| Displaying 1-9 of 9 results found.
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1, 4, 9, 12, 25, 36, 49, 48, 72, 100, 121, 108, 169, 196, 225, 192, 289, 288, 361, 300, 441, 484, 529, 432, 600, 676, 648, 588, 841, 900, 961, 768, 1089, 1156, 1225, 864, 1369, 1444, 1521, 1200, 1681, 1764, 1849, 1452, 1800, 2116, 2209, 1728, 2352, 2400
COMMENTS
The Dirichlet convolution of a(n) and sigma(n) is sigma(n^2).
FORMULA
a(n) = n^2 * Sum_{d^2 | n} (moebius(d) / d^2).
Multiplicative with a(p) = p^2; a(p^e) = p^(2e) - p^(2e-2), for e > 1.
Dirichlet g.f.: zeta(s-2) / zeta(2s-2).
Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + 1/p^2 + 1/(p^2 - 1)^2) = 1.681923034881403168503816690236967736500606659628336043348190538886262268... - Vaclav Kotesovec, Sep 20 2020
PROG
(PARI) a(n) = n^2*sumdiv(n, d, if (issquare(d), moebius(sqrtint(d))/d)); \\ Michel Marcus, Feb 10 2015
For every positive integer m, let u(m) = (d(1),d(2),...,d(k)) be the unitary divisors of m. The sequence (a(n)) consists of successive numbers m which d(k)/d(1) + d(k-1)/d(2) + ... + d(k)/d(1) is an integer.
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1, 10, 65, 130, 260, 340, 1105, 1972, 2210, 4420, 8840, 9860, 15650, 20737, 32045, 41474, 44200, 51272, 55250, 64090, 75140, 82948, 103685, 128180, 207370, 207553, 221000, 256360, 352529, 414740, 415106, 512720, 532100, 705058, 759025, 813800, 829480, 830212
COMMENTS
Also numbers m such that the sum of the squares of the unitary divisors of m is divisible by m (the unitary version of A046762). - Amiram Eldar, Jun 16 2018
EXAMPLE
The first 10 sums: 1, 5/2, 10/3, 17/4, 26/5, 25/3, 50/7, 65/8, 82/9, 13, so that a(1) = 1 and a(10) = 13.
MATHEMATICA
z = 1000; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &]; k[n_] := Length[r[n]];
t[n_] := Table[r[n][[k[n] + 1 - i]]/r[n][[k[1] + i - 1]], {i, 1, k[n]}];
s = Table[Plus @@ t[n], {n, 1, z}]; a[n_] := If[IntegerQ[s[[n]]], 1, 0]; u = Table[a[n], {n, 1, z}]; Flatten[Position[u, 1]] (* A229996 *) usigma2[n_] := If[n == 1, 1, Times @@ (1 + Power @@@ FactorInteger[n]^2)]; seqQ[n_] := Divisible[usigma2[n], n]; Select[Range[10^6], seqQ] (* Amiram Eldar, Jun 16 2018 *)
PROG
(PARI) is(n) = {my(f = factor(n)); !(prod(i = 1, #f~, f[i, 1]^(2*f[i, 2]) + 1) % n); } \\ Amiram Eldar, Jun 16 2024
Square array A(n,k), n>=0, k>=1, read by antidiagonals, where row n is the sum of n-th powers of unitary divisors of k (divisors d such that gcd(d, k/d) = 1).
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1, 2, 1, 2, 3, 1, 2, 4, 5, 1, 2, 5, 10, 9, 1, 4, 6, 17, 28, 17, 1, 2, 12, 26, 65, 82, 33, 1, 2, 8, 50, 126, 257, 244, 65, 1, 2, 9, 50, 252, 626, 1025, 730, 129, 1, 4, 10, 65, 344, 1394, 3126, 4097, 2188, 257, 1, 2, 18, 82, 513, 2402, 8052, 15626, 16385, 6562, 513, 1, 4, 12, 130, 730, 4097, 16808, 47450, 78126, 65537, 19684, 1025, 1
COMMENTS
For row r > 0, Sum_{k=1..n} A(r,k) ~ zeta(r+1) * n^(r+1) / ((r+1) * zeta(r+2)). - Vaclav Kotesovec, May 20 2021
FORMULA
Dirichlet g.f. of row n: zeta(s)*zeta(s-n)/zeta(2*s-n).
EXAMPLE
Square array begins:
1, 2, 2, 2, 2, 4, ...
1, 3, 4, 5, 6, 12, ...
1, 5, 10, 17, 26, 50, ...
1, 9, 28, 65, 126, 252, ...
1, 17, 82, 257, 626, 1394, ...
1, 33, 244, 1025, 3126, 8052, ...
The sum of unitary divisors of the exponentially odd numbers ( A268335).
