A370050 -id:A370050 - cp's OEIS frontend

A297402 a(n) = gcd_{k=1..n} (prime(k+1)^n-1)/2.

1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 32, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 64, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 32, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 128, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 32, 1, 4, 1, 8, 1, 4, 1, 16, 1, 4, 1, 8, 1, 4, 1, 64, 1, 4, 1, 8

Offset: 1

Frank M Jackson, Dec 29 2017

If p is an odd prime and p^n is the length of the odd leg of a primitive Pythagorean triangle it constrains the other leg and hypotenuse to be (p^(2n)-1)/2 and (p^(2n)+1)/2. The resulting triangle has a semiperimeter of p^n(p^n+1)/2, an area of (p^n-1)p^n(p^n+1)/4 and an inradius of (p^n-1)/2. a(n) equals the GCD of the inradius terms (p^n-1)/2 for at least the first n odd primes.
Conjecture: a(n) equals the GCD of the inradius terms (p^n-1)/2 for all odd primes, i.e. a(n) = GCD_{k=1..oo} (prime(k+1)^n-1)/2.
From David A. Corneth, Dec 29 2017: (Start)
All terms are powers of 2. Proof: suppose p | a(n) for some odd prime p. Then p | (p^n - 1) / 2 and so p | (p^n - 1) which isn't the case.
If n is odd then a(n) = 1. Proof: 2 | (p^k - 1) for all k and odd primes p. 3^n - 1 = 3 * 9^k - 1 = 3 - 1 = 2 (mod 4), so 3^n - 1 is of the form 2*m for some odd m, hence the GCD of all (p^n - 1) / 2 is 1 for odd n. (End)
This is the even bisection of A059159. - Rémy Sigrist, Dec 30 2017
a(n) is the size of the group Z_2*/(Z_2*)^n, where Z_2 is the ring of 2-adic integers. We have that Z_2*/(Z_2*)^n is the inverse limit of (Z/2^iZ)*/((Z/2^iZ)*)^n as i tends to infinity. If n is odd, then the group is trivial. If n = 2^e * n' is even, where n' is odd, then the group is the product of a cyclic group of order 2^e and a cyclic group of order 2. See A370050. - Jianing Song, May 12 2024

			a(4)=8 because for n=4 and for the first 4 odd primes {3, 5, 7, 11}, the term (p^n-1)/2 gives {40, 312, 1200, 7320} with a GCD of 8.

Frank M Jackson, Table of n, a(n) for n = 1..10000

Cf. A001620, A006519, A059159, A074723, A127922, A171977, A202318.

a[n_] := GCD @@ Array[(Prime[# +1]^n -1)/2 &, n]; Array[a, 90] (* slightly modified by Robert G. Wilson v, Jan 01 2018 *)
a[n_] := If[EvenQ[n], 2^(FactorInteger[n][[1]][[2]] + 1), 1]; Array[a, 90] (* Frank M Jackson, Jul 28 2018 *)

a(n) = gcd(vector(n, i, (prime(i+1)^n-1)/2)) \\ Iain Fox, Dec 29 2017

a(n)=if(n%2,1,2)<Charles R Greathouse IV, Jan 06 2018

It appears that for k > 0, a(2^k) = 2^(k+1).
a(n) = A006519(2n) for even n and a(n) = 1 for odd n. - David A. Corneth, Dec 29 2017
a(n) = A074723(n)/2. - Iain Fox, Dec 30 2017
Multiplicative with a(2^e) = 2^(e+1), a(p^e) = 1 for odd prime p. - Andrew Howroyd, Jul 25 2018
It appears that for m > 0, a(2m-1) = 1 (proved in comments) and a(2m) = 2^(k+1) where k is the exponent of the even prime in the prime factorization of 2m. - Frank M Jackson, Jul 28 2018
From Amiram Eldar, Nov 24 2023: (Start)
Dirichlet g.f.: zeta(s) * (1 + 1/2^s + 1/(2^(s-1) - 1)).
Sum_{k=1..n} a(k) ~ (n/log(2)) * (log(n) + gamma + log(2) - 1), where gamma is Euler's constant (A001620). (End)

A370067 Square array read by ascending antidiagonals: T(n,k) is the size of the group Q_p/(Q_p)^k, where p = prime(n), and Q_p is the field of p-adic numbers.

