cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 51-56 of 56 results.

A332530 Numbers k such that k and k + 1 has the same value of A319445, the equivalent of the Euler totient function in the ring of Eisenstein integers.

Original entry on oeis.org

34, 51, 152, 679, 1065, 1845, 6525, 12122, 12970, 15656, 38607, 48398, 175473, 272935, 401505, 953342, 1035895, 1210054, 1222988, 1406665, 1589245, 1607095, 2108186, 2116975, 2272425, 2500615, 2751160, 3399591, 4542225, 5298559, 5412986, 6813585, 6898736, 7115553
Offset: 1

Views

Author

Amiram Eldar, Feb 15 2020

Keywords

Examples

			34 is a term since A319445(34) = A319445(35) = 864.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[p == 3, 2*3^(2*e - 1), Switch[Mod[p, 3], 1, (p - 1)^2*p^(2*e - 2), 2, (p^2 - 1)*p^(2*e - 2)]]; eisPhi[1] = 1; eisPhi[n_] := Times @@ f @@@ FactorInteger[n]; seq = {}; e1 = eisPhi[1]; Do[e2 = eisPhi[n]; If[e1 == e2, AppendTo[seq, n - 1]]; e1 = e2, {n, 2, 10^6}]; seq

A336673 Numbers k such that A063659(k) = A063659(k+1).

Original entry on oeis.org

3, 49, 1681, 18490, 23762, 656914, 843637, 5606230, 35558770, 46297822, 59006794, 114594493, 132859642, 138852445, 157906534, 289405462, 299441785, 536671282, 813736930, 1175272581, 1276553470, 1655870629, 5086602202, 5429407657, 6549516022, 8645559934, 10373399185
Offset: 1

Views

Author

Amiram Eldar, Jul 29 2020

Keywords

Comments

Analogous to A001274 as A063659 is analogous to Euler's phi function (A000010).

Examples

			3 is a term since A063659(3) = A063659(4) = 3.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, p, p^e - p^(e-2)]; s[n_] := Times @@ f @@@ FactorInteger[n]; s1 = 1; seq = {}; Do[s2 = s[n]; If[s1 == s2, AppendTo[seq, n-1]]; s1 = s2, {n, 2, 10^5}]; seq

A385743 Numbers k such that A384247(k) = A384247(k+1).

Original entry on oeis.org

1, 20, 27, 35, 63, 64, 104, 143, 194, 208, 740, 836, 1220, 1299, 1419, 1803, 1892, 2625, 3255, 3705, 3716, 3843, 4096, 5184, 5186, 5635, 5695, 7868, 10659, 13365, 16904, 17948, 18507, 18914, 21007, 22935, 25388, 25545, 27675, 30380, 31599, 32304, 32864, 34595
Offset: 1

Views

Author

Amiram Eldar, Jul 08 2025

Keywords

Comments

63 is the only number k below 10^11 such that A384247(k) = A384247(k+1) = A384247(k+2). Are there any other such terms?

Examples

			1 is a term since A384247(1) = A384247(2) = 1.
20 is a term since A384247(20) = A384247(21) = 12.
		

Crossrefs

Cf. A384247.
Similar sequences: A001274, A287055, A293184, A301866, A326403, A349307.

Programs

  • Mathematica
    f[p_, e_] := p^e*(1 - 1/p^(2^(IntegerExponent[e, 2]))); iphi[1] = 1; iphi[n_] := iphi[n] = Times @@ f @@@ FactorInteger[n]; Select[Range[35000], iphi[#] == iphi[# + 1] &]
  • PARI
    iphi(n) = {my(f = factor(n)); n * prod(i = 1, #f~, (1 - 1/f[i, 1]^(1 << valuation(f[i, 2], 2)))); }
    list(lim) = {my(s1 = iphi(1), s2); for(k = 2, lim, s2 = iphi(k); if(s1 == s2, print1(k-1, ", ")); s1 = s2);}

A063100 Compute the cototient function for the g(n) = p(n+1)-p(n)-1 composite numbers between two consecutive primes. Let the number of distinct cototient values be c(n). Then, a(n) = g(n)-c(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Labos Elemer, Aug 07 2001

Keywords

Comments

a(n) = 0 means that all cototients in the gap are different, while 1, 2, or more means that 1 or more times inside the gap equal cototients occur.
Unlike totient, where phi(n+1) = phi(n) may occur (see A001274), cototients of consecutive numbers are different for n<1000000. At cases x, of A001274, cototient(x+1) = 1+cototient(x).

