A297402 -id:A297402 - cp's OEIS frontend

A370050 Square array read by ascending antidiagonals: T(n,k) is the size of the group Z_p/(Z_p)^k, where p = prime(n), and Z_p is the ring of p-adic integers.

1, 1, 4, 1, 2, 1, 1, 2, 3, 8, 1, 2, 1, 2, 1, 1, 2, 3, 4, 1, 4, 1, 2, 1, 2, 5, 6, 1, 1, 2, 3, 2, 1, 2, 1, 16, 1, 2, 1, 4, 5, 6, 1, 2, 1, 1, 2, 3, 4, 1, 2, 7, 4, 9, 4, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 3, 10, 1, 8, 1, 2, 3, 4, 1, 6, 1, 4, 1, 2, 1, 6, 1

Offset: 1

Jianing Song, Apr 30 2024

We have that Z_p*/(Z_p*)^k is the inverse limit of (Z/p^iZ)*/((Z/p^iZ)*)^k as i tends to infinity. Write k = p^e * k' with k' not being divisible by p. If p is odd, then the group is cyclic of order p^e * gcd(p-1,k'). If p = 2 and k is odd, then the group is trivial. If p = 2 and k is even, then the group is the product of a cyclic group of order 2^e and a cyclic group of order 2.

Each row is multiplicative.

			Table reads
  1, 4, 1, 8, 1, 4, 1, 16, 1, 4
  1, 2, 3, 2, 1, 6, 1, 2, 9, 2
  1, 2, 1, 4, 5, 2, 1, 4, 1, 10
  1, 2, 3, 2, 1, 6, 7, 2, 3, 2
  1, 2, 1, 2, 5, 2, 1, 2, 1, 10
  1, 2, 3, 4, 1, 6, 1, 4, 3, 2
  1, 2, 1, 4, 1, 2, 1, 8, 1, 2
  1, 2, 3, 2, 1, 6, 1, 2, 9, 2
  1, 2, 1, 2, 1, 2, 1, 2, 1, 2
  1, 2, 1, 4, 1, 2, 7, 4, 1, 2
For p = prime(1) = 2 and k = 2, we have Z_p*/(Z_p*)^k = Z_2*/(1+8Z_2) = (Z/8Z)*/(1+8Z) = C_2 X C_2, so T(1,2) = 4.
For p = prime(2) = 3 and k = 3, we have Z_p*/(Z_p*)^k = Z_3*/((1+9Z_3) U (8+9Z_3)) = (Z/9Z)*/((1+9Z) U (8+9Z)) = C_3, so T(2,3) = 3.

Jianing Song, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals)

Cf. A370067. Row 1-4: A297402, A370180, A370181, A370182.

T(n,k) = my(p = prime(n), e = valuation(k,p)); p^e*gcd(p-1,k/p^e) * if(p==2 && e>=1, 2, 1)

Write k = p^e * k' with k' not being divisible by p, and p = prime(n). If p is odd, then T(n,k) = p^e * gcd(p-1,k'). If p = 2 and k is odd, then T(n,k) = 1. If p = 2 and k is even, then T(n,k) = 2^(e+1).

A370180 Size of the group Z_3/(Z_3)^n, where Z_3 is the ring of 3-adic integers.

Original entry on oeis.org

1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 2, 3, 2, 1, 18, 1, 2, 3, 2, 1, 6, 1, 2, 27, 2, 1, 6, 1, 2, 3, 2, 1, 18, 1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 2, 3, 2, 1, 54, 1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 2, 3, 2, 1, 18, 1, 2, 3, 2, 1, 6, 1, 2, 81, 2, 1, 6, 1, 2, 3, 2, 1, 18

Offset: 1

Jianing Song, Apr 30 2024

We have that Z_3*/(Z_3*)^n is the inverse limit of (Z/3^iZ)*/((Z/3^iZ)*)^n as i tends to infinity. Write n = 3^e * n' with n' not being divisible by 3, then the group is cyclic of order 3^e * gcd(2,n'). See A370050.

