A045939 -id:A045939 - cp's OEIS frontend

A077657 Least number with exactly n consecutive successors, all having the same number of prime factors (counted with multiplicity).

1, 2, 33, 603, 602, 2522, 211673, 3405123, 3405122, 49799889, 202536181, 3195380868, 5208143601, 85843948321, 97524222465

Offset: 0

Reinhard Zumkeller, Nov 13 2002

A001222(a(n))=A001222(a(n)+k) for k<=n;

A077655(a(n))=n and A077655(k)

			a(0)=A077656(1)=1; a(1)=A045920(1)=2; a(2)=A045939(1)=33; a(3)=A045940(2)=603; a(4)=A045941(1)=602; a(5)=A045942(1)=2522.

Cf. A045984.

a(n)=A045984(n+1)+A077655(A045984(n+1))-n - Martin Fuller, Nov 21 2006

More terms from Martin Fuller, Nov 21 2006

A338453 Starts of runs of 3 consecutive numbers with the same total binary weight of their divisors (A093653).

Original entry on oeis.org

3, 242, 243, 1837, 2361, 3693, 3728, 6061, 6457, 9782, 11181, 11721, 13855, 15177, 20017, 22591, 28021, 31461, 31887, 33098, 33993, 38137, 52016, 52112, 60321, 76897, 78542, 78745, 80461, 108394, 116017, 119541, 124453, 125493, 127117, 127417, 145369, 151805, 154113

Offset: 1

Amiram Eldar, Oct 28 2020

Numbers k such that A093653(k) = A093653(k+1) = A093653(k+2).

			3 is a term since A093653(3) = A093653(4) = A093653(5) = 3.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A093653.
Subsequence of A338452.
Similar sequences: A005238, A006073, A045939.

f[n_] := DivisorSum[n, DigitCount[#, 2, 1] &]; s = {}; m = 3; fs = f /@ Range[m]; Do[If[Equal @@  fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 155000}]; s
SequencePosition[Table[Total[DigitCount[Divisors[n],2,1]],{n,160000}],{x_,x_,x_}][[All,1]] (* Harvey P. Dale, Feb 04 2023 *)

A077655 Number of consecutive successors of n having the same number of prime factors as n (counted with multiplicity).

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Reinhard Zumkeller, Nov 13 2002

nonn

If a(n) > 0 then a(n+1) = a(n)-1.

			33=3*11 has only two successors also with two factors: 34=2*17 and 35=5*7 (whereas 33+3=36=2*2*3*3), therefore a(33)=2.

Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A001222, A077656, A045920, A045939, A045940, A045941, A045942, A077657.

snpf[n_]:=Module[{f=PrimeOmega[n],k=0},While[f==PrimeOmega[n+k],k++];k]; Array[snpf,110]-1 (* Harvey P. Dale, Aug 01 2021 *)

A077655(n) = { my(k=n+1,w=bigomega(n)); while(bigomega(k)==w,k++); (k-n)-1; }; \\ Antti Karttunen, Jan 22 2020

A355711 Starts of runs of 3 consecutive numbers with the same number of 5-smooth divisors.

Original entry on oeis.org

33, 85, 93, 145, 213, 265, 393, 445, 453, 475, 505, 633, 685, 753, 805, 813, 865, 933, 985, 993, 1045, 1113, 1165, 1293, 1345, 1353, 1405, 1430, 1533, 1585, 1624, 1653, 1705, 1713, 1765, 1833, 1885, 1893, 1945, 2013, 2065, 2193, 2245, 2253, 2275, 2305, 2433, 2485

Offset: 1

Amiram Eldar, Jul 15 2022

nonn

Numbers k such that A355583(k) = A355583(k+1) = A355583(k+2).

			33 is a term since A355583(33) = A355583(34) = A355583(35) = 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A355583.
Subsequence of A355710.
A355712 is a subsequence.
Similar sequences: A005238, A006073, A045939, A332313, A332387.

f[n_] := Times @@ (1 + IntegerExponent[n, {2, 3, 5}]); s = {}; m = 3; fs = f /@ Range[m]; Do[If[Equal @@ fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 2500}]; s

s(n) = (valuation(n, 2) + 1) * (valuation(n, 3) + 1) * (valuation(n, 5) + 1);
s1 = s(1); s2 = s(2); for(k = 3, 2500, s3 = s(k); if(s1 == s2 && s2 == s3, print1(k-2,", ")); s1 = s2; s2 = s3);

A278311 Numbers n such that n-1 and n+1 have the same number of prime factors as n (with multiplicity).

