A175647 -id:A175647 - cp's OEIS frontend

A369564 Powerful numbers whose prime factors are all of the form 4*k + 3.

1, 9, 27, 49, 81, 121, 243, 343, 361, 441, 529, 729, 961, 1089, 1323, 1331, 1849, 2187, 2209, 2401, 3087, 3249, 3267, 3481, 3969, 4489, 4761, 5041, 5929, 6241, 6561, 6859, 6889, 8649, 9261, 9747, 9801, 10609, 11449, 11907, 11979, 12167, 14283, 14641, 16129, 16641

Offset: 1

Amiram Eldar, Jan 26 2024

nonn

Closed under multiplication.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for sequences related to powerful numbers.

Intersection of A001694 and A004614.
Similar sequence: A352492, A369563, A369565, A369566.
Cf. A002145, A013661, A175647, A334426.

q[n_] := n == 1 || AllTrue[FactorInteger[n], Mod[First[#], 4] == 3 && Last[#] > 1 &]; Select[Range[20000], q]

is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i, 1]%4 != 3 || f[i, 2] == 1, return(0))); 1;}

Sum_{n>=1} 1/a(n) = Product_{primes p == 3 (mod 4)} (1 + 1/(p*(p-1))) = 3*A013661*A334426/(4*A175647) = 1.2161513254... .

A204617 Multiplicative with a(p^e) = p^(e-1)*H(p). H(2) = 1, H(p) = p - 1 if p == 1 (mod 4) and H(p) = p + 1 if p == 3 (mod 4).

Original entry on oeis.org

1, 1, 4, 2, 4, 4, 8, 4, 12, 4, 12, 8, 12, 8, 16, 8, 16, 12, 20, 8, 32, 12, 24, 16, 20, 12, 36, 16, 28, 16, 32, 16, 48, 16, 32, 24, 36, 20, 48, 16, 40, 32, 44, 24, 48, 24, 48, 32, 56, 20, 64, 24, 52, 36, 48, 32, 80, 28, 60, 32, 60, 32, 96, 32, 48, 48, 68, 32

Offset: 1

José María Grau Ribas, Jan 17 2012

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A000010, A072437.
Cf. A175647, A243381.
Cf. A079458, A062570.

with(numtheory):
a := n->add(jacobi(-1,d)*mobius(d)*n/d, d in divisors(n)):
seq(a(n), n = 1..60); # Peter Bala, Dec 26 2023

ar[p_,s_] := Which[Mod[p,4]==1, p^(s-1)*(p-1), Mod[p,4]==3, p^(s-1)*(p+1), True,p^(s-1)]; arit[1] = 1; arit[n_] := Product[ar[FactorInteger[n][[i,1]], FactorInteger[n][[i,2]]], {i, Length[FactorInteger[n]]}]; Array[arit, 100]

A204617(n) = { my(f=factor(n),p); prod(i=1, #f~, p=f[i, 1]; (p^(f[i, 2]-1)) * if(2==p,1,if(1==(p%4),p-1,p+1))); }; \\ Antti Karttunen, Nov 16 2021

a(n) = phi(n) if n is in A072437.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (3/8) * Product_{primes p == 1 (mod 4)} (1 - 1/p^2) * Product_{primes p == 3 (mod 4)} (1 + 1/p^2) = 3*A243381/(8*A175647) = 0.409404... . - Amiram Eldar, Dec 24 2022
a(n) = n*Product_{primes p, p | n} (1 - A034947(p)/p) = Sum_{d | n} A034947(d)* mobius(d)*n/d. Cf. A000010(n) = Sum_{d | n} mobius(d)*n/d. - Peter Bala, Dec 26 2023
a(n) = A079458(n)/A062570(n). - Ridouane Oudra, Jun 04 2024

A344123 Decimal expansion of Sum_{i > 0} 1/A001481(i)^2.

Original entry on oeis.org

1, 4, 2, 6, 5, 5, 6, 0, 6, 3, 5, 1, 2, 5, 9, 2, 8, 7, 8, 6, 9, 6, 8, 0, 9, 3, 1, 6, 1, 5, 5, 0, 8, 1, 6, 3, 6, 1, 2, 7, 6, 6, 9, 3, 6, 3, 6, 7, 7, 0, 3, 9, 0, 2, 8, 8, 7, 9, 9, 2, 2, 3, 0, 4, 4, 1, 2, 9, 6, 0, 4, 5, 2, 8, 6, 1, 5, 1, 9, 0, 1, 9, 1, 4, 6, 7

Offset: 1

A.H.M. Smeets, May 09 2021

This constant can be considered as an equivalent of zeta(2) (= Pi^2/6 = A013661), where Euler's zeta(2) is over all positive integers, with prime elements in A000040, while this constant is over all positive integers that can be written as the sum of two squares (A001481) with prime elements given in A055025.

Close to the value of e^(3/2)/Pi.

			1.4265560635125928786968093161550816361276693636770...

R. J. Mathar, Table of Dirichlet L-series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015.

Cf. A000040, A001481, A055025, A175647, A243379, A334448.

Cf. A344124, A344125.

Equals Sum_{i > 0} 1/A001481(i)^2.
Equals Product_{i > 0} 1/(1-A055025(i)^-2).
Equals 1/(1-prime(1)^(-2)) * Product_{i>1 and prime(i) == 1 (mod 4)} 1/(1-prime(i)^(-2)) * Product_{i>1 and prime(i) == 3 (mod 4)} 1/(1-prime(i)^(-4)), where prime(n) = A000040(n).
Equals (4/3)/(A243379*A334448).
Equals zeta_{2,0} (2) * zeta_{4,1} (2) * zeta_{4,3} (4), where zeta_{4,1} (2)  = A175647 and zeta_{2,0} (s) = 2^s/(2^s - 1).

A115076 Number of 2 X 2 symmetric matrices over Z(n) having determinant 1.

Original entry on oeis.org

1, 4, 6, 12, 30, 24, 42, 48, 54, 120, 110, 72, 182, 168, 180, 192, 306, 216, 342, 360, 252, 440, 506, 288, 750, 728, 486, 504, 870, 720, 930, 768, 660, 1224, 1260, 648, 1406, 1368, 1092, 1440, 1722, 1008, 1806, 1320, 1620, 2024, 2162, 1152, 2058, 3000

Offset: 1

T. D. Noe, Jan 12 2006

a(1)=1 because the matrix of all zeros has determinant 0, but 0=1 (mod 1).

Andrew Howroyd, Table of n, a(n) for n = 1..2500

Cf. A000056 (order of the group SL(2, Z_n)), A175647, A243380.

Table[cnt=0; Do[m={{a, b}, {b, c}}; If[Det[m, Modulus->n]==1, cnt++ ], {a, 0, n-1}, {b, 0, n-1}, {c, 0, n-1}]; cnt, {n, 50}]
f[p_, e_] := If[Mod[p, 4] == 1, (p+1)*p^(2*e-1), (p-1)*p^(2*e-1)]; f[2, 1] = 4; f[2, e_] := 3*2^(2*e-2); a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Aug 28 2023 *)

a(n)={my(v=vector(n)); for(i=0, n-1, for(j=0, n-1, v[i*j%n+1]++)); sum(i=0, n-1, v[(i^2+1)%n+1])} \\ Andrew Howroyd, Jul 04 2018

a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); p^(2*e-1)*if(p==2, if(e==1, 2, 3/2), if(p%4==1, p+1, p-1)))} \\ Andrew Howroyd, Jul 04 2018

Multiplicative with a(2^1) = 4, a(2^e) = 3*2^(2*e-2) for e > 1, a(p^e) = (p+1)*p^(2*e-1) for p mod 4 == 1, a(p^e) = (p-1)*p^(2*e-1) for p mod 4 == 3. - Andrew Howroyd, Jul 04 2018

Sum_{k=1..n} a(k) ~ c * n^3, where c = (5/(2*Pi^2)) * A175647 * A243380 = 0.282098596071... . - Amiram Eldar, Aug 28 2023

A327122 Expansion of Sum_{k>=1} sigma(k) * x^k / (1 + x^(2*k)), where sigma = A000203.

Original entry on oeis.org

1, 3, 3, 7, 7, 9, 7, 15, 10, 21, 11, 21, 15, 21, 21, 31, 19, 30, 19, 49, 21, 33, 23, 45, 38, 45, 30, 49, 31, 63, 31, 63, 33, 57, 49, 70, 39, 57, 45, 105, 43, 63, 43, 77, 70, 69, 47, 93, 50, 114, 57, 105, 55, 90, 77, 105, 57, 93, 59, 147, 63, 93, 70, 127, 105

Offset: 1

Ilya Gutkovskiy, Sep 14 2019

Inverse Moebius transform of A050469.

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

Cf. A000203, A050469, A101455, A167181 (fixed points), A262209, A288417, A327123.

Cf. A072691, A175647, A243381.

nmax = 65; CoefficientList[Series[Sum[DivisorSigma[1, k] x^k/(1 + x^(2 k)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
A050469[n_] := DivisorSum[n, # &, MemberQ[{1}, Mod[n/#, 4]] &] - DivisorSum[n, # &, MemberQ[{3}, Mod[n/#, 4]] &]; a[n_] := DivisorSum[n, A050469[#] &]; Table[a[n], {n, 1, 65}]
f[p_, e_] := If[Mod[p, 4] == 1, (p^(e+2)-(e+2)*p+e+1)/(p-1)^2, (2*p^(e+2) + ((-1)^e-1)*p - ((-1)^e+1))/(2*(p^2-1))]; f[2, e_] := 2^(e+1)-1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 70] (* Amiram Eldar, Aug 28 2023 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; if(p == 2, 2^(e+1)-1, if(p%4 == 1, (p^(e+2)-(e+2)*p+e+1)/(p-1)^2, (2*p^(e+2) + ((-1)^e-1)*p - ((-1)^e+1))/(2*(p^2-1))))); } \\ Amiram Eldar, Aug 28 2023

a(n) = Sum_{d|n} A050469(d).
From Amiram Eldar, Aug 28 2023: (Start)
Multiplicative with a(2^e) = 2^(e+1)-1, and if p is an odd prime a(p^e) = (p^(e+2)-(e+2)*p+e+1)/(p-1)^2 if p == 1 (mod 4) and (2*p^(e+2) + ((-1)^e-1)*p - ((-1)^e+1))/(2*(p^2-1)) otherwise.
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/12 * (A175647/A243381) = 0.753351504961... . (End)

A340857 Decimal expansion of constant K5 = 29log(2+sqrt(5))(Product_{primes p == 1 (mod 5)} (1-4(2p-1)/(p(p+1)^2)))/(15Pi^2).

Original entry on oeis.org

2, 6, 2, 6, 5, 2, 1, 8, 8, 7, 2, 0, 5, 3, 6, 7, 6, 6, 6, 7, 5, 9, 6, 2, 0, 1, 1, 4, 7, 2, 0, 8, 8, 3, 4, 6, 5, 3, 0, 2, 0, 4, 3, 9, 3, 0, 6, 4, 7, 4, 4, 7, 3, 9, 1, 0, 6, 8, 2, 5, 5, 1, 0, 5, 8, 7, 0, 9, 2, 6, 6, 8, 3, 8, 6, 9, 0, 2, 2, 7, 4, 1, 7, 9, 4, 1, 9, 3, 8, 3, 6, 5, 5, 2, 3, 5, 0, 0, 2, 0, 1, 0, 0, 8, 9, 1

Offset: 0

Artur Jasinski, Jan 24 2021

Finch and Sebah, 2009, p. 7 (see link) call this constant K_5. K_5 is related to the Mertens constant C(5,1) (see A340839). For more references see the links in A340711. Finch and Sebah give the following definition:

Consider the asymptotic enumeration of m-th order primitive Dirichlet characters mod n. Let b_m(n) denote the count of such characters. There exists a constant 0 < K_m < oo such that Sum_{n <= N} b_m(n) ∼ K_m*N*log(N)^(d(m) - 2) as N -> oo, where d(m) is the number of divisors of m.

			0.262652188720536766675962011472088346530204393064744739106825510587...

Steven Finch and Pascal Sebah, Residue of a Mod 5 Euler Product, arXiv:0912.3677 [math.NT], 2009 p. 10.

Cf. A340878 (K3), A340879 (K4).
Cf. A340004, A340794, A340665, A340127.
Cf. A340629, A340710, A340711, A340628.
Cf. A175646, A301429, A333240.
Cf. A175647, A248930, A248938, A335963.
Cf. A340576, A340577, A340578, A334826.

$MaxExtraPrecision = 1000; digits = 121; f[p_] := (1 - 4*(2*p-1)/(p*(p+1)^2));
coefs = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, 1000}], x]];
S[m_, n_, s_] := (t = 1; sums = 0; difs = 1; While[Abs[difs] > 10^(-digits - 5) || difs == 0, difs = (MoebiusMu[t]/t) * Log[If[s*t == 1, DirichletL[m, n, s*t], Sum[Zeta[s*t, j/m]*DirichletCharacter[m, n, j]^t, {j, 1, m}]/m^(s*t)]]; sums = sums + difs; t++]; sums);
P[m_, n_, s_] := 1/EulerPhi[m] * Sum[Conjugate[DirichletCharacter[m, r, n]]*S[m, r, s], {r, 1, EulerPhi[m]}] + Sum[If[GCD[p, m] > 1 && Mod[p, m] == n, 1/p^s, 0], {p, 1, m}];
m = 2; sump = 0; difp = 1; While[Abs[difp] > 10^(-digits - 5) || difp == 0, difp = coefs[[m]]*P[5, 1, m]; sump = sump + difp; PrintTemporary[m]; m++];
RealDigits[Chop[N[29*Log[2+Sqrt[5]]/(15*Pi^2) * Exp[sump], digits]], 10, digits-1][[1]] (* Vaclav Kotesovec, Jan 25 2021, took over 50 minutes *)

Equals (29/25)*(Product_{primes p} (1-1/p)^2*(1+gcd(p-1,5)/(p-1))) [Finch and Sebah, 2009, p. 10].

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A369564 Powerful numbers whose prime factors are all of the form 4*k + 3.

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A204617 Multiplicative with a(p^e) = p^(e-1)*H(p). H(2) = 1, H(p) = p - 1 if p == 1 (mod 4) and H(p) = p + 1 if p == 3 (mod 4).

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A344123 Decimal expansion of Sum_{i > 0} 1/A001481(i)^2.

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A115076 Number of 2 X 2 symmetric matrices over Z(n) having determinant 1.

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A327122 Expansion of Sum_{k>=1} sigma(k) * x^k / (1 + x^(2*k)), where sigma = A000203.

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A340857 Decimal expansion of constant K5 = 29*log(2+sqrt(5))*(Product_{primes p == 1 (mod 5)} (1-4*(2*p-1)/(p*(p+1)^2)))/(15*Pi^2).

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A340857 Decimal expansion of constant K5 = 29log(2+sqrt(5))(Product_{primes p == 1 (mod 5)} (1-4(2p-1)/(p(p+1)^2)))/(15Pi^2).