Displaying 1-10 of 49 results found.
Irregular triangle T(n,m) read by rows: smallest power e of n that is divisible by m = term k in row n of A162306.
+20
4
0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 2, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 2, 1, 3, 1, 0, 1, 0, 1, 1, 1, 1, 2, 2, 1, 0, 1, 0, 1, 2, 1, 3, 1, 0, 1, 1, 2, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 2, 1, 3, 1, 2, 4, 1, 0, 1, 0, 1, 1, 1, 2, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1, 2, 3, 1, 4, 1, 0, 1, 0, 1, 1, 1, 1, 1
COMMENTS
This table eliminates the negative values in row n of A279907. Let k = A162306(n,m), i.e., the value in column m of row n. T(n,1) = 0 since 1 | n^0.
T(n,p) = 1 for prime divisors p of n since p | n^1.
T(n,d) = 1 for divisors d > 1 of n since d | n^1.
Row n for prime p have two terms, {0,1}, the maximum value 1, since all k < p are coprime to p, and k | p^1 only when k = p.
Row n for prime power p^i have (i+1) terms, one zero and i ones, since all k that appear in corresponding row n of A162306 are divisors d of p^i. Values greater than 1 pertain only to composite k of composite n > 4, but not in all cases. T(n,k) = 1 for squarefree kernels k of composite n.
Numbers k > 1 coprime to n and numbers that are products of at least one prime q coprime to n and one prime p | n do not appear in A162306; these do not divide n^e evenly. T(n,k) is nonnegative for all numbers k for which n^k (mod k) = 0, i.e., all the prime divisors p of k also divide n.
The largest possible value s in row n of T = floor(log_2(n)), since the largest possible multiplicity of any number m <= n pertains to perfect powers of 2, as 2 is the smallest prime. This number s first appears at T(2^s + 2, 2^s) for s > 1.
1/k terminates T(n,k) digits after the radix point in base n for values of k that appear in row n of A162306. T(a*b,c*d) = max(T(a,c),T(b,d)) if GCD(a,b)=1, GCD(b,d)=1,T(a,c)>=0 and T(b,d)>=0.
T(n,a*b) = max(T(n,a),T(n,b)) if GCD(a,b)=1 and T(n,a)>=0 and T(n,b)>=0.
(End)
EXAMPLE
Triangle T(n,m) begins: Triangle A162306(n,k): 1: 0 1
2: 0 1 1 2
3: 0 1 1 3
4: 0 1 1 1 2 4
5: 0 1 1 5
6: 0 1 1 2 1 1 2 3 4 6
7: 0 1 1 7
8: 0 1 1 1 1 2 4 8
9: 0 1 1 1 3 9
10: 0 1 2 1 3 1 1 2 4 5 8 10
...
MATHEMATICA
Table[SelectFirst[Range[0, #], PowerMod[n, #, k] == 0 &] /. m_ /; MissingQ@ m -> Nothing &@ Floor@ Log2@ n, {n, 24}, {k, n}] // Flatten (* Version 10.2, or *)
DeleteCases[#, -1] & /@ Table[If[# == {}, -1, First@ #] &@ Select[Range[0, #], PowerMod[n, #, k] == 0 &] &@ Floor@ Log2@ n, {n, 24}, {k, n}] // Flatten (* or *)
DeleteCases[#, -1] & /@ Table[Boole[k == 1] + (Boole[#[[-1, 1]] == 1] (-1 + Length@ #) /. 0 -> -1) &@ NestWhileList[Function[s, {#1/s, s}]@ GCD[#1, #2] & @@ # &, {k, n}, And[First@# != 1, ! CoprimeQ @@ #] &], {n, 24}, {k, n}] // Flatten
2, 4, 4, 4, 5, 7, 5, 8, 8, 2, 10, 8, 12, 11, 13, 6, 13, 6, 6, 9, 8, 11, 4, 8, 16, 5, 6, 7, 13, 12, 7, 10, 19, 15, 16, 17, 9, 6, 15, 10, 3, 11, 8, 18, 28, 14, 14, 10, 30, 28, 15, 4, 20, 33, 13, 12, 6, 22, 18, 21, 12, 11, 29, 12, 11, 8, 24, 18, 8, 14, 17, 32, 33
COMMENTS
Let r(x) = A010846(x), the number of m <= x such that rad(m) | x, where rad = A007947. Let row k of A162306 contain { m : rad(m) | k, m <= k }. Thus r(k) is the length of row k of A162306. Let T(k,j) represent the j-th term in row k of irregular triangle A162306. a(n) = j is the position of b(n) in row s(n-1) of A162306. b(n) = T(s(n-1), a(n)).
