A134922 -id:A134922 - cp's OEIS frontend

A236255 Prime numbers s for which there exist primes p, q, r such that phi(pqs)=phi(rs^3), sigma(pqs)=sigma(rs^3).

2, 3, 5, 7, 31, 43, 139, 157, 191, 269, 293, 337, 463, 557, 593, 683, 709, 757, 769, 983, 1021, 1567, 1583, 2293, 2309, 2689, 2707, 2801, 2917, 3319, 3323, 3583, 3823, 4271, 5507, 5557, 6037, 6043, 6079, 6151, 6469, 6779, 6959, 6977, 7207, 7963, 8419, 8429, 8521, 8627, 8663, 8861, 8887, 9677, 9769, 10163, 10613, 10847, 11003

Offset: 1

Vladimir Letsko, Jan 21 2014

nonn

Obviously tau(pqs) = tau(rs^3). So we have pairs of terms of A134922.

s = 593 is the least number such that there are just two matching pairs: (593*381187517*703949, 593^3*763079633) and (593*3911429*780389, 593^3*8680337). And for s = 853693 there are as many as 3 matching pairs.

			2 is in the sequence because for p = 11, q = 29, r = 71 we have phi(pqs)=phi(rs^3) and sigma(pqs)=sigma(rs^3).

Vladimir Letsko, Math Marathon, problem 163 (in Russian)

cf. A134922, A000010, A000203

is_A236255:=proc(s::prime) local f,Q, c,d,cc,p,q,r;
f:=false:c:=2*s^2+1:
cc:=(c^2-1)/2;
Q:=numtheory[divisors](cc):
for d in Q do q:=d+c:
if isprime(q) then
p:=c+cc/(q-c): if pA236255(s) then print(s) fi od:

A306430 Twelve-column table read by rows: Primitive distinct 12-tuples that have the same value of phi, sigma, and tau.

Original entry on oeis.org

38848397472, 38850486624, 39955028960, 41002131936, 41266440252, 41271882804, 41273232372, 42433832980, 42439669580, 42441951860, 43545898068, 44591929866

Offset: 1

Jud McCranie, Feb 14 2019

The terms are consecutive 12-tuples, ordered so that (A) a(12i-11) < a(11i-10) < ... < a(12i) for i > 0, and (B) a(12i+1) < a(12i+13) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all of the terms in a primitive 12-tuple, then multiplying the terms in that 12-tuple by k gives another solution.

			38848397472, 38850486624, 39955028960, 41002131936, 41266440252, 41271882804, 41273232372, 42433832980, 42439669580, 42441951860, 43545898068, and 44591929866 have the same value of phi (12070840320), sigma (121347072000), and tau (384), so these twelve numbers are in the sequence.

Cf. A134922, A322687, A322688, A322689, A322690, A322691, A322692, A322693, A322694, A322695, A322696, A322697.

A336485 The least positive integer k for which there exist primes p, q, r such that phi(pqs^k) = phi(rs^(2k+1)) and sigma(pqs^k) = sigma(rs^(2k+1)), where s is the n-th prime.

Original entry on oeis.org

1, 1, 1, 1, 17, 3, 29, 4, 4, 4, 1, 5, 4, 1, 20, 32, 2, 38, 12, 29, 9, 4, 26, 20, 8, 14, 2, 14, 8, 41, 4

Offset: 1

Vladimir Letsko, Jul 23 2020

This sequence is an inversion of A336486.
A236255 contains the primes s = prime(m) for which a(m) = 1.
Let "a and b are similar" mean that for positive integers a and b we have phi(a) = phi(b), tau(a) = tau(b) and sigma(a) = sigma(b).
Conjecture: For each prime s there are infinitely many positive integers k such that numbers p*q*s^k and r*s^(2k+1) are similar for some primes p, q, r.

			a(6) = 3 because:
1. For the 6th prime, s = 13, k = 3 and with primes p = 62807837, q = 57149, r = 125672849 we have phi(p*q*s^k) = phi(r*s^(2k+1)) and sigma(p*q*s^k) = sigma(r*s^(2k+1)).
2. There is no such equality for s = 13 and k less than 3.