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1, 3, 4, 6, 12, 8, 9, 18, 12, 14, 24, 24, 18, 20, 32, 36, 24, 36, 42, 28, 30, 72, 32, 33, 48, 54, 48, 38, 60, 56, 54, 42, 96, 44, 72, 48, 72, 54, 84, 72, 72, 80, 90, 60, 62, 96, 84, 144, 68, 96, 144, 72, 74, 114, 96, 168, 80, 126, 84, 108, 132, 120, 108, 90, 112
FORMULA
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = (zeta(4)/d^2) * Product_{p prime} (1 - 2/p^4 + 1/p^5) = 1.92835521961603199612..., d = A065463 is the asymptotic density of the exponentially odd numbers. The asymptotic mean of the unitary abundancy index of the exponentially odd numbers: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/ A268335(k) = c * d = 1.35841479521454692063... .
MATHEMATICA
s[n_] := Module[{f = FactorInteger[n], e}, e = f[[;; , 2]]; If[AllTrue[e, OddQ], Times @@ (1 + Power @@@ f), Nothing]]; s[1] = 1; Array[s, 100]
PROG
(PARI) lista(max) = for(k = 1, max, my(f = factor(k), e = f[, 2], isexpodd = 1); for(i = 1, #e, if(!(e[i] % 2), isexpodd = 0; break)); if(isexpodd, print1(prod(i = 1, #e, 1 + f[i, 1]^e[i]), ", ")));
The sum of unitary divisors of the cubefree numbers ( A004709).
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1, 3, 4, 5, 6, 12, 8, 10, 18, 12, 20, 14, 24, 24, 18, 30, 20, 30, 32, 36, 24, 26, 42, 40, 30, 72, 32, 48, 54, 48, 50, 38, 60, 56, 42, 96, 44, 60, 60, 72, 48, 50, 78, 72, 70, 54, 72, 80, 90, 60, 120, 62, 96, 80, 84, 144, 68, 90, 96, 144, 72, 74, 114, 104, 100, 96
FORMULA
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(3)^2 * Product_{p prime} (1 + 1/p^2 - 2/p^3 + 1/p^4 - 1/p^5) = 1.665430860774244601005... .
The asymptotic mean of the unitary abundancy index of the cubefree numbers: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/ A004709(k) = c / zeta(3) = 1.38548421160152785073... .
MATHEMATICA
s[n_] := Module[{f = FactorInteger[n], e}, e = f[[;; , 2]]; If[AllTrue[e, # < 3 &], Times @@ (1 + Power @@@ f), Nothing]]; s[1] = 1; Array[s, 100]
PROG
(PARI) lista(max) = for(k = 1, max, my(f = factor(k), e = f[, 2], iscubefree = 1); for(i = 1, #e, if(e[i] > 2, iscubefree = 0; break)); if(iscubefree, print1(prod(i = 1, #e, 1 + f[i, 1]^e[i]), ", ")));
(Python)
from sympy.ntheory.factor_ import udivisor_sigma
from sympy import mobius, integer_nthroot
def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x, 3)[0]+1))
m, k = n, f(n)
while m != k:
m, k = k, f(k)
The sum of unitary divisors of the exponentially 2^n-numbers ( A138302).
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1, 3, 4, 5, 6, 12, 8, 10, 18, 12, 20, 14, 24, 24, 17, 18, 30, 20, 30, 32, 36, 24, 26, 42, 40, 30, 72, 32, 48, 54, 48, 50, 38, 60, 56, 42, 96, 44, 60, 60, 72, 48, 68, 50, 78, 72, 70, 54, 72, 80, 90, 60, 120, 62, 96, 80, 84, 144, 68, 90, 96, 144, 72, 74, 114, 104
COMMENTS
Also the sum of infinitary divisors of the terms of A138302, since A138302 is also the sequence of numbers whose sets of unitary divisors ( A077610) and infinitary divisors ( A077609) coincide.
FORMULA
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = (1/d^2) * Product_{p prime} f(1/p) = 1.58107339851877782285..., d = A271727 is the asymptotic density of A138302, and f(x) = 1 + x^2 + 2 * Sum_{k>=2} (x^(2^k)-x^(2^k+1)). The asymptotic mean of the unitary abundancy index of A138302: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/ A138302(k) = c * d = 1.37948208055913856387... .
MATHEMATICA
s[n_] := Module[{f = FactorInteger[n], e}, e = f[[;; , 2]]; If[AllTrue[e, # == 2^IntegerExponent[#, 2] &], Times @@ (1 + Power @@@ f), Nothing]]; s[1] = 1; Array[s, 100]
PROG
(PARI) lista(max) = for(k = 1, max, my(f = factor(k), e = f[, 2], is = 1); for(i = 1, #e, if(e[i] >> valuation(e[i], 2) > 1, is = 0; break)); if(is, print1(prod(i = 1, #e, 1 + f[i, 1]^e[i]), ", ")));
a(n) is the sum of the squares of the unitary divisors of n that are exponentially odd numbers ( A268335).