EXAMPLE
a(3) = 2 since b(3) = 3 is the 2nd term in row s(3) = 3 of A162306, {1, [3]}. a(4) = 4 since b(4) = 4 is the 4th term in row s(4) = 6 of A162306, {1, 2, 3, [4], 6}. a(5) = 4 since b(5) = 5 is T(s(n-1), 4) = T(10, 4), {1, 2, 4, [5], 8, 10}.
a(6) = 4 since b(6) = 9 is T(s(n-1), 4) = T(15, 4), {1, 3, 5, [9], 15}.
a(7) = 5 since b(7) = 6 is T(s(n-1), 5) = T(24, 5), {1, 2, 3, 4, [6], 8, 9, 12, 16, 18, 24}, etc.
n b(n) s(n-1) a(n) r(n) row s(n-1) of A162306 ---------------------------------------------------------------------
3 3 3 2 2 {1, [3]}
4 4 6 4 5 {1, 2, 3, [4], 6}
5 5 10 4 6 {1, 2, 4, [5], 8, 10}
6 9 15 4 5 {1, 3, 5, [9], 15}
7 6 24 5 11 {1, 2, 3, 4, [6], ..., 24}
8 8 30 7 18 {1, 2, 3, 4, 5, 6, [8], ..., 30}
9 16 38 5 8 {1, 2, 4, 8, [16], 19, 32, 38}
10 12 54 8 16 {1, 2, 3, 4, 6, 8, 9, [12], ..., 54}
11 11 66 8 22 {1, 2, 3, 4, 6, 8, 9, [11], ..., 66}
12 7 77 2 5 {1, [7], 11, 49, 77}
13 14 84 10 28 {1, 2, 3, 4, ..., 12, [14], ..., 84}
14 28 98 8 13 {1, 2, 4, 7, ..., 16, [28], ..., 98}
MATHEMATICA
nn = 75; c[_] := False;
rad[x_] := rad[x] = Times @@ FactorInteger[x][[All, 1]];
f[x_] := Select[Range[x], Divisible[x, rad[#]] &];
Array[Set[{a[#], c[#]}, {#, True}] &, 2]; s = a[1] + a[2];
Reap[Do[r = f[s]; k = SelectFirst[r, ! c[#] &];
Sow[FirstPosition[r, k][[1]]]; c[k] = True;
s += k, {i, 3, nn}] ][[-1, 1]]
Number of numbers <= n whose set of prime factors is a subset of the set of prime factors of n.
+10
68
1, 2, 2, 3, 2, 5, 2, 4, 3, 6, 2, 8, 2, 6, 5, 5, 2, 10, 2, 8, 5, 7, 2, 11, 3, 7, 4, 8, 2, 18, 2, 6, 6, 8, 5, 14, 2, 8, 6, 11, 2, 19, 2, 9, 8, 8, 2, 15, 3, 12, 6, 9, 2, 16, 5, 11, 6, 8, 2, 26, 2, 8, 8, 7, 5, 22, 2, 10, 6, 20, 2, 18, 2, 9, 9, 10, 5, 23, 2, 14, 5, 9, 2, 28, 5, 9, 7, 11, 2, 32, 5, 10
COMMENTS
This function of n appears in an ABC-conjecture by Andrew Granville. See Goldfeld. - T. D. Noe, Jun 30 2009
FORMULA
a(n) = Sum_{j = 1..n} Product_{primes p | j} delta(n mod p,0) where delta is the Kronecker delta. - Robert Israel, Feb 09 2015 a(n) = Sum_{1<=k<=n,(n,k)=1} mu(k)*floor(n/k). - Benoit Cloitre, May 07 2016 a(n) = Sum_{k=1..n} floor(n^k/k)-floor((n^k -1)/k). - Anthony Browne, May 28 2016
EXAMPLE
a(1) = 1 because the empty set is a subset of any set.