Original entry on oeis.org

34, 86, 94, 122, 142, 171, 202, 214, 218, 245, 285, 302, 394, 429, 435, 446, 507, 603, 604, 605, 634, 638, 698, 842, 922, 963, 1042, 1075, 1084, 1085, 1131, 1138, 1245, 1262, 1275, 1310, 1346, 1402, 1413, 1431, 1435, 1449, 1491, 1533, 1557, 1587, 1605, 1635, 1642, 1676, 1762, 1772, 1838, 1886, 1894, 1925, 1942

Offset: 1

Ely Golden, Nov 17 2016

nonn

			a(1) = 34, as 33, 34, and 35 all have 2 prime factors.
a(2) = 86, as 85, 86, and 87 all have 2 prime factors.

Ely Golden, Table of n, a(n) for n = 1..10000

Intersection of A045920 and A278291.

a(n) = A045939(n) + 1.

public class A278311{
public static void main(String[] args)throws Exception{
    long dim0=numberOfPrimeFactors(2);//note that this method must be manually implemented by the user
    long dim1=numberOfPrimeFactors(3);
    long dim2;
    long counter=4;
    long index=1;
    while(index<=10000){
      dim2=numberOfPrimeFactors(counter);
      if(dim2==dim1&&dim1==dim0){System.out.println(index+" "+(counter-1));index++;}
      dim0=dim1;
      dim1=dim2;
      counter++;
    }
  }
}

isok(n) = (bigomega(n-1) == bigomega(n)) && (bigomega(n) == bigomega(n+1)); \\ Michel Marcus, Nov 17 2016

def bigomega(x):
    s=0;
    f=list(factor(x));
    for c in range(len(f)):
        s+=f[c][1]
    return s;
dim0=bigomega(2);
dim1=bigomega(3);
counter=4
index=1
while(index<=10000):
    dim2=bigomega(counter);
    if(dim2==dim1&dim1==dim0):
        print(str(index)+" "+str(counter-1))
        index+=1;
    dim0=dim1;
    dim1=dim2;
    counter+=1;

A176292 Numbers k such that the prime factorizations of composite(k) and composite(k+1) have the same number of primes (including multiplicities).

Original entry on oeis.org

1, 4, 7, 10, 12, 15, 17, 18, 21, 22, 25, 28, 29, 40, 47, 53, 61, 62, 64, 68, 69, 72, 85, 87, 90, 91, 93, 100, 102, 106, 107, 110, 112, 114, 116, 120, 125, 130, 131, 132, 133, 136, 151, 154, 155, 158, 165, 166, 169, 170, 179, 181, 190, 191, 198, 212, 221, 222, 223

Offset: 1

Juri-Stepan Gerasimov, Apr 14 2010

nonn

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A045920, A045939.

A001222 := proc(n) numtheory[bigomega](n) ; end proc:
A002808 := proc(n) if n = 1 then return 4; else for a from procname(n-1)+1 do if not isprime(a) then return a; end if; end do; end if; end proc:
for n from 1 to 400 do if A001222(A002808(n)) = A001222(A002808(n+1)) then printf("%d,",n) ; end if; end do: # R. J. Mathar, Apr 20 2010

SequencePosition[PrimeOmega/@Select[Range[300],CompositeQ],{x_,x_}][[;;,1]] (* Harvey P. Dale, Jun 21 2023 *)

A001222(A002808(a(n))) = A001222(A002808(a(n)+1)).

Corrected (86 replaced with 87, 89 removed, many terms after 92 replaced) by R. J. Mathar, Apr 20 2010

A374023 Numbers m such that m .. m+11 all have the same number of prime factors, counted with multiplicity.