Cf. A000005, A000010, A000203, A134922, A336486.

with(NumberTheory):
SK := []; for ii to 31 do s := ithprime(ii); tf := false;
for k do c := 2*s^(k+1)+1; cc := (c^2-1)*(1/2); Q := Divisors(cc);
for d in Q do q := d+c; if isprime(q) then p := c+cc/(q-c); if p < q then break end if;
if isprime(p) then r := 2*(p+q)-c; if isprime(r) then print([s, [p, q], r], k); SK := [op(SK), [s, k]]; tf := true; break end if end if end if end do;
if tf then break end if end do end do; SK

is(t, u, x, y) = ispseudoprime(t*x+1) && ispseudoprime(u*y/t+1) && ispseudoprime(x*y+1);
a(n) = {my(s=prime(n), t, u); for(k=1, oo, for(i=0, 1+k\2, t=s^i; fordiv(2*(1+u=s^(k+1)), d, if(is(t, u, 2*u/t+d, 2*t+(2*u+2)/d) || is(t, u, 2*u/t-d, 2*t-(2*u+2)/d), return(k))))); } \\ Jinyuan Wang, Sep 30 2020

A336486 The least prime s for which there exist primes p, q, r such that phi(pqs^n) = phi(rs^(2n+1)) and sigma(pqs^n) = sigma(rs^(2n+1)).

Original entry on oeis.org

2, 2, 2, 2, 5, 37, 13, 7, 2, 19, 7, 7, 2, 19, 4447, 2, 2, 2, 2, 5, 73, 23, 37, 2, 2, 19, 19, 2, 2

Offset: 1

Vladimir Letsko, Jul 23 2020

This sequence is an inversion of A336485.
Let "a and b are similar" mean that for positive integers a and b we have phi(a) = phi(b), tau(a) = tau(b) and sigma(a) = sigma(b).
Conjecture: For each positive integer n there are infinitely many primes s such that numbers p*q*s^n and r*s^(2n+1) are similar for some primes p, q, r.

			a(2) = 2 because for n = 2 and prime s = 2, a = 19*89*s^n and b = 199*s^(2n+1), we have phi(a) = phi(b) and sigma(a) = sigma(b).

Cf. A000005, A000010, A000203, A134922, A336485.

with(NumberTheory):
KS := []; for k to 29 do tf := false;
for ii do s := ithprime(ii); c := 2*s^(k+1)+1; cc := (c^2-1)*(1/2); Q := Divisors(cc);
for d in Q do q := d+c; if isprime(q) then p := c+cc/(q-c); if p < q then break end if;
if isprime(p) then r := 2*(p+q)-c; if isprime(r) then print([k, [p, q], r], s); KS := [op(KS), s]; tf := true; break end if end if end if end do;
if tf then break end if end do end do; KS

is(t, u, x, y) = ispseudoprime(t*x+1) && ispseudoprime(u*y/t+1) && ispseudoprime(x*y+1);
a(n) = {my(s=1, t, u); while(s=nextprime(s+1), for(i=0, 1+n\2, t=s^i; fordiv(2*(1+u=s^(n+1)), d, if(is(t, u, 2*u/t+d, 2*t+(2*u+2)/d) || is(t, u, 2*u/t-d, 2*t-(2*u+2)/d), return(s))))); } \\ Jinyuan Wang, Sep 30 2020

cp's OEIS Frontend

A236255 Prime numbers s for which there exist primes p, q, r such that phi(pqs)=phi(rs^3), sigma(pqs)=sigma(rs^3).

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A306430 Twelve-column table read by rows: Primitive distinct 12-tuples that have the same value of phi, sigma, and tau.

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A336485 The least positive integer k for which there exist primes p, q, r such that phi(pqs^k) = phi(rs^(2k+1)) and sigma(pqs^k) = sigma(rs^(2k+1)), where s is the n-th prime.

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A336486 The least prime s for which there exist primes p, q, r such that phi(pqs^n) = phi(rs^(2n+1)) and sigma(pqs^n) = sigma(rs^(2n+1)).

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cp's OEIS Frontend

A236255 Prime numbers s for which there exist primes p, q, r such that phi(pqs)=phi(rs^3), sigma(pqs)=sigma(rs^3).

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Maple

A306430 Twelve-column table read by rows: Primitive distinct 12-tuples that have the same value of phi, sigma, and tau.

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Author

Keywords

Comments

Examples

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A336485 The least positive integer k for which there exist primes p, q, r such that phi(p*q*s^k) = phi(r*s^(2k+1)) and sigma(p*q*s^k) = sigma(r*s^(2k+1)), where s is the n-th prime.

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A336486 The least prime s for which there exist primes p, q, r such that phi(p*q*s^n) = phi(r*s^(2n+1)) and sigma(p*q*s^n) = sigma(r*s^(2n+1)).

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A336485 The least positive integer k for which there exist primes p, q, r such that phi(pqs^k) = phi(rs^(2k+1)) and sigma(pqs^k) = sigma(rs^(2k+1)), where s is the n-th prime.

A336486 The least prime s for which there exist primes p, q, r such that phi(pqs^n) = phi(rs^(2n+1)) and sigma(pqs^n) = sigma(rs^(2n+1)).