a(6) = 5 from the five numbers: 1 with the empty set, 2 with the set {2}, 3 with {3}, 4 with {2} and 6 with {2,3}, which are all subsets of {2,3}. 5 is out because {5} is not a subset of {2,3}. (End)
Let p# be the product of primes up to p, A002110. Then, a(13#) = 1161
a(17#) = 4843
a(19#) = 19985
a(23#) = 83074
a(29#) = 349670
a(31#) = 1456458
a(37#) = 6107257
a(41#) = 25547835
(End)
MAPLE
A:= proc(n) local F, S, s, j, p;
F:= numtheory:-factorset(n);
S:= {1};
for p in F do
S:= {seq(seq(s*p^j, j=0..floor(log[p](n/s))), s=S)}
od;
nops(S)
end proc;
MATHEMATICA
pf[n_] := If[n==1, {}, Transpose[FactorInteger[n]][[1]]]; SubsetQ[lst1_, lst2_] := Intersection[lst1, lst2]==lst1; Table[pfn=pf[n]; Length[Select[Range[n], SubsetQ[pf[ # ], pfn] &]], {n, 100}] (* T. D. Noe, Jun 30 2009 *) Table[Total[MoebiusMu[#] Floor[n/#] &@ Select[Range@ n, CoprimeQ[#, n] &]], {n, 92}] (* Michael De Vlieger, May 08 2016 *)
PROG
(PARI) a(n, f=factor(n)[, 1])=if(#f>1, my(v=f[1..#f-1], p=f[#f], s); while(n>0, s+=a(n, v); n\=p); s, if(#f&&n>0, log(n+.5)\log(f[1])+1, n>0)) \\ Charles R Greathouse IV, Jun 27 2013 (PARI) a(n) = sum(k=1, n, if(gcd(n, k)-1, 0, moebius(k)*(n\k))) \\ Benoit Cloitre, May 07 2016 (PARI) a(n, f=factor(n)[, 1])=if(#f<2, return(if(#f, valuation(n, f[1])+1, 0))); my(v=f[1..#f-1], p=f[#f], s); while(n, s+=a(n, v); n\=p); s \\ Charles R Greathouse IV, Nov 03 2021 (Python)
def A010846(n): return sum((m:=n**k)//k-(m-1)//k for k in range(1, n+1)) # Chai Wah Wu, Aug 15 2024
Number of k < n such that rad(k) | n but k does not divide n, where rad = A007947.
+10
33
0, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 2, 0, 2, 1, 0, 0, 4, 0, 2, 1, 3, 0, 3, 0, 3, 0, 2, 0, 10, 0, 0, 2, 4, 1, 5, 0, 4, 2, 3, 0, 11, 0, 3, 2, 4, 0, 5, 0, 6, 2, 3, 0, 8, 1, 3, 2, 4, 0, 14, 0, 4, 2, 0, 1, 14, 0, 4, 2, 12, 0, 6, 0, 5, 3, 4, 1, 15, 0, 4, 0, 5, 0, 16, 1, 5, 3, 3, 0, 20, 1, 4, 3, 5, 1, 8, 0, 7, 2, 6
COMMENTS
Former name: number of "semidivisors" of n, numbers m < n that do not divide n but divide n^e for some integer e > 1. See ACM Inroads paper.
