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
A077655 Number of consecutive successors of n having the same number of prime factors as n (counted with multiplicity).
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
Keywords
Comments
If a(n) > 0 then a(n+1) = a(n)-1.
Examples
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.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..100000
Programs
-
Mathematica
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 *) -
PARI
A077655(n) = { my(k=n+1,w=bigomega(n)); while(bigomega(k)==w,k++); (k-n)-1; }; \\ Antti Karttunen, Jan 22 2020
A374023 Numbers m such that m .. m+11 all have the same number of prime factors, counted with multiplicity.
3195380868, 5208143601, 5208143602, 5327400945, 5604994082, 5604994083, 6940533603, 6940533604, 7109053186, 7112231268, 19355940562, 22180594465, 24073076004, 24155988484, 29495293764, 30997967601, 41999754228, 42322452483, 42322452484, 45479198003, 46553917683
Offset: 1
Keywords
Comments
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.
Examples
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.
Links
- Martin Ehrenstein, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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PARI
isok(m) = #Set(apply(bigomega, vector(11, i, m+i-1))) == 1; \\ Michel Marcus, Jul 11 2024
Extensions
Missing term inserted by, and more terms from Martin Ehrenstein, Jul 11 2024
A375239 Numbers k such that k, k+1, ..., k+5 all have 3 prime factors (counted with multiplicity).
2522, 4921, 18241, 25553, 27290, 40313, 90834, 95513, 98282, 98705, 117002, 120962, 136073, 136865, 148682, 153794, 181441, 181554, 185825, 211673, 211674, 212401, 215034, 216361, 231002, 231665, 234641, 236041, 236634, 266282, 281402, 285410, 298433, 298434, 330473, 331985, 346505, 381353
Offset: 1
Keywords
Comments
All terms == 1 or 2 (mod 8).
One of the numbers k, k+1, ..., k+5 is a Zumkeller number (A083207), since it is of the form 2*3*p, where p is prime > 3. - Ivan N. Ianakiev, Aug 08 2024
Examples
a(3) = 18241 is a term because 18241 = 17 * 29 * 37 18242 = 2 * 7 * 1303 18243 = 3^2 * 2027 18244 = 2^2 * 4561 18245 = 5 * 41 * 89 18246 = 2 * 3 * 3041 are all products of 3 primes (counted with multiplicity).
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
-
Maple
R:= NULL: count:= 0: p:= 1: while count < 100 do p:= nextprime(p); x:= 4*p; if andmap(t -> numtheory:-bigomega(t)=3, [x-2,x-1,x+1,x+2]) then if numtheory:-bigomega(x-3) = 3 then R:= R, x-3; count:= count+1; fi; if numtheory:-bigomega(x+3) = 3 then R:= R, x-2; count:= count+1; fi; fi; od: R; -
Mathematica
s = {}; Do[If[{3, 3, 3, 3, 3, 3} == PrimeOmega[Range[k, k + 5]], AppendTo[s, k]], {k, 1000000}]; s
A375160 Square array T(n, k), n >= 2 and k >= 1, read by antidiagonals in ascending order, give the smallest number that starts a sequence of exactly k consecutive numbers each having exactly n prime factors (counted with multiplicity), or -1 if no such number exists.
4, 8, 9, 16, 27, 33, 32, 135, 170, -1, 64, 944, 1274, 603, -1, 128, 5264, 15470, 4023, 602, -1, 256, 29888, 33614, 57967, 12122, 2522, -1, 512, 50624, 3145310, 8706123, 632148, 204323, 211673, -1
Offset: 2
Comments
All positive terms are composite.
Examples
T(2,3) = 33 = 3*11, because both 34 and 35 have the same number of prime factors. Thus, 33 is the starting number of a run of 3 numbers that each have 2 prime factors (counted with multiplicity). No lesser number has this property, so T(2,3) = 33. Table begins (upper left corner = T(2,1)): 4 9 33 -1 ... 8 27 170 603 ... 16 135 1274 4023 ... 32 944 15470 57967 ... ... ... ... ... ...
Comments
Examples
Crossrefs
Formula
Extensions