at least three prime factors, not exactly two, when duplicates are counted.
a(3) > 10^63. - Hiroaki Yamanouchi, Sep 25 2014

Examples

			a(0) = 2 (prime) is the smallest number with one prime factor. a(1) = 11 as 10 (=2*5), 11 (prime) and 12 (=2^2*3) have 2,1,2 distinct prime factors (A001221), respectively and there is no smaller center of such a run. a(2) = 2917 as 2915 (=5*11*53), 2916 (=2^2*3^6), 2917 (prime), 2918 (=2*1459) and 2919 (=3*7*139) have 3,2,1,2,3 distinct prime factors and there is no smaller such run.
		

Crossrefs

Cf. A072664 (smallest finish with run pattern n, ..., 2, 1), A086560 (smallest start with run pattern 1, 2, ..., n), A001221 (omega).

Extensions

Comment expanded and small typos fixed by Rick L. Shepherd, Jun 22 2017

A136106 a(n) is the smallest prime p such that in the sequence of n numbers p, p+1, p+2, ..., p+n-1, the i-th term has exactly i distinct prime factors, for i = 1, ..., n.

Original entry on oeis.org

2, 5, 103, 1867, 491851, 17681491, 35565206671, 43194825904693
Offset: 1

Views

Author

Enoch Haga, Dec 14 2007

Keywords

Examples

			a(4) = 1867 because it begins with the prime 1867 followed by 1868 with two distinct prime factors, 2 and 467; then 1869 with three distinct prime factors, 3, 7 and 89; then 1870 with four distinct prime factors, 2, 5, 11 and 17.
		

Crossrefs

Programs

  • Mathematica
    Table[First[Select[Prime@Range@100000,(n=1; k=#; While[Length[First/@FactorInteger@k]==n, k++; n++]; n-1==t)&]],{t, 5}]  (* Giorgos Kalogeropoulos, May 07 2019 *)
  • PARI
    /* a brute force program */ a136106(st,ed,ct)={ forprime(x=st,ed, if ((x%6)!=1,next); goodFlag = 1; c = 1; while(goodFlag, if (!(c%2) && isprime(x+c), goodFlag=0, v = factor(x+c); if (length(v[,2]) == c+1, c+=1; if (c > ct,print("Level = ",c," at ",x+c-1,"=",v); ct+=1), goodFlag = 0 ) ) ) ); } \\ Fred Schneider, Dec 18 2007

Formula

a(n) >= A086560(n). - R. J. Mathar, Feb 05 2008
A001221(a(n)+k) = k+1 for 0 <= k <= n-1. - Pontus von Brömssen, Jan 09 2023

Extensions

Edited by N. J. A. Sloane, Dec 23 2007
a(5)-a(6) from Fred Schneider, Dec 18 2007
a(7) from Donovan Johnson, Sep 19 2009
a(8) from Donovan Johnson, Jul 19 2011
Name clarified by Pontus von Brömssen, Jan 09 2023

A279518 Start of first run of n successive numbers in which the sum of aliquot parts of the i-th number has exactly i prime factors, for i = 1..n.

Original entry on oeis.org

4, 8, 8, 1909, 558031, 783975, 185363811, 1584002413
Offset: 1

Views

Author

Paolo P. Lava, Dec 14 2016

Keywords

Examples

			sigma(1909) - 1909 = 107 that is a prime number;
sigma(1910) - 1910 = 1546 = 2*773;
sigma(1911) - 1911 = 1281 = 3*7*61;
sigma(1912) - 1912 = 1688 = 2*2*2*211.
No other number < 1909 has this property and therefore a(4) = 1909.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,d,i,j,k,ok,n; d:=1;
    for k from 1 to q do for n from d to q do ok:=1; for j from 1 to k do
    b:=ifactors(sigma(n+j-1)-n-j+1)[2]; if add(b[i][2],i=1..nops(b))<>j then ok:=0; break; fi; od;
    if ok=1 then d:=n; print(n); break; fi; od; od; end: P(10^12);

Extensions

a(7)-a(8) from Giovanni Resta, Dec 14 2016

A279520 Start of first run of n successive numbers in which the arithmetic derivative of the i-th number has exactly i prime factors, for i = 1..n.