FORMULA
a(n) = number of terms in row n of A272618. a(n) = sum of row n of A304570. (End)
EXAMPLE
a(10) = 2 since S(10) \ D(10) = {1, 2, 4, 5, 8, 10} \ {1, 2, 5, 10} = {4, 8}.a(16) = 0 since S(16) \ D(16) = {1, 2, 4, 8, 16} \ {1, 2, 4, 8, 16} is null, etc.Table of a(n) and S(n) \ D(n):
---------------------------
6 1 {4}
10 2 {4, 8}
12 2 {8, 9}
14 2 {4, 8}
15 1 {9}
18 4 {4, 8, 12*, 16}
20 2 {8, 16}
21 1 {9}
22 3 {4, 8, 16}
24 3 {9, 16, 18*}
26 3 {4, 8, 16}
28 2 {8, 16}
30 10 {4, 8, 9, 12, 16, 18, 20, 24, 25, 27}
Terms in A272618 marked with an asterisk are counted by A355432. All other terms are counted by A361235. (End)
MATHEMATICA
Table[Count[Range[n], _?(And[Divisible[n, Times @@ FactorInteger[#][[All, 1]]], ! Divisible[n, #]] &)], {n, 120}] (* Michael De Vlieger, Aug 11 2024 *)
CROSSREFS
Cf. A000005, A000961, A024619, A027750, A010846, A045763, A162306, A243823, A272618, A304570, A355432, A361235.
Irregular array read by rows: n-th row contains (in ascending order) the nondivisors 1 <= k < n such that all the prime divisors p of k also divide n.
+10
29
0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 8, 0, 8, 9, 0, 4, 8, 9, 0, 0, 4, 8, 12, 16, 0, 8, 16, 9, 4, 8, 16, 0, 9, 16, 18, 0, 4, 8, 16, 0, 8, 16, 0, 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 0, 0, 9, 27, 4, 8, 16, 32, 25, 8, 16, 24, 27, 32, 0, 4, 8, 16, 32, 9, 27, 16, 25, 32, 0, 4, 8, 9, 12, 16, 18, 24, 27, 28, 32
COMMENTS
The k are the "semidivisors" or nondivisor regular numbers of n as counted by A243822(n). All nonzero terms k are composite and pertain to composite rows n. This is because prime k must either divide or be coprime to n, and k = 1 is both a divisor of and coprime to n.
Row n for prime p contains zero, since numbers 1 <= k < p must either divide or be coprime to prime p.
Row n for prime powers p^e contains zero, since there is only one prime divisor p of p^e and every power 1 <= m <= e of p divides p^e.
Row n = 4 is a special case of composite n that contains zero. This is because 4 is the smallest composite number; there are no composites k < n.
Thus rows n for composite n > 4 contain at least 1 nonzero value.
In base n, 1/a(n) has a terminating expansion with at least 2 places.
REFERENCES
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, pp. 144-145, Theorem 136.
EXAMPLE
For n = 12, the numbers 1 <= k < n such that the prime divisors p of k also divide n are {2, 3, 4, 6, 8, 9}; {2, 3, 4, 6} divide n = 12, thus row n = 12 is {8, 9}.
n: k
1: 0
2: 0
3: 0
4: 0
5: 0
6: 4
7: 0
8: 0
9: 0
10: 4 8
11: 0
12: 8 9
13: 0
14: 4 8
15: 9
16: 0
17: 0
18: 4 8 12 16
19: 0
20: 8 16
MATHEMATICA
Table[With[{r = First /@ FactorInteger@ n}, Select[Range@ n,
And[SubsetQ[r, Map[First, FactorInteger@ #]], ! Divisible[n, #]] &]], {n, 30}] /. {} -> 0 // Flatten (* Michael De Vlieger, May 03 2016 *)
a(n) = number of k < n such that rad(k) = rad(n) and k does not divide n, where rad(k) = A007947(k).
