Reinhard Muehlfeld - cp's OEIS frontend

A245599 Numbers m with A030101(m) XOR A030109(m) = m for the binary representation of m.

1, 11, 91, 731, 5851, 46811, 374491, 2995931, 23967451, 191739611, 1533916891, 12271335131, 98170681051, 785365448411, 6282923587291, 50263388698331, 402107109586651, 3216856876693211, 25734855013545691, 205878840108365531, 1647030720866924251, 13176245766935394011

Offset: 1

Reinhard Muehlfeld, Jul 27 2014

Sequence consists of all numbers with binary representation 1(011)*.

			A030101(11) = 13,  A030109(11) = 6, and 13 XOR 6 = (1101)_2 XOR (0110)_2 = (1011)_2 = 11, so 11 is in the sequence.

Colin Barker, Table of n, a(n) for n = 1..1001
Index entries for linear recurrences with constant coefficients, signature (9,-8).

Cf. A007088, A030109, A030101.

a[n_] := (5*8^n - 12)/28; Array[a, 20] (* Giovanni Resta, Apr 25 2020 *)

Vec(x*(1 + 2*x) / ((1 - x)*(1 - 8*x)) + O(x^20)) \\ Colin Barker, Apr 25 2020

a(n) = 1(011)^(n-1) in binary representation.
a(n) = (5*8^n - 12)/28. - Giovanni Resta, Apr 25 2020
From Colin Barker, Apr 25 2020: (Start)
G.f.: x*(1 + 2*x) / ((1 - x)*(1 - 8*x)).
a(n) = 9*a(n-1) - 8*a(n-2) for n>2.
(End)

More terms from Giovanni Resta, Apr 25 2020

A245471 If n is odd, then a(n) = A065621(n+1). If n is even, then a(n) = n/2.

Original entry on oeis.org

2, 1, 4, 2, 14, 3, 8, 4, 26, 5, 28, 6, 22, 7, 16, 8, 50, 9, 52, 10, 62, 11, 56, 12, 42, 13, 44, 14, 38, 15, 32, 16, 98, 17, 100, 18, 110, 19, 104, 20, 122, 21, 124, 22, 118, 23, 112, 24, 82, 25, 84, 26, 94, 27, 88, 28, 74, 29, 76, 30, 70, 31, 64, 32, 194, 33, 196, 34, 206, 35, 200, 36, 218, 37, 220, 38, 214, 39, 208, 40, 242, 41, 244, 42, 254, 43, 248, 44, 234, 45, 236, 46, 230, 47, 224, 48, 162, 49, 164, 50, 174, 51, 168, 52, 186, 53, 188, 54, 182, 55, 176, 56, 146, 57, 148, 58, 158, 59, 152, 60

Offset: 1

Reinhard Muehlfeld, Jul 23 2014

A Collatz-like function: the difference is that for odd n the term 3n+1 is calculated without overflow, only using xor operations (n xor(2n+1)). It is known that for each argument the iterated function always ends up in a cycle which contains 1 (namely 1-2-1).

Reinhard Zumkeller, Table of n, a(n) for n = 1..8192

Cf. A006370, A065621.

import Data.List (transpose)
a245471 n = a245471_list !! (n-1)
a245471_list = concat $ transpose [odds a065621_list, [1..]]
   where odds [] = []; odds [x] = []; odds (_:x:xs) = x : odds xs
-- Reinhard Zumkeller, Jul 27 2014

def A245471(n): return (m:=n+1)^ (m&~-m)<<1 if n&1 else n>>1 # Chai Wah Wu, Jun 29 2022

Definition corrected by Chai Wah Wu, Jun 29 2022

A245202 Numbers k such that tau(k) + phi(k) is a perfect square.

Original entry on oeis.org

3, 9, 21, 24, 26, 30, 51, 72, 77, 84, 90, 93, 100, 119, 122, 162, 168, 174, 194, 210, 213, 221, 276, 282, 291, 301, 381, 384, 386, 408, 414, 437, 469, 510, 527, 533, 564, 594, 597, 616, 723, 731, 744, 770, 791, 794, 858, 869, 896, 917, 930, 948, 952, 954

Offset: 1

Reinhard Muehlfeld, Jul 13 2014

nonn

Numbers k such that A000010(k) + A000005(k) is a perfect square.

			3 is in the sequence because phi(3) + tau(3) = 2 + 2 = 4^2.
9 is in the sequence because phi(9) + tau(9) = 6 + 3 = 3^2.
15 is not in the sequence because phi(15) + tau(15) = 8 + 4 = 12 = 2^2 * 3, which is not a perfect square.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Chai Wah Wu)

Cf. A000005, A000010.

Select[Range[1000], IntegerQ[Sqrt[DivisorSigma[0, #] + EulerPhi[#]]] &] (* Amiram Eldar, Apr 27 2024 *)

isok(n) = issquare(numdiv(n) + eulerphi(n)); \\ Michel Marcus, Jul 23 2014

from sympy import totient, divisor_count
from gmpy2 import is_square
[n for n in range(1,10**4) if is_square(int(divisor_count(n)+totient(n)))] # Chai Wah Wu, Aug 04 2014

A245199 Numbers n where phi(n) and tau(n) are perfect squares.

