A100978 - OEIS

A100978

Number of all extensions over Q_5 with degree n in the algebraic closure of Q_5.

1, 3, 4, 7, 106, 12, 8, 15, 13, 1818, 12, 28, 14, 24, 12424, 31, 18, 39, 20, 109242, 32, 36, 24, 60, 8281131, 42, 40, 56, 30, 4687272, 32, 63, 48, 54, 15624848, 91, 38, 60, 56, 146484090, 42, 96, 44, 84, 634765378, 72, 48, 124, 57, 443847655893, 72, 98, 54, 120, 14648437272, 120, 80, 90

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

REFERENCES

M. Krasner, Le nombre des surcorps primitifs d'un degré donné et le nombre des surcorps métagaloisiens d'un degré donné d'un corps de nombres p-adiques. Comptes Rendus Hebdomadaires, Académie des Sciences, Paris 254, 255, 1962

LINKS

Robin Visser, Table of n, a(n) for n = 1..5000

Jordi Guàrdia-Rúbies, John W. Jones, Kevin Keating, Sebastian Pauli, David P. Roberts, and David Roe, Families of p-adic fields, arXiv:2507.02360 [math.NT], 2025.

FORMULA

a(n) = (Sum_{d|h}d)*(Sum_{s=0..m} (p^(m+s+1)-p^(2*s))/(p-1)*(p^(eps(s)*n)-p^(eps(s-1)*n))), where p=5, n=h*p^m, with gcd(h, p)=1, eps(-1)=-oo, eps(0)=0 and eps(s)=Sum_{i=1..s} 1/(p^i).

EXAMPLE

a(2)=3: There are 2 ramified extensions with minimal polynomials x^2-5, x^2-10, and one unramified extension with minimal polynomial x^2+4*x+2.

MAPLE

p:=5; eps:=proc()local p, s, i, sum; p:=args[1]; s:=args[2]; if s=-1 then return -infinity; fi; if s=0 then return 0; fi; sum:=0; for i from 1 to s do sum:=sum+1/p^i; od; return sum; end: ppart:=proc() local p, n; p:=args[1]; n:=args[2]; return igcd(n, p^n); end: qpart:=proc() local p, n; p:=args[1]; n:=args[2]; return n/igcd(n, p^n); end: logp:=proc() local p, pp; p:=args[1]; pp:=args[2]; if op(ifactors(pp))[2]=[] then return 0; else return op(op(ifactors(pp))[2])[2]; fi; end: summe:=0; m:=logp(p, ppart(p, n)); h:=qpart(p, n); for s from 0 to m do summe:=summe+(p^(m+s+1)-p^(2*s))/(p-1)*(p^(eps(p, s)*n)-p^(eps(p, s-1)*n)); od; a(n):=sigma(h)*summe;

PROG

(SageMath)

def a(n):

ans = 0

for e in Integer(n).divisors():

s = Integer(e).valuation(5); ans += e

for i in range(1, s+1):

ans += e*5^(i+sum([n/(5^j) for j in range(1, i+1)]))

ans -= e*5^(i+sum([n/(5^j) for j in range(1, i)]))

return ans # Robin Visser, Nov 30 2025

CROSSREFS

Cf. A100976, A100977, A100979, A100980, A100981, A100983, A100984, A100985, A100986.