+10
16
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1
COMMENTS
a(n) = 0 for prime powers and squarefree numbers.
LINKS
Michael De Vlieger, Plot (k, n) at (x, -y), k = 1..n, n = 1..54, showing k in A126706 in dark blue, n in A360768 in dark red, and for n and nondivisor k such that rad(k) = rad(n), we highlight in large black dots. This sequence counts the number of black dots in row n.
EXAMPLE
a(1) = 18, since 18/6 >= 3. We note that rad(12) = rad(18) = 6, yet 12 does not divide 18.
a(2) = 24, since 24/6 >= 3. rad(18) = rad(24) = 6 and 24 mod 18 = 6.
a(3) = 36, since 36/6 >= 3. rad(24) = rad(36) = 6 and 36 mod 24 = 12.
a(6) = 54, since 54/6 >= 3. m in {12, 24, 36, 48} are such that rad(m) = rad(54) = 6, but none divides 54, etc.
MATHEMATICA
rad[n_] := rad[n] = Times @@ FactorInteger[n][[All, 1]]; Table[Which[PrimePowerQ[n], 0, SquareFreeQ[n], 0, True, r = rad[n]; Count[Select[Range[n], Nor[PrimePowerQ[#], SquareFreeQ[#]] &], _?(And[rad[#] == r, Mod[n, #] != 0] &)]], {n, 120}]
PROG
(PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947a(n) = my(rn=rad(n)); sum(k=1, n-1, if (n % k, rad(k)==rn)); \\ Michel Marcus, Feb 23 2023
CROSSREFS
Cf. A005361, A007947, A008479, A010846, A013929, A020639, A024619, A027750, A126706, A162306, A243822, A272618, A360589, A360768.
Largest integer k < n such that any prime factor of k is also a prime factor of n.
+10
13
1, 1, 2, 1, 4, 1, 4, 3, 8, 1, 9, 1, 8, 9, 8, 1, 16, 1, 16, 9, 16, 1, 18, 5, 16, 9, 16, 1, 27, 1, 16, 27, 32, 25, 32, 1, 32, 27, 32, 1, 36, 1, 32, 27, 32, 1, 36, 7, 40, 27, 32, 1, 48, 25, 49, 27, 32, 1, 54, 1, 32, 49, 32, 25, 64, 1, 64, 27, 64, 1, 64, 1, 64, 45, 64, 49, 72, 1, 64, 27
COMMENTS
The function a(n) complements Euler's phi-function: 1) a(n)+phi(n) = n if n is a power of a prime (actually, in A285710). 2) It seems also that a(n)+phi(n) >= n for "almost all numbers" (see A285709, A208815). 3) a(2n) = n+1 if and only if n is a Mersenne prime. 4) Lim a(n^k)/n^k =1 if n has at least two prime factors and k goes to infinity. In other words, largest integer k < n such that k | n^e with integer e >= 0.
Penultimate term of row n in A162306. (The last term of row n in A162306 is n.) For prime p, a(p) = 1. More generally, for n with omega(n) = 1, that is, a prime power p^e with e > 0, a(p^e) = p^(e - 1).
For n with omega(n) > 1, a(n) does not divide n. If n = pq with q = p + 2, then p^2 < n though p^2 does not divide n, yet p^2 | n^e with e > 1. If n has more than 2 distinct prime divisors p, powers p^m of these divisors will appear in the range (1..n-1) such that p^m > n/lpf(n) (lpf(n) = A020639(n)). Since a(n) is the largest of these, a(n) is not a divisor of n. If a(n) does not divide n, then a(n) appears last in row n of A272618. (End)
FORMULA
Largest k < n with rad(kn) = rad(n), where rad = A007947.
EXAMPLE
a(10)=8 since 8 is the largest integer< 10 that can be written using only the primes 2 and 5. a(78)=72 since 72 is the largest number less than 78 that can be written using only the primes 2, 3 and 13. (78=2*3*13).