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1, 5, 10, 1, 26, 50, 50, 65, 1, 130, 122, 10, 170, 250, 260, 1, 290, 5, 362, 26, 500, 610, 530, 650, 1, 850, 730, 50, 842, 1300, 962, 1025, 1220, 1450, 1300, 1, 1370, 1810, 1700, 1690, 1682, 2500, 1850, 122, 26, 2650, 2210, 10, 1, 5, 2900, 170, 2810, 3650, 3172
COMMENTS
The number of unitary divisors of n that are exponentially odd is A055076(n) and their sum is A358346(n).
FORMULA
a(n) >= 1 with equality if and only if n is a square ( A000290). a(n) <= A374537(n) with equality if and only if n is squarefree ( A005117). Multiplicative with a(p^e) = p^(2*e) + 1 if e is odd, and 1 otherwise.
Dirichlet g.f.: zeta(s) * zeta(2*s-4) * Product_{p prime} (1 + 1/p^(s-2) - 1/p^(2*s-4) - 1/p^(2*s-2)).
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = zeta(2) * zeta(3) * Product_{p prime} (1 - 2/p^2 + 1/p^3 - 1/p^4 + 1/p^5) = 0.79482441214759383925... .
MATHEMATICA
f[p_, e_] := 1 + If[OddQ[e], p^(2*e), 0]; a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100]
PROG
(PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, 1 + if(f[i, 2]%2, f[i, 1]^(2*f[i, 2]), 0)); }
1, 1, 2, 1, 0, 3, 1, 2, 0, 3, 1, 0, 0, 0, 5, 1, 2, 3, 0, 0, 6, 1, 0, 0, 0, 0, 0, 7, 1, 2, 0, 3, 0, 0, 0, 6, 1, 0, 3, 0, 0, 0, 0, 0, 8, 1, 2, 0, 0, 5, 0, 0, 0, 0, 10, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 1, 2, 3, 3, 0, 6, 0, 0, 0, 0, 0, 9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 1, 2, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 14
COMMENTS
Right border = A063659, (1, 2, 3, 3, 5, 6, 7, 6, 8, 10, ...), the Moebius transform of A001615: (1, 3, 4, 6, 6, 12, 8, 12, 12, ...). A130106 * (1, 2, 3, ...) = A034676: (1, 5, 10, 17, 26, 50, 50, ...).
A034676^(-1) * (1,2,3,...) = 1/1, 1/2, 2/3, 2/3, 4/5, 2/6, 6/7, 4/6, 6/8, 4/10, ...; where the numerators = phi(n), A000010: (1, 1, 2, 2, 4, 2, 6, 4, ...); and the denominators = A063659, the right border of the triangle: (1, 2, 3, 3, 5, 6, 7, 8, 10, ...).
FORMULA
Inverse Moebius transform of an infinite lower triangular matrix with A063659, (1, 2, 3, 3, 5, 6, 7, 6, 8, 10, ...) in the main diagonal and the rest zeros.
EXAMPLE
First few rows of the triangle:
1;
1, 2;
1, 0, 3;
1, 2, 0, 3;
1, 0, 0, 0, 5;
1, 2, 3, 0, 0, 6;
1, 0, 0, 0, 0, 0, 7,
1, 2, 0, 3, 0, 0, 0, 6;
1, 0, 3, 0, 0, 0, 0, 0, 8;
...
MATHEMATICA
m = 14;
A051731 = Table[If[Mod[n, k] == 0, 1, 0], {n, m}, {k, m}];
A063659 = Table[Sum[MoebiusMu[GCD[n, k]]^2, {k, n}], {n, m}] // DiagonalMatrix;
The sum of the squares of the infinitary divisors of n.
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1
1, 5, 10, 17, 26, 50, 50, 85, 82, 130, 122, 170, 170, 250, 260, 257, 290, 410, 362, 442, 500, 610, 530, 850, 626, 850, 820, 850, 842, 1300, 962, 1285, 1220, 1450, 1300, 1394, 1370, 1810, 1700, 2210, 1682, 2500, 1850, 2074, 2132, 2650, 2210, 2570, 2402, 3130, 2900
COMMENTS
Also the sum of the infinitary divisors of n^2.
FORMULA
Multiplicative with a(p^e) = Product{k>=1, e_k=1} (p^(2^(k+1)) + 1), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = Product_{P} (1 + 1/(P^2*(P+1))) = 1.14142906130350119631..., and P are numbers of the form p^(2^k) where p is prime and k >= 0 ( A050376).
MATHEMATICA
f[p_, e_] := p^(2^Position[Reverse@IntegerDigits[e, 2], _?(# == 1 &)]); a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]
PROG
(PARI) a(n) = {my(f = factor(n), b); prod(i = 1, #f~, b = binary(2 * f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1))); }
(Python)
from math import prod
from sympy import factorint
def A374539(n): return prod(p**(1<<i)+1 for p, e in factorint(n).items() for i, j in enumerate(bin(e)[-1:1:-1], 1) if j=='1') # Chai Wah Wu, Jul 11 2024
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