Examples

			Case 1: a(n) = 0; primes = {229, 233}; primes and gap = {229, 230, 231, 232, 233}; cototients = {1, 142, 111, 120, 1}, all cototients inside gap are different, thus a(n) = 0 for p(n) = p(40) = 229 prime.
Case 2: a(n) = 1; primes = {113, 127}; gap = {113, 114, 115, ..., 125, 126, 127}; cototients = {1, 78, 27, 60, 45, 60, 23, 88, 11, 62, 43, 64, 25, 90, 1}; seemingly 60 occurs twice, so a(n) = a(30) = g(n)-c(n) = 13-12 = 1.
Case 3: a(n) = 3, primes = {2861, 2879}, gap = {2861, 2862, ..., 2878, 2879}; cototients = {1, 1926, 415, 1440, 1345, 1434, 107, 1916, 169, 1910, 1191, 1440, 377, 1918, 675, 1440, 1245, 1440, 1}; observe that 1440 occurs four times, so a(n) = 3.
		

Crossrefs

Cf. A051953 (cototient function), A000010, A001274, A061106.

A139766 A number n is included in the sequence if and only if the n-th integer from among those positive integers which are coprime to n+1 = the (n+1)-st integer from among those positive integers which are coprime to n.

Original entry on oeis.org

3, 15, 104, 164, 255, 2625, 2834, 11715, 18315, 48704, 49215, 64004, 65535, 73124, 131144, 215775, 491535, 525986, 546272, 568815, 952575, 1925564, 5781434, 5861583, 13496384, 14409548, 17646615, 17949434, 20171384, 21475124, 22632285
Offset: 1

Views

Author

Leroy Quet, Nov 07 2007

Keywords

Comments

So far it appears that this is a proper subset of A001274.

Crossrefs

Formula

Indices n such that A126356(n) = A126357(n). - Ray Chandler

Extensions

More terms from Stefan Steinerberger, Nov 07 2007
a(8) onwards from Ray Chandler, Jul 01 2009

A303820 a(n) begins the first run of least n consecutive numbers whose phi values have the same set of distinct prime divisors.

Original entry on oeis.org

1, 1, 3, 3, 35, 43570803, 22154517001
Offset: 1

Views

Author

Amiram Eldar, May 01 2018

Keywords

Comments

a(n) is the least k such that rad(phi(k)) = rad(phi(k+1)) = ... = rad(phi(k+n-1)), where rad(n) is the squarefree kernel of n (A007947) and phi(n) is the Euler totient function of n (A000010).

Examples

			a(5) = 35 since it is the least number such that
phi(35) = 24 = 2^3 * 3^1,
phi(36) = 12 = 2^2 * 3^1,
phi(37) = 36 = 2^2 * 3^2,
phi(38) = 18 = 2^1 * 3^2,
phi(39) = 24 = 2^3 * 3^1,
all having the same set of prime divisors: 2 and 3.
		

Crossrefs

Programs

  • Mathematica
    rad[n_] := Times @@ (First@# & /@ FactorInteger@n); radphi[n_] := rad[ EulerPhi[n] ]; Seq[n_, q_] := Map[radphi, Range[n, n + q - 1]]; findConsec[q_, nmin_, nmax_] := Module[{}, s = Seq[1, q]; n = q + 1; Do[If[CountDistinct[s] == 1, Break[]]; s = Rest[AppendTo[s, radphi[n]]]; n++, {k, nmin, nmax}]; n - q]; seq = {1}; nmax = 10^10; Do[n1 = Last[ seq ]; s1 = findConsec[m, n1, nmax]; AppendTo[seq, s1], {m, 2, 6}]; seq

Extensions

a(7) from Giovanni Resta, May 08 2018
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