MATHEMATICA
Table[If[n == 2, 1, Module[{k = n - 2, e = Floor@ Log2@ n}, While[PowerMod[n, e, k] != 0, k--]; k]], {n, 2, 81}] (* Michael De Vlieger, Apr 26 2017 *)
PROG
(PARI) a(n) = {forstep(k = n - 1, 2, -1, f = factor(k); okk = 1; for (i=1, #f~, if ((n % f[i, 1]) != 0, okk = 0; break; )); if (okk, return (k)); ); return (1); } \\ Michel Marcus, Jun 11 2013 (PARI)
(Python)
from sympy import divisors
from sympy.ntheory.factor_ import core
def a007947(n): return max(d for d in divisors(n) if core(d) == d)
def a(n):
k=n - 1
while True:
if a007947(k*n) == a007947(n): return k
else: k-=1
CROSSREFS
Cf. A000010, A007947, A051953, A162306, A208815, A272618, A285328, A285699, A285707, A285709, A285710, A285711.
AUTHOR
Istvan Beck (istbe(AT)online.no), Feb 07 2003
Array T(n,k) read by antidiagonals (downward): T(1,k) = A005117(k+1) (squarefree numbers > 1); for n > 1, columns are nonsquarefree numbers (in ascending order) with exactly the same prime factors as T(1,k).
+10
11
2, 3, 4, 5, 9, 8, 6, 25, 27, 16, 7, 12, 125, 81, 32, 10, 49, 18, 625, 243, 64, 11, 20, 343, 24, 3125, 729, 128, 13, 121, 40, 2401, 36, 15625, 2187, 256, 14, 169, 1331, 50, 16807, 48, 78125, 6561, 512, 15, 28, 2197, 14641, 80, 117649, 54, 390625, 19683, 1024
COMMENTS
A permutation of the natural numbers > 1.
T(1,k)= A005117(m) with m > 1; terms in column k = T(1,k) * A162306(T(1,k)) only not bounded by T(1,k). Let T(1,k) = b. Terms in column k are multiples of b and numbers c such that c | b^e with e >= 0. Alternatively, terms in column k are multiples bc with c those numbers whose prime divisors p also divide b. - Michael De Vlieger, Mar 25 2017
EXAMPLE
Array starts:
2 3 5 6 7 10 11 13 14 15
4 9 25 12 49 20 121 169 28 45
8 27 125 18 343 40 1331 2197 56 75
16 81 625 24 2401 50 14641 371293 98 135
32 243 3125 36 16807 80 161051 4826809 112 225
64 729 15625 48 117649 100 1771561 62748517 196 375
128 2187 78125 54 823543 160 19487171 815730721 224 405
Column 6 is: T(1,6) = 2*5; T(2,6) = 2^2*5; T(3,6) = 2^3*5; T(4,6) = 2*5^2; T(5,6) = 2^4*5, etc.
MATHEMATICA
f[n_, k_: 1] := Block[{c = 0, sgn = Sign[k], sf}, sf = n + sgn; While[c < Abs[k], While[! SquareFreeQ@ sf, If[sgn < 0, sf--, sf++]]; If[sgn < 0, sf--, sf++]; c++]; sf + If[sgn < 0, 1, -1]] (* after Robert G. Wilson v at A005117 *); T[n_, k_] := T[n, k] = Which[And[n == 1, k == 1], 2, k == 1, f@ T[n - 1, k], PrimeQ@ T[n, 1], T[n, 1]^k, True, Module[{j = T[n, k - 1]/T[n, 1] + 1}, While[PowerMod[T[n, 1], j, j] != 0, j++]; j T[n, 1]]]; Table[T[n - k + 1, k], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Mar 25 2017 *)
PROG
(Scheme)
(define (A284311bi row col) (if (= 1 row) ( A005117 (+ 1 col)) ( A065642 (A284311bi (- row 1) col))))
CROSSREFS
Cf. A284457 (this sequence read by antidiagonals upwards), A285321 (a similar array, but columns come in different order). Cf. A008479 (index of the row where n is located), A285329 (of the column).