Original entry on oeis.org

1, 8, 10, 34, 57, 74, 85, 125, 185, 202, 219, 394, 451, 456, 489, 505, 514, 546, 570, 629, 640, 679, 680, 802, 985, 1000, 1026, 1057, 1154, 1285, 1354, 1365, 1387, 1417, 1480, 1717, 1752, 1938, 2005, 2016, 2047, 2176, 2190, 2340, 2457, 2509, 2565, 2594, 2649

Offset: 1

Reinhard Muehlfeld, Jul 13 2014

Numbers n such that A000005(n) and A000010(n) are perfect squares.

Intersection of A036436 and A039770. - Michel Marcus, Jul 15 2014

			8 is in the sequence because phi(8) = 4, tau(8) = 4, and 4 is a perfect square.
12 is not in the sequence because tau(12) = 6 is not a square.

Robert Israel, Table of n, a(n) for n = 1..8000

Cf. A000005, A000010, A036436, A039770.

filter:= proc(n) uses numtheory; issqr(phi(n)) and issqr(tau(n)) end proc:
select(filter, [$1..1000]); # Robert Israel, Jul 27 2014

fQ[n_] := IntegerQ@ Sqrt@ EulerPhi[n] && IntegerQ@ Sqrt@ DivisorSigma[0, n]; Select[ Range@ 3000, fQ] (* Robert G. Wilson v, Jul 21 2014 *)
Select[Range[3000],AllTrue[{Sqrt[EulerPhi[#]],Sqrt[DivisorSigma[0, #]]}, IntegerQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Apr 01 2018 *)

isok(n) = issquare(numdiv(n)) && issquare(eulerphi(n)); \\ Michel Marcus, Jul 15 2014

from sympy import totient, divisor_count
from gmpy2 import is_square
[n for n in range(1,10**4) if is_square(int(divisor_count(n))) and is_square(int(totient(n)))] # Chai Wah Wu, Aug 04 2014

A245047 Numbers n where phi(n)|n or tau(n)|n.

Original entry on oeis.org

1, 2, 4, 6, 8, 9, 12, 16, 18, 24, 32, 36, 40, 48, 54, 56, 60, 64, 72, 80, 84, 88, 96, 104, 108, 128, 132, 136, 144, 152, 156, 162, 180, 184, 192, 204, 216, 225, 228, 232, 240, 248, 252, 256, 276, 288, 296, 324, 328, 344, 348, 360, 372, 376, 384, 396, 424, 432, 441, 444, 448, 450

Offset: 1

Reinhard Muehlfeld, Jul 13 2014

nonn

Phi(n) is Euler totient function (A000010); tau(n) is the number of divisors of n (A000005).

Union of A007694 and A033950. - Michel Marcus, Jul 15 2014

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000005, A000010, A007694, A033950, A235353.

select(t -> (t mod numtheory:-phi(t) = 0) or (t mod numtheory:-tau(t) = 0), [$1..1000]); # Robert Israel, Jul 15 2014

Select[Range[500],AnyTrue[{#/EulerPhi[#],#/DivisorSigma[0,#]},IntegerQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 12 2020 *)

isok(n) = !((n % eulerphi(n)) && (n % numdiv(n))); \\ Michel Marcus, Jul 15 2014

A245197 Numbers n where tau(n) and n-tau(n) are perfect squares, with tau(n) the number of divisors of n (A000005).

Original entry on oeis.org

1, 8, 85, 125, 365, 445, 533, 629, 965, 1685, 1800, 1853, 2340, 2605, 2813, 3029, 3973, 4765, 5045, 5220, 5629, 5933, 6245, 6893, 8285, 8653, 11029, 11453, 11885, 12773, 14165, 15133, 16645, 17165, 17460, 17693, 20453, 21029, 22205, 22805, 23413, 24653, 27229

Offset: 1

Reinhard Muehlfeld, Jul 13 2014

nonn

Subsequence of A036436. - Michel Marcus, Jul 15 2014

Intersection of A036436 and A245388. - Robert Israel, Jul 20 2014

Jens Kruse Andersen, Table of n, a(n) for n = 1..2989

Cf. A000005 (tau(n)), A049820 (n - tau(n)), A036436 (n such that tau(n) is square), A245388 (n such that n - tau(n) is square).

filter:= proc(n) local t;
  t:= numtheory:-tau(n);
  issqr(t) and issqr(n-t)
end proc;
select(filter, [$1..10^5]); # Robert Israel, Jul 20 2014

okQ[n_] := With[{t = DivisorSigma[0, n]}, IntegerQ@Sqrt[t] && IntegerQ@Sqrt[n-t]];
Select[Range[10^5], okQ