Original entry on oeis.org

6, 105, 1001, 2945, 240485, 1671414, 22551962, 22551962
Offset: 1

Views

Author

Paolo P. Lava, Dec 14 2016

Keywords

Examples

			2945' = 839 that is a prime number;
2946' = 2461 = 23*107;
2947' = 428 = 2*2*107;
2948' = 3260 = 2*2*5*163.
No other number < 2945 has this property and therefore a(4) = 2945.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,b,d,i,j,k,ok,n; d:=1;
    for k from 1 to q do for n from d to q do ok:=1; for j from 1 to k do
    b:=ifactors(sigma(n+j-1))[2]; if add(b[i][2],i=1..nops(b))<>j then ok:=0; break; fi; od;
    if ok=1 then d:=n; print(n); break; fi; od; od; end: P(10^12);

Extensions

a(7)-a(8) from Giovanni Resta, Dec 14 2016

A373618 Least prime starting a run of n consecutive primes p_i, i=1..n, such that p_i + 1 is squarefree and p_(n+1) + 1 is not squarefree.

Original entry on oeis.org

2, 37, 397, 389, 11617, 11597, 11593, 2048509, 2772409, 5193997, 33933701, 125624813, 125624809, 432787781, 432787777, 4762221193, 4762221181, 182839149373, 547414016069, 551900822513
Offset: 1

Views

Author

Jean-Marc Rebert, Jun 11 2024

Keywords

Examples

			a(1) = 2, because 2 is the least prime starting a run of 1 prime such that 2+1 is squarefree and 3+1 = 4 = 2^2 is not squarefree.
For n=4 the first run of 4 squarefree p+1 starts at a(4) = 389, and no run of n=3 so a(3) = 397 is the ending 3 of this run.
  p              = 389, 397, 401, 409, 419
  p+1 squarefree = yes  yes  yes  yes  no
  n=4 run          \----------------/
  n=3 run               \-----------/
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Module[{k=1}, While[pr=Product[Boole[SquareFreeQ[Prime[k+i-1]+1]], {i, n}]==0||pr&& Boole[SquareFreeQ[Prime[k+n]+1]]==1, k++]; Prime[k]]; Array[a, 8] (* Stefano Spezia, Jun 11 2024 *)

A373533 Least starting prime of exactly n consecutive primes p_i (i = 1..n) such that omega(p_i + 1) = 1 + i.

Original entry on oeis.org

5, 23, 499, 13093, 501343, 162598021, 25296334003
Offset: 1

Views

Author

Jean-Marc Rebert, Jun 08 2024

Keywords

Examples

			a(1) = 5, because omega(5+1) = 2, and no lesser number has this property.
For n=3, the primes starting at a(3) = 499 are as follows and are a run of exactly 3 omega = i+1,
  i          =  1    2    3
  p          = 499, 503, 509, 521
  omega(p+1) =  2,   3,   4,   3
               \-----------/
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Module[{k=1},While[Product[Boole[PrimeNu[Prime[k+i-1]+1]==1+i],{i,n}]==0, k++]; Prime[k]]; Array[a,5] (* Stefano Spezia, Jun 10 2024 *)

A373626 Least prime of a run of n consecutive primes p_i, i = 1..n, such that bigomega(p_i + 1) = omega(p_i + 1) + i and bigomega(p_(n+1) + 1) <> omega(p_(n+1) + 1) + n + 1, or -1 if no such prime exists.

Original entry on oeis.org

3, 19, 739, 76913, 4510333, 746264059, 290623032907
Offset: 1

Views

Author

Jean-Marc Rebert, Jun 11 2024

Keywords

Examples

			19 starts a run of 2 consecutive primes 19 and 23, bigomega(19+1) = 2 = omega(19+1) + 1, bigomega(23+1) = 4 = omega(23+1) + 2 and bigomega(29+1) = 3 <> omega(29+1) + 3. So a(2) = 19.
Let i, p, b and w be the indices, the primes p_i, bigomega(p_i + 1) and omega(p_i + 1).
i: [ 1  2  3]
p: [19 23 29]
b: [ 3  4  3]
w: [ 2  2  3]
a(2) = 19
i: [  1   2   3   4]
p: [739 743 751 757]
b: [  4   5   5   2]
w: [  3   3   2   2]
a(3) = 739
		

Crossrefs

Showing 1-10 of 10 results.