Product of numbers m with 2 <= m <= n whose prime divisors all divide n.
+10
10
1, 2, 3, 8, 5, 144, 7, 64, 27, 3200, 11, 124416, 13, 6272, 2025, 1024, 17, 35831808, 19, 1024000, 3969, 247808, 23, 859963392, 125, 346112, 729, 2809856, 29, 261213880320000000, 31, 32768, 264627, 18939904, 30625, 26748301344768, 37, 23658496, 369603, 32768000000, 41
COMMENTS
This sequence is the product of n-regular numbers.
A number m is said to be "regular" to n or "n-regular" if all the prime factors p of m also divide n.
The divisor is a special case of a regular m such that m also divides n in addition to all of its prime factors p | n.
Analogous to A007955 (Product of divisors of n). If n is 1 or prime, a(n) = n.
If n is a prime power, a(n) = A007955(n). Note: b-file ends at n = 4619, because a(4620) has more than 1000 decimal digits.
Product of the numbers 1 <= k <= n such that (floor(n^k/k) - floor((n^k - 1)/k)) = 1. - Michael De Vlieger, May 26 2016
FORMULA
a(n) = product of terms of n-th row of irregular triangle A162306(n,k). a(n) = Product_{k=1..n} k^( floor(n^k/k)-floor((n^k -1)/k) ). - Anthony Browne, Jul 06 2016 a(n) = Product_{k=2..n, A123275(n,k)=1} k. (End)
EXAMPLE
a(12) = 124416 since 1 * 2 * 3 * 4 * 6 * 8 * 9 * 12 = 124416. These numbers are products of prime factors that are the distinct prime divisors of 12 = {2, 3}.
Let p# be the product of primes up to p, A002110. Then a(13#) ~= 8.3069582 * 10 ^ 4133
a(17#) ~= 1.3953000 * 10 ^ 22689
a(19#) ~= 3.8258936 * 10 ^ 117373
a(23#) ~= 6.7960327 * 10 ^ 594048
a(29#) ~= 1.3276817 * 10 ^ 2983168
a(31#) ~= 2.8152792 * 10 ^ 14493041
a(37#) ~= 1.9753840 * 10 ^ 69927040
Up to n = 11# already in the table.
(End)
MAPLE
A:= proc(n) local F, S, s, j, p;
F:= numtheory:-factorset(n);
S:= {1};
for p in F do
S:= {seq(seq(s*p^j, j=0..floor(log[p](n/s))), s=S)}
od;
convert(S, `*`)
end proc:
MATHEMATICA
regularQ[m_Integer, n_Integer] := Module[{omega = First /@ FactorInteger @ m }, If[Length[Select[omega, Divisible[n, #] &]] == Length[omega], True, False]]; a20140819[n_Integer] := Times @@ Flatten[Position[Thread[regularQ[Range[1, n], n]], True]]; a20140819 /@ Range[41]
regulars[n_] := Block[{f, a}, f[x_] := First /@ FactorInteger@ x; a = f[n]; {1}~Join~Select[Range@ n, SubsetQ[a, f@ #] &]]; Array[Times @@ regulars@ # &, 12] (* Michael De Vlieger, Feb 09 2015 *) Table[Times @@ Select[Range@ n, (Floor[n^#/#] - Floor[(n^# - 1)/#]) == 1 &], {n, 41}] (* Michael De Vlieger, May 26 2016 *)
PROG
(PARI) lista(nn) = {vf = vector(nn, n, Set(factor(n)[, 1])); vector(nn, n, prod(i=1, n, if (setintersect(vf[i], vf[n]) == vf[i], i, 1))); } \\ Michel Marcus, Aug 23 2014 (PARI) for(n=1, 100, print1(prod(k=1, n, k^(floor(n^k/k) - floor((n^k - 1)/k))), ", ")) \\ Indranil Ghosh, Mar 22 2017 (Python)
from sympy import primefactors
y, pf = 1, set(primefactors(n))
for m in range(2, n+1):
if set(primefactors(m)) <= pf:
y *= m
(Scheme)
;; A naive implementation, code for A123275bi given under A123275: (define ( A243103 n) (let loop ((k n) (m 1)) (cond ((= 1 k) m) ((= 1 (A123275bi n k)) (loop (- k 1) (* m k))) (else (loop (- k 1) m)))))
CROSSREFS
Cf. A162306 (irregular triangle of regular numbers of n), A010846 (number of regular numbers of n), A244974 (sum of regular numbers of n), A007955, A244052 (record transform of regular numbers of n).
