A383177
Sphenic numbers k such that floor(log(k)/log(lpf(k))) = 1+floor(log(k)/log(p)) for all primes p | k such that p > lpf(k), where lpf = A020639(k).
Original entry on oeis.org
1001, 1309, 1547, 1729, 2093, 2261, 3553, 4199, 4301, 4807, 5681, 6061, 6479, 7337, 7843, 8671, 9269, 9361, 9889, 10373, 10879, 11063, 11339, 11687, 11803, 11891, 12121, 12617, 13079, 13717, 13949, 13981, 14911, 15283, 15457, 16211, 16523, 17081, 17329, 17719
Offset: 1
Let s(n) = A010846(a(n)).
Table of a(n) for n = 1..12, showing prime factors of a(n) and
n a(n) facs(a(n)) s(n)
---------------------------
1 1001 7*11*13 15
2 1309 7*11*17 15
3 1547 7*13*17 15
4 1729 7*13*19 15
5 2093 7*13*23 15
6 2261 7*17*19 15
7 3553 11*17*19 15
8 4199 13*17*19 15
9 4301 11*17*23 15
10 4807 11*19*23 15
11 5681 13*19*23 15
12 6061 11*19*29 15
Let f(p,k) = floor(log(k)/log(p)) and let w be the list of f(p,k) across the sorted list of distinct prime factors of k.
30 = 2*3*5 is not in the sequence since f(30,2) = 4, f(30,3) = 3, f(30,5) = 2.
a(1) = 1001 = 7*11*13; f(7,1001) = 3, f(11,1001) = 2, f(13,1001) = 2.
a(2) = 1309 = 7*11*17; w(1309) = {3,2,2}, etc.
Pattern of numbers in row a(n) of A275280:
Level r^0 Level r^1 Level r^2
1, p, p^2, p^3 | r, p*r, p^2*r | r^2
q, p*q, p^2*q | q*r, p*q*r |
q^2, p*q^2; |
Example: k = 1001 = 7*11*13
1, 7, 49, 343 | 13, 91, 637 | 169
11, 77, 539 | 143, 1001 |
121, 847 |
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Michael De Vlieger, Hasse diagram of R(1001) with logarithmic vertical scale. Gray represents the empty product, red represents primes, gold represents proper prime powers, green squarefree composites, and blue numbers that are neither squarefree nor prime powers.
- Michael De Vlieger, Three dimensional diagram of R(a(n)), labeling exponents along axes, showing p^3, q^2, and r^2, and using the color scheme above.
- Michael De Vlieger, Plot prime(i) | a(n) at (x,y) = (n,i) for n = 1..2048, 8X vertical exaggeration. The green bar at the bottom of the graph emphasizes the x axis that rides on the top edge of the bar.
-
f[om_, lm_ : 0] := Block[{f, i, j, k, nn, w}, i = Abs[om]; j = 1;
If[lm == 0, nn = Times @@ Prime@ Range[i], nn = Abs[lm]]; w = ConstantArray[1, i];
Union@ Reap[Do[
While[Set[k, Times @@ Map[Prime, Accumulate@w]]; k <= nn,
If[Or[k == 1, Union[#2] == #1 - 1 & @@
TakeDrop[Map[Floor@Log[#, k] &, FactorInteger[k][[All, 1]] ], 1] ],
Sow[k]];
j = 1; w[[-j]]++];
If[j == i, Break[], j++; w[[-j]]++;
w = PadRight[w[[;; -j]], i, 1]], {n, Infinity}] ][[-1, 1]] ];
f[3, 20000]
A381736
Integers k = p*q*r, where p < q < r are distinct primes and p*q > r.
Original entry on oeis.org
30, 70, 105, 154, 165, 182, 195, 231, 273, 286, 357, 374, 385, 399, 418, 429, 442, 455, 494, 561, 595, 598, 627, 646, 663, 665, 715, 741, 759, 782, 805, 874, 897, 935, 957, 969, 986, 1001, 1015, 1023, 1045, 1054, 1085, 1102, 1105, 1131, 1173, 1178, 1209
Offset: 1
30 = 2*3*5 and 2*3 > 5, so 30 is in the sequence.
70 = 2*5*7 and 2*5 > 7, so 70 is in the sequence.
110 = 2*5*11 but 2*5 < 11, so 110 is not in the sequence.