			We have Z_3*/(Z_3*)^3 = Z_3* / ((1+9Z_3) U (8+9Z_3)) = (Z/9Z)*/((1+9Z) U (8+9Z)) = C_3, so a(3) = 3.
We have Z_3*/(Z_3*)^6 = Z_3* / (1+9Z_3) = (Z/9Z)*/(1+9Z) = C_6, so a(6) = 6.

Jianing Song, Table of n, a(n) for n = 1..10000

Row 2 of A370050. Cf. A001620, A297402, A370181, A370182.

Cf. A370565.

a[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3]}, 2^If[e2 == 0, 0, 1] * 3^e3]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

a(n,{p=3}) = my(e = valuation(n, p)); p^e*gcd(p-1, n/p^e)

Multiplicative with a(3^e) = 3^e, a(2^e) = 2 and a(p^e) = 1 for primes p != 2, 3.
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: (1 + 1/2^s) * ((1 - 1/3^s)/(1 - 1/3^(s-1))) * zeta(s).
Sum_{k=1..n} a(k) ~ (n/log(3)) * (log(n) + gamma - 1 + log(3) - log(2)/3), where gamma is Euler's constant (A001620). (End)

A370181 Size of the group Z_5/(Z_5)^n, where Z_5 is the ring of 5-adic integers.

Original entry on oeis.org

1, 2, 1, 4, 5, 2, 1, 4, 1, 10, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 25, 2, 1, 4, 1, 10, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 50, 1, 4, 1, 2, 5, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 10, 1, 4, 1, 2, 25, 4, 1, 2, 1, 20, 1, 2, 1, 4, 5, 2, 1, 4, 1, 10

Offset: 1

Jianing Song, Apr 30 2024

We have that Z_5*/(Z_5*)^n is the inverse limit of (Z/5^iZ)*/((Z/5^iZ)*)^n as i tends to infinity. Write n = 5^e * n' with n' not being divisible by 5, then the group is cyclic of order 5^e * gcd(4,n'). See A370050.

			We have Z_5*/(Z_5*)^5 = Z_5* / ((1+25Z_5) U (7+25Z_5) U (18+25Z_5) U (24+25Z_5)) = (Z/25Z)*/((1+25Z) U (7+25Z) U (18+25Z) U (24+25Z)) = C_5, so a(5) = 5.
We have Z_5*/(Z_5*)^10 = Z_5* / ((1+25Z_5) U (24+25Z_5)) = (Z/25Z)*/((1+25Z) U (25+25Z)) = C_10, so a(10) = 10.

Jianing Song, Table of n, a(n) for n = 1..10000

Row 3 of A370050. Cf. A001620, A297402, A370180, A370182.

Cf. A370566.

a[n_] := Module[{e2 = IntegerExponent[n, 2], e5 = IntegerExponent[n, 5]}, 2^Min[e2, 2] * 5^e5]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

a(n,{p=5}) = my(e = valuation(n, p)); p^e*gcd(p-1, n/p^e)

Multiplicative with a(5^e) = 5^e, a(2) = 2, a(2^e) = 4 for e >= 2 and a(p^e) = 1 for primes p != 2, 5.
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: (1 + 1/2^s + 1/2^(2*s-1)) * ((1 - 1/5^s)/(1 - 1/5^(s-1))) * zeta(s).
Sum_{k=1..n} a(k) ~ (8*n/(5*log(5))) * (log(n) + gamma - 1 + (3/4)*log(5/2)), where gamma is Euler's constant (A001620). (End)

A370182 Size of the group Z_7/(Z_7)^n, where Z_7 is the ring of 7-adic integers.

Original entry on oeis.org

1, 2, 3, 2, 1, 6, 7, 2, 3, 2, 1, 6, 1, 14, 3, 2, 1, 6, 1, 2, 21, 2, 1, 6, 1, 2, 3, 14, 1, 6, 1, 2, 3, 2, 7, 6, 1, 2, 3, 2, 1, 42, 1, 2, 3, 2, 1, 6, 49, 2, 3, 2, 1, 6, 1, 14, 3, 2, 1, 6, 1, 2, 21, 2, 1, 6, 1, 2, 3, 14, 1, 6, 1, 2, 3, 2, 7, 6, 1, 2, 3, 2, 1, 42, 1, 2, 3, 2, 1, 6

Offset: 1

Jianing Song, Apr 30 2024

We have that Z_7*/(Z_7*)^n is the inverse limit of (Z/7^iZ)*/((Z/7^iZ)*)^n as i tends to infinity. Write n = 7^e * n' with n' not being divisible by 7, then the group is cyclic of order 7^e * gcd(7,n'). See A370050.