Original entry on oeis.org

1, 1, 8, 1, 4, 3, 1, 4, 9, 32, 1, 4, 3, 8, 5, 1, 4, 9, 16, 5, 24, 1, 4, 3, 8, 25, 36, 7, 1, 4, 9, 8, 5, 12, 7, 128, 1, 4, 3, 16, 25, 36, 7, 16, 9, 1, 4, 9, 16, 5, 12, 49, 32, 81, 40, 1, 4, 3, 8, 5, 36, 7, 16, 9, 20, 11, 1, 4, 3, 8, 5, 12, 7, 16, 27, 100, 11, 96, 1, 4, 9, 16, 5, 36, 7, 32, 9, 20, 11, 72, 13

Offset: 1

Jianing Song, Apr 30 2024

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

Each row is multiplicative.

			Table reads
  1, 8, 3, 32, 5, 24, 7, 128, 9, 40
  1, 4, 9, 8, 5, 36, 7, 16, 81, 20
  1, 4, 3, 16, 25, 12, 7, 32, 9, 100
  1, 4, 9, 8, 5, 36, 49, 16, 27, 20
  1, 4, 3, 8, 25, 12, 7, 16, 9, 100
  1, 4, 9, 16, 5, 36, 7, 32, 27, 20
  1, 4, 3, 16, 5, 12, 7, 64, 9, 20
  1, 4, 9, 8, 5, 36, 7, 16, 81, 20
  1, 4, 3, 8, 5, 12, 7, 16, 9, 20
  1, 4, 3, 16, 5, 12, 49, 32, 9, 20

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

Cf. A370050. Row 1-4: A370564, A370565, A370566, A370567.

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

T(n,k) = k * A370050(n,k).

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

A370565 Size of the group Q_3/(Q_3)^n, where Q_3 is the field of 3-adic numbers.

Original entry on oeis.org

1, 4, 9, 8, 5, 36, 7, 16, 81, 20, 11, 72, 13, 28, 45, 32, 17, 324, 19, 40, 63, 44, 23, 144, 25, 52, 729, 56, 29, 180, 31, 64, 99, 68, 35, 648, 37, 76, 117, 80, 41, 252, 43, 88, 405, 92, 47, 288, 49, 100, 153, 104, 53, 2916, 55, 112, 171, 116, 59, 360, 61, 124, 567, 128

Offset: 1

Jianing Song, Apr 30 2024

We have Q_3* = 3^Z X Z_3*, so Q_3*/(Q_3*)^k = (3^Z/3^(kZ)) X (Z_p*/(Z_3*)^k). Note that 3^Z/3^(kZ) is a cyclic group of order k. For the group structure of (Z_3*/(Z_3*)^k), see A370050.

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

Row 2 of A370067. Cf. A001620, A370050, A370564, A370566, A370567.

Cf. A370180.

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

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

Write n = 3^e * n' with k' not being divisible by 3, then a(n) = n * 3^e * gcd(2,n').
Multiplicative with a(3^e) = 3^(2*e), a(2^e) = 2^(e+1) and a(p^e) = p^e for primes p != 2, 3.
a(n) = n * A370180(n).
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: ((1 + 1/2^(s-1)) * (1 - 1/3^(s-1))/(1 - 1/3^(s-2))) * zeta(s-1).
Sum_{k=1..n} a(k) ~ (n^2/(2*log(3))) * (log(n) + gamma - 1/2 + log(3) - log(2)/3), where gamma is Euler's constant (A001620). (End)

A370566 Size of the group Q_5/(Q_5)^n, where Q_5 is the field of 5-adic numbers.

Original entry on oeis.org

1, 4, 3, 16, 25, 12, 7, 32, 9, 100, 11, 48, 13, 28, 75, 64, 17, 36, 19, 400, 21, 44, 23, 96, 625, 52, 27, 112, 29, 300, 31, 128, 33, 68, 175, 144, 37, 76, 39, 800, 41, 84, 43, 176, 225, 92, 47, 192, 49, 2500, 51, 208, 53, 108, 275, 224, 57, 116, 59, 1200, 61, 124, 63, 256

Offset: 1

Jianing Song, Apr 30 2024

We have Q_5* = 5^Z X Z_5*, so Q_5*/(Q_5*)^k = (5^Z/5^(kZ)) X (Z_p*/(Z_5*)^k). Note that 5^Z/5^(kZ) is a cyclic group of order k. For the group structure of (Z_5*/(Z_5*)^k), see A370050.