-
N:= 2000: # for terms < N
P:= select(isprime, [2,seq(i,i=3..isqrt(N),2)]):
R:= NULL:
for k from 1 to nops(P) do
for i from 1 to k-2 while P[i]*P[i+1]*P[k] < N do
jmin:= max(i+1,ListTools:-BinaryPlace(P,P[k]/P[i])+1);
jmax:= min(k-1,ListTools:-BinaryPlace(P,N/(P[i]*P[k])));
R:= R, seq(P[i]*P[j]*P[k],j=jmin .. jmax);
od od:
sort([R]); # Robert Israel, Mar 28 2025
-
q[n_] := Module[{f = FactorInteger[n]}, f[[;; , 2]] == {1, 1, 1} && f[[1, 1]]*f[[2, 1]] > f[[3, 1]]]; Select[Range[1500], q] (* Amiram Eldar, Mar 20 2025 *)
-
is_a381736(n) = my(F=factor(n)); omega(F)==3 && bigomega(F)==3 && F[1,1]*F[2,1]>F[3,1] \\ Hugo Pfoertner, Mar 08 2025
-
from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A381736(n):
def bisection(f,kmin=0,kmax=1):
while f(kmax) > kmax: kmax <<= 1
kmin = kmax >> 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x): return n+x-sum(primepi(min(x//(p*q),p*q-1))-b for a,p in enumerate(primerange(integer_nthroot(x,3)[0]+1),1) for b,q in enumerate(primerange(p+1,isqrt(x//p)+1),a+1))
return bisection(f,n,n) # Chai Wah Wu, Mar 28 2025
A381250
a(n) = least k with n distinct prime factors such that floor(log_q(k)) = floor(log_p(k))-1, where p is the smallest prime factor of k, and q is any other distinct prime factor of k.
Original entry on oeis.org
1, 2, 6, 1001, 81719, 101007559, 84248643949, 78464111896111, 997804397813471821, 1314665322768473913751, 25030469300030639321689313, 93516019518175801382127421211, 1873482639168918364977596279806547, 60958708904928776821774364389940352443, 1089851191947047137351117158610882538395561
Offset: 0
Let lpf = A020639, slpf = A119288, and gpf = A006530.
Table of a(n), n=0..12, listing the indices of the smallest, second smallest, and greatest prime factors, the latter 2 pertaining to n >= 2 and n >= 3, respectively.
prime indices
n a(n) lpf slpf-gpf prime factors
-------------------------------------------------------------------------
0 1 0 -
1 2 1 2
2 6 1 2 2*3
3 1001 4 5-6 7*11*13
4 81719 5 7-9 11*17*19*23
5 101007559 9 13-16 23*41*43*47*53
6 84248643949 12 19-23 etc.
7 78464111896111 17 25-30
8 997804397813471821 26 41-47
9 1314665322768473913751 32 48-55
10 25030469300030639321689313 47 69-77
11 93516019518175801382127421211 56 83-92
12 1873482639168918364977596279806547 73 108-118
Let f(p,k) = floor(log_p k) and let w be the list of f(p,k) across the sorted list of distinct prime factors of k.
a(0) = 1 since 1 is the only number that does not have prime factors.
a(1) = 2 since prime numbers have just 1 prime factor, and 2 is the smallest prime.
a(2) = 6 since f(2,6) = 2 and f(3,6) = 1; 6 is the smallest squarefree semiprime.
a(3) = 1001 since w(1001) = {3,2,2} and is the smallest sphenic number with this property.
30 is not in the sequence since w(30) = {4,3,2}; 42 is not in since w(42) = {5,3,1}, etc.
a(4) = 81719 since w(81719) = {4,3,3,3} and is the smallest number with 4 distinct prime factors with this property, etc.
-
f[om_, lm_] := Block[{f, i, j, k, nn, p, q, w, z},
i = Abs[om]; z = i - 1; j = z; nn = Abs[lm]; w = ConstantArray[1, i];
Catch@ Do[
While[Set[{k, p, q}, {Times @@ #, #[[1]], #[[2]]}] &@
Map[Prime, Accumulate@ w]; k <= nn,
If[And[q^i > k, p^(i + 1) > k], Throw[k]];
j = z; w[[-j]]++];
If[j == i, Break[], j++; w[[-j]]++;
w = PadRight[w[[;; -j]], i, 1]], {ii, Infinity}] ];
{1, 2}~Join~Table[f[n, 2^(11*n + 2)], {n, 2, 16}]
Showing 1-3 of 3 results.
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