			We have Z_7*/(Z_7*)^7 = Z_7* / ((1+49Z_7) U (18+49Z_7) U (19+49Z_7) U (30+49Z_7) U (31+49Z_7) U (48+49Z_7)) = (Z/49Z)*/((1+49Z) U (18+49Z) U (19+49Z) U (30+49Z) U (31+49Z) U (48+49Z)) = C_7, so a(7) = 7.
We have Z_7*/(Z_7*)^14 = Z_7* / ((1+49Z_7) U (18+49Z_7) U (30+49Z_7)) = (Z/49Z)*/((1+49Z) U (18+49Z) U (30+49Z)) = C_14, so a(14) = 14.

Jianing Song, Table of n, a(n) for n = 1..10000

Row 4 of A370050. Cf. A001620, A297402, A370180, A370181.

Cf. A370567.

a[n_] := Module[{e2 = IntegerExponent[n, 2], e3 = IntegerExponent[n, 3], e7 = IntegerExponent[n, 7]}, 2^Min[e2, 1] * 3^Min[e3, 1] * 7^e7]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

a(n,{p=7}) = my(e = valuation(n, p)); p^e*gcd(p-1, n/p^e)

Multiplicative with a(7^e) = 7^e, a(2^e) = 2, a(3^e) = 3 and a(p^e) = 1 for primes p != 2, 3, 7.
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: (1 + 1/2^s) * (1 + 2/3^s) * ((1 - 1/7^s)/(1 - 1/7^(s-1))) * zeta(s).
Sum_{k=1..n} a(k) ~ (15*n/(7*log(7))) * (log(n) + gamma - 1 + 2*log(7)/3 - 2*log(3)/5 - log(2)/3), where gamma is Euler's constant (A001620). (End)

A370564 Size of the group Q_2/(Q_2)^n, where Q_2 is the field of 2-adic numbers.

Original entry on oeis.org

1, 8, 3, 32, 5, 24, 7, 128, 9, 40, 11, 96, 13, 56, 15, 512, 17, 72, 19, 160, 21, 88, 23, 384, 25, 104, 27, 224, 29, 120, 31, 2048, 33, 136, 35, 288, 37, 152, 39, 640, 41, 168, 43, 352, 45, 184, 47, 1536, 49, 200, 51, 416, 53, 216, 55, 896, 57, 232, 59, 480, 61, 248, 63, 8192

Offset: 1

Jianing Song, Apr 30 2024

We have Q_2* = 2^Z X Z_2*, so Q_2*/(Q_2*)^k = (2^Z/2^(kZ)) X (Z_p*/(Z_2*)^k). Note that 2^Z/2^(kZ) is a cyclic group of order k. For the group structure of (Z_2*/(Z_2*)^k), see A370050.

Jianing Song, Table of n, a(n) for n = 1..10000

Row 1 of A370067. Cf. A001620, A370050, A370565, A370566, A370567.

Cf. A297402.

a[n_] := Module[{e = IntegerExponent[n, 2]}, 2^If[e == 0, 0, e + 1] * n]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

a(n) = my(e = valuation(n, 2)); n * 2^e * if(e>=1, 2, 1)

If n is odd, then a(n) = n. If n = 2^e * n' is even, where n' is odd, then a(n) = n * 2^(e+1).
Multiplicative with a(2^e) = 2^(2*e+1).
a(n) = n * A297402(n).
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: ((1 - 1/2^(s-1)) * (1 + 1/2^(s-2)) / (1 - 1/2^(s-2))) * zeta(s-1).
Sum_{k=1..n} a(k) ~ (n^2/(2*log(2))) * (log(n) + gamma - 1/2 + log(2)), where gamma is Euler's constant (A001620). (End)

A074723 Largest power of 2 dividing F(3n) where F(k) is the k-th Fibonacci number.