Irregular triangle where row n lists m such that rad(m) | n and bigomega(m) <= bigomega(n), where rad = A007947 and bigomega = A001222.
+10
10
1, 1, 2, 1, 3, 1, 2, 4, 1, 5, 1, 2, 3, 4, 6, 9, 1, 7, 1, 2, 4, 8, 1, 3, 9, 1, 2, 4, 5, 10, 25, 1, 11, 1, 2, 3, 4, 6, 8, 9, 12, 18, 27, 1, 13, 1, 2, 4, 7, 14, 49, 1, 3, 5, 9, 15, 25, 1, 2, 4, 8, 16, 1, 17, 1, 2, 3, 4, 6, 8, 9, 12, 18, 27, 1, 19, 1, 2, 4, 5, 8, 10, 20, 25, 50, 125
COMMENTS
Analogous to A162306 regarding m such that rad(m) | n, but instead of taking m <= n, we take m such that bigomega(m) <= bigomega(n). Row n is a finite set of products of prime power factors p^k (i.e., p^k | n) such that Sum_{p|n} k <= bigomega(n).
For prime power n = p^k, k >= 0 (i.e., n in A000961), row p^k of this sequence is the same as row p^k of A027750 and A162306. Therefore, for prime p, row p of this sequence is the same as row p of A027750 and A162306: {1, p}. For n in A024619, row n of this sequence does not match row n of A162306, since the former contains gpf(n)^bigomega(n) = A006530(n)^ A001222(n), which is larger than n.
FORMULA
Row n of this sequence is { m : rad(m) | n, bigomega(m) <= bigomega(n) }.
A376567(n) = binomial(bigomega(n) + omega(n)) = Length of row n, where omega = A001221.
EXAMPLE
Triangle begins:
n row n of this sequence:
-------------------------------------------
1: 1;
2: 1, 2;
3: 1, 3;
4: 1, 2 4;
5: 1, 5;
6: 1, 2, 3, 4, 6, 9;
7: 1, 7;
8: 1, 2, 4, 8;
9: 1, 3, 9;
10: 1, 2, 4, 5, 10, 25;
11: 1, 11;
12: 1, 2, 3, 4, 6, 8, 9, 12, 18, 27;
...
Row n = 10 of this sequence, presented according to 2^k, k = 0..bigomega(n) by columns, 5^i, i = 0..bigomega(n) by rows, showing terms m > n with an asterisk. The remaining m and the parenthetic 8 are in row 10 of A162306: 1 2 4 (8)
5 10
25*
Row n = 12 of this sequence, presented according to 2^k, k = 0..bigomega(n) by columns, 3^i, i = 0..bigomega(n) by rows, showing terms m > n with an asterisk. The remaining m are in row 12 of A162306: 1 2 4 8
3 6 12
9 18*
27*
MATHEMATICA
Table[Clear[p]; MapIndexed[Set[p[First[#2]], #1] &, FactorInteger[n][[All, 1]]]; k = PrimeOmega[n]; w = PrimeNu[n]; Union@ Map[Times @@ MapIndexed[p[First[#2]]^#1 &, #] &, Select[Tuples[Range[0, k], w], Total[#] <= k &] ], {n, 120}]
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