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

Row 3 of A370067. Cf. A001620, A370050, A370564, A370565, A370567.

Cf. A370181.

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

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

Write n = 5^e * n' with k' not being divisible by 5, then a(n) = n * 5^e * gcd(4,n').
Multiplicative with a(5^e) = 5^(2*e), a(2) = 4, a(2^e) = 2^(e+2) for e >= 2 and a(p^e) = p^e for primes p != 2, 5.
a(n) = n * A370181(n).
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: ((1 + 1/2^(s-1) + 1/2^(2*s-3)) * (1 - 1/5^(s-1))/(1 - 1/5^(s-2))) * zeta(s-1).
Sum_{k=1..n} a(k) ~ (4*n^2/(5*log(5))) * (log(n) + gamma - 1/2 + 3*log(5/2)/4), where gamma is Euler's constant (A001620). (End)

A370567 Size of the group Q_7/(Q_7)^n, where Q_7 is the field of 7-adic numbers.

Original entry on oeis.org

1, 4, 9, 8, 5, 36, 49, 16, 27, 20, 11, 72, 13, 196, 45, 32, 17, 108, 19, 40, 441, 44, 23, 144, 25, 52, 81, 392, 29, 180, 31, 64, 99, 68, 245, 216, 37, 76, 117, 80, 41, 1764, 43, 88, 135, 92, 47, 288, 2401, 100, 153, 104, 53, 324, 55, 784, 171, 116, 59, 360, 61, 124, 1323, 128

Offset: 1

Jianing Song, Apr 30 2024

We have Q_7* = 7^Z X Z_7*, so Q_7*/(Q_7*)^k = (7^Z/7^(kZ)) X (Z_p*/(Z_7*)^k). Note that 7^Z/7^(kZ) is a cyclic group of order k. For the group structure of (Z_7*/(Z_7*)^k), see A370050.

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

Row 4 of A370067. Cf. A001620, A370050, A370564, A370565, A370566.

Cf. A370182.

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 * n]; Array[a, 100] (* Amiram Eldar, May 20 2024 *)

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

Write n = 7^e * n' with k' not being divisible by 7, then a(n) = n * 7^e * gcd(6,n').
Multiplicative with a(7^e) = 7^(2*e), a(2^e) = 2^(e+1), a(3^e) = 3^(e+1) and a(p^e) = p^e for primes p != 2, 3, 7.
a(n) = n * A370182(n).
From Amiram Eldar, May 20 2024: (Start)
Dirichlet g.f.: ((1 + 1/2^(s-1)) * (1 + 2/3^(s-1)) * (1 - 1/7^(s-1))/(1 - 1/7^(s-2))) * zeta(s-1).
Sum_{k=1..n} a(k) ~ (15*n^2/(14*log(7))) * (log(n) + gamma - 1/2 + 2*log(7)/3 - 2*log(3)/5 - log(2)/3), where gamma is Euler's constant (A001620). (End)

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)

cp's OEIS Frontend

A297402 a(n) = gcd_{k=1..n} (prime(k+1)^n-1)/2.

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A370067 Square array read by ascending antidiagonals: T(n,k) is the size of the group Q_p*/(Q_p*)^k, where p = prime(n), and Q_p is the field of p-adic numbers.

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A370565 Size of the group Q_3*/(Q_3*)^n, where Q_3 is the field of 3-adic numbers.

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A370566 Size of the group Q_5*/(Q_5*)^n, where Q_5 is the field of 5-adic numbers.

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A370567 Size of the group Q_7*/(Q_7*)^n, where Q_7 is the field of 7-adic numbers.

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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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Programs

Mathematica

A370067 Square array read by ascending antidiagonals: T(n,k) is the size of the group Q_p/(Q_p)^k, where p = prime(n), and Q_p is the field of p-adic numbers.

A370565 Size of the group Q_3/(Q_3)^n, where Q_3 is the field of 3-adic numbers.

A370566 Size of the group Q_5/(Q_5)^n, where Q_5 is the field of 5-adic numbers.

A370567 Size of the group Q_7/(Q_7)^n, where Q_7 is the field of 7-adic numbers.

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.