Original entry on oeis.org

2, 8, 2, 16, 2, 8, 2, 32, 2, 8, 2, 16, 2, 8, 2, 64, 2, 8, 2, 16, 2, 8, 2, 32, 2, 8, 2, 16, 2, 8, 2, 128, 2, 8, 2, 16, 2, 8, 2, 32, 2, 8, 2, 16, 2, 8, 2, 64, 2, 8, 2, 16, 2, 8, 2, 32, 2, 8, 2, 16, 2, 8, 2, 256, 2, 8, 2, 16, 2, 8, 2, 32, 2, 8, 2, 16, 2, 8, 2, 64, 2, 8, 2, 16, 2, 8, 2, 32, 2, 8, 2

Offset: 1

Benoit Cloitre, Sep 04 2002

nonn

If m == 1 or 2 (mod 3) then F(m) is odd.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A001620, A006519, A014445, A297402.

seq(`if`(n::odd,2,2^(2+padic:-ordp(n,2))),n=1..100); # Robert Israel, Oct 10 2016

Table[2^(Length@ NestWhileList[#/2 &, Fibonacci[3 n], IntegerQ] - 2), {n, 120}] (* Michael De Vlieger, Oct 10 2016 *)
a[n_] := If[EvenQ[n], 2^(FactorInteger[n][[1]][[2]] + 2), 2]; Array[a, 90] (* Frank M Jackson, Jul 28 2018 *)

a(n) = 2^valuation(fibonacci(3*n), 2); \\ Michel Marcus, Oct 10 2016

It appears that 4 never appears and : a(2k+1)=2 a(2^m*(2k+1))=2^(m+2) for k>=0 and m >=1.
From Robert Israel, Oct 10 2016: (Start)
a(2k+1)=2 follows from F(n+6) = 5 F(n) + 8 F(n+1) == F(n) mod 4.
a(2*(2k+1))=8 follows from F(n+12) = 89 F(n) + 144 F(n+1) == 9 F(n) mod 16.
a(2^m*(2k+1)) = 2^(m+2) for m > 2 follows from F(2n) = F(n) (2 F(n-1) + F(n)).
G.f. 2*x/(1-x^2) + Sum_{m>=1} 2^(m+2)*x^(2^m)/(1 - x^(2^(m+1))). (End)
a(n) = A006519(A014445(n)). - Michel Marcus, Oct 10 2016
As proved above, for m > 0, a(2m-1) = 2 and a(2m) = 2^(k+2) where k is the exponent of the even prime in the prime factorization of 2m. Also a(n) = 2*A297402(n). - Frank M Jackson, Jul 28 2018
Sum_{k=1..n} a(k) ~ (2*n/log(2)) * (log(n) + gamma + log(2) - 1), where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 27 2023

cp's OEIS Frontend

A370050 Square array read by ascending antidiagonals: T(n,k) is the size of the group Z_p*/(Z_p*)^k, where p = prime(n), and Z_p is the ring of p-adic integers.

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A370180 Size of the group Z_3*/(Z_3*)^n, where Z_3 is the ring of 3-adic integers.

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A370181 Size of the group Z_5*/(Z_5*)^n, where Z_5 is the ring of 5-adic integers.

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A370182 Size of the group Z_7*/(Z_7*)^n, where Z_7 is the ring of 7-adic integers.

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A370564 Size of the group Q_2*/(Q_2*)^n, where Q_2 is the field of 2-adic numbers.

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A074723 Largest power of 2 dividing F(3n) where F(k) is the k-th Fibonacci number.

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A370050 Square array read by ascending antidiagonals: T(n,k) is the size of the group Z_p/(Z_p)^k, where p = prime(n), and Z_p is the ring of p-adic integers.

A370180 Size of the group Z_3/(Z_3)^n, where Z_3 is the ring of 3-adic integers.

A370181 Size of the group Z_5/(Z_5)^n, where Z_5 is the ring of 5-adic integers.

A370182 Size of the group Z_7/(Z_7)^n, where Z_7 is the ring of 7-adic integers.

A370564 Size of the group Q_2/(Q_2)^n, where Q_2 is the field of 2-adic numbers.