Original entry on oeis.org

3195380868, 5208143601, 5208143602, 5327400945, 5604994082, 5604994083, 6940533603, 6940533604, 7109053186, 7112231268, 19355940562, 22180594465, 24073076004, 24155988484, 29495293764, 30997967601, 41999754228, 42322452483, 42322452484, 45479198003, 46553917683

Offset: 1

Zak Seidov and Robert Israel, Jun 25 2024

nonn

Since a(3) = a(2) + 1, a(6) = a(5) + 1 and a(8) = a(7) + 1, a(2) = 5208143601, a(5) = 5604994082 and a(7) = 6940533603 are the first three m such that m .. m+12 have the same number of prime factors, counted with multiplicity.

For n <= 12, A001222(a(n)) = 4. It must always be at least 4 because at least one of a(n) .. a(n)+11 is divisible by 8.

			5208143601 is a term because
  5208143601 = 3 * 139 * 2153 * 5801
  5208143602 = 2 * 47 * 4261 * 13003
  5208143603 = 13 * 103 * 419 * 9283
  5208143604 = 2^2 * 3 * 434011967
  5208143605 = 5 * 7^2 * 21257729
  5208143606 = 2 * 37 * 109 * 645691
  5208143607 = 3^2 * 647 * 894409
  5208143608 = 2^3 * 651017951
  5208143609 = 73^2 * 367 * 2663
  5208143610 = 2 * 3 * 5 * 173604787
  5208143611 = 11 * 29 * 1129 * 14461
  5208143612 = 2^2 * 7 * 186005129
all have 4 prime factors, counted with multiplicity.

Martin Ehrenstein, Table of n, a(n) for n = 1..10000

Subsequence of A033987.
Cf. A001222.
Numbers m through m+k have the same value of A001222: A045920 (k=1), A045939 (k=2), A045940 (k=3), A045941 (k=4), A045942 (k=5), A123103 (k=6), A123201 (k=7), A358017 (k=8), A358018 (k=9), A358019 (k=10).

isok(m) = #Set(apply(bigomega, vector(11, i, m+i-1))) == 1; \\ Michel Marcus, Jul 11 2024

A001222(a(n)) = A001222(a(n)+1) = ... = A001222(a(n)+11).

Missing term inserted by, and more terms from Martin Ehrenstein, Jul 11 2024

A334583 Numbers m such that m, m + 1 and m + 2 each have exactly eight prime factors, not necessarily distinct.

Original entry on oeis.org

40909374, 71410624, 87278750, 126237375, 152439488, 161590624, 166450624, 209140623, 227929624, 243409374, 267308990, 267639470, 290696768, 291513248, 292088510, 295644734, 307885374, 310314158, 319874750, 321890750, 331690624, 336958622, 343030624, 352749248, 354109374, 356269374, 366681248, 391390623, 401375168, 407590623

Offset: 1

Zak Seidov, May 06 2020

nonn

			40909374 = 2 * 3^4 * 11^2 * 2087, 40909375 = 5^5 * 13 * 19 * 53, and 40909376 = 2^6 * 179 * 3571.

Intersection of A045939 and A046310.

Cf. A001222, A113752, A113789.

list(lim)=my(v=List(), k, o); forfactored(n=40909374, lim\1+2, o=bigomega(n); if(o==8, if(k++>2, listput(v, n[1]-2)), k=0)); Vec(v) \\ Charles R Greathouse IV, May 07 2020

A001222(a(n)+i) = 8 for i in {0,1,2}.

Previous Showing 11-20 of 21 results. Next

cp's OEIS Frontend

A358019 Numbers m such that the factorizations of m..m+10 have the same number of primes (including multiplicities).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

A045984 a(n) = smallest number m such that factorizations of n consecutive integers starting at m have same number of primes (counted with multiplicity).

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions

A077657 Least number with exactly n consecutive successors, all having the same number of prime factors (counted with multiplicity).

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A338453 Starts of runs of 3 consecutive numbers with the same total binary weight of their divisors (A093653).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A077655 Number of consecutive successors of n having the same number of prime factors as n (counted with multiplicity).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A355711 Starts of runs of 3 consecutive numbers with the same number of 5-smooth divisors.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A278311 Numbers n such that n-1 and n+1 have the same number of prime factors as n (with multiplicity).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Java

PARI

SageMath

A176292 Numbers k such that the prime factorizations of composite(k) and composite(k+1) have the same number of primes (including multiplicities).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions