A066498 -id:A066498 - cp's OEIS frontend

A276919 Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.

1, 8, 27, 64, 125, 216, 336, 512, 1296, 1000, 1331, 1728, 1794, 2688, 3375, 4096, 4913, 10368, 7410, 8000, 9072, 10648, 12167, 13824, 15625, 14352, 34992, 21504, 24389, 27000, 30225, 32768, 35937, 39304, 42000, 82944, 48396, 59280, 48438, 64000, 68921, 72576, 77529, 85184, 162000, 97336

Offset: 1

José María Grau Ribas, Sep 22 2016

It appears that a(n) = n^3 for n in A088232. See also A066498. - Michel Marcus, Oct 11 2016

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A000189, A047726, A060839, A063454, A087412, A087786, A254073, A276920.

JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 1], {n, 1, 50}]

a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 1)))); \\ Michel Marcus, Oct 11 2016

qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i,t+1);t=0));r*=binomial(#v,t+1)}
a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1, 4, Mod(v[i], n)^3)==1, print1(v", "); t+=qperms(v)),1);t} \\ David A. Corneth, Oct 11 2016

def A276919(n):
    ndict = {}
    for i in range(n):
        i3 = pow(i,3,n)
        for j in range(i+1):
            j3 = pow(j,3,n)
            m = (i3+j3) % n
            if m in ndict:
                if i == j:
                    ndict[m] += 1
                else:
                    ndict[m] += 2
            else:
                if i == j:
                    ndict[m] = 1
                else:
                    ndict[m] = 2
    count = 0
    for i in ndict:
        j = (1-i) % n
        if j in ndict:
            count += ndict[i]*ndict[j]
    return count # Chai Wah Wu, Jun 06 2017

A076989 Smallest cube of the form n*k + 1 with k>0.

Original entry on oeis.org

8, 27, 64, 125, 216, 343, 8, 729, 64, 1331, 1728, 2197, 27, 729, 4096, 4913, 5832, 343, 343, 9261, 64, 12167, 13824, 15625, 17576, 27, 1000, 729, 27000, 29791, 125, 35937, 39304, 42875, 1331, 2197, 1000, 343, 4096, 68921, 74088, 15625, 216, 91125, 4096

Offset: 1

Amarnath Murthy, Oct 25 2002

			a(9) = 64 as 64 = 7*9 + 1.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A066498, A076947.

a[1] := 8:for n from 2 to 150 do j := 2:while((j^3 mod n)<>1)do j := j+1:od: a[n] := j^3:od:seq(a[k],k=1..150);
# Alternative
f:=proc(n) local R;
  R:= sort([numtheory:-rootsunity(3,n)] mod n);
  if nops(R)=1 then (n+1)^3 else R[2]^3 fi
end proc:
map(f, [$1..150]); # Robert Israel, Mar 30 2018

sc[n_]:=Module[{k=1},While[!IntegerQ[Surd[n*k+1,3]],k++];n*k+1]; Array[ sc,50] (* Harvey P. Dale, Mar 30 2018 *)

first(n) = my(res = vector(n)); {res[1] = 8; for(i = 2, n + 1, i3 = i ^ 3-1; d = divisors(i3); j = 2; while(j <= #d && d[j] <= n, if(res[d[j]] == 0, res[d[j]] = i3 + 1); j++)); res} \\ David A. Corneth, Mar 30 2018

a(n) <= (n+1)^3. In particular, a(n) < (n+1)^3 if n is in A066498. - David A. Corneth, Mar 30 2018

a(n) = A076947(n)*n + 1. - Altug Alkan, Mar 30 2018

More terms from Sascha Kurz, Jan 26 2003

A143973 Distances between multiples of 3 in A000010.

Original entry on oeis.org

2, 4, 1, 4, 1, 2, 5, 1, 1, 3, 4, 1, 1, 1, 1, 3, 1, 2, 4, 3, 2, 2, 1, 4, 1, 1, 2, 2, 3, 2, 1, 1, 2, 1, 1, 1, 2, 3, 2, 4, 1, 2, 2, 2, 1, 1, 4, 1, 1, 3, 1, 2, 1, 2, 3, 2, 3, 2, 2, 1, 2, 1, 3, 1, 1, 4, 1, 3, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 5, 1, 2, 1

Offset: 1

J. Perry (johnandruth(AT)jrperry.orangehome.co.uk), Sep 06 2008

nonn

Inspired by the question: is n == 1 mod phi(n) for n composite?

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

Cf. A000010, A008585, A066498.

Differences @ Position[EulerPhi[Range[200]], ?(Divisible[#, 3] &)] // Flatten (* _Amiram Eldar, Feb 29 2020 *)

a(1) = 2 since the first two numbers k such that 3 | phi(k) are 7 and 9, and 9 - 7 = 2.

Offset corrected and more terms added by Amiram Eldar, Feb 29 2020

A343783 a(n) is the largest primorial number (A002110) which divides phi(n).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 6, 2, 6, 2, 2, 2, 6, 6, 2, 2, 2, 6, 6, 2, 6, 2, 2, 2, 2, 6, 6, 6, 2, 2, 30, 2, 2, 2, 6, 6, 6, 6, 6, 2, 2, 6, 6, 2, 6, 2, 2, 2, 6, 2, 2, 6, 2, 6, 2, 6, 6, 2, 2, 2, 30, 30, 6, 2, 6, 2, 6, 2, 2, 6, 2, 6, 6, 6, 2, 6, 30, 6, 6, 2, 6, 2, 2, 6, 2, 6

Offset: 1

Amiram Eldar, Apr 29 2021

nonn

			a(3) = 2 since phi(3) = 2 and 2 = A002110(1).
a(5) = 2 since phi(5) = 4 and 2 = A002110(1) is the largest primorial dividing 4.
a(7) = 6 since phi(7) = 6 and 6 = A002110(2).

Amiram Eldar, Table of n, a(n) for n = 1..10000
Paul Pollack and Carl Pomerance, Phi, primorials, and Poisson, Illinois J. Math., Vol. 64, No. 3 (2020), pp. 319-330; alternative link.

Cf. A000010, A002110, A034386, A053589.

prim[n_] := Times @@ Prime[Range[n]]; gp[n_] := Module[{k = 1}, While[Divisible[n, prim[k]], k++]; prim[k - 1]]; a[n_] := gp[EulerPhi[n]]; Array[a, 100]

f(n) = my(s=1); forprime(p=2, , if(n%p, return(s), s *= p)); \\ A053589
a(n) = f(eulerphi(n)); \\ Michel Marcus, May 01 2021

a(n) = A053589(A000010(n)).
Let pr(n) be the largest prime divisor of a(n) (i.e., a(n) = pr(n)# = A034386(pr(n))). Then pr(n) ~ log(log(n))/log(log(log(n))) on a set of integers of asymptotic density 1 (Pollack and Pomerance, 2020).
From Bernard Schott, May 05 2021: (Start)
a(2n) = a(n) for n>=1.
a(n) = 1 iff n = 1 or n = 2.
a(n) = 2 iff 3 does not divide phi(n) (A088232)
a(n) >= 6 iff 3 divides phi(n) (A066498). (End)

A358043 Numbers k such that phi(k) is a multiple of 8.

Original entry on oeis.org

15, 16, 17, 20, 24, 30, 32, 34, 35, 39, 40, 41, 45, 48, 51, 52, 55, 56, 60, 64, 65, 68, 70, 72, 73, 75, 78, 80, 82, 84, 85, 87, 88, 89, 90, 91, 95, 96, 97, 100, 102, 104, 105, 110, 111, 112, 113, 115, 116, 117, 119, 120, 123, 128, 130, 132, 135, 136, 137, 140, 143

Offset: 1

Darío Clavijo, Oct 26 2022

nonn

Cf. A000010 (phi), A053574 (its 2-adic valuation), A037074 (a subsequence).

Totient multiples: A066498 (3), A172019 (4), A066500 (5), A066502 (7), A332512 (12).

Select[Range[150], Divisible[EulerPhi[#], 8] &] (* Amiram Eldar, Oct 27 2022 *)

isok(k) = Mod(eulerphi(k), 8) == 0; \\ Michel Marcus, Oct 27 2022

from sympy.ntheory import totient
def isok(n): return totient(n) % 8 == 0

A000010(a(n)) == 0 (mod 8).

cp's OEIS Frontend

A276919 Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.

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A076989 Smallest cube of the form n*k + 1 with k>0.

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A143973 Distances between multiples of 3 in A000010.

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A343783 a(n) is the largest primorial number (A002110) which divides phi(n).

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Mathematica

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A358043 Numbers k such that phi(k) is a multiple of 8.

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Author

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Crossrefs

Programs

Mathematica

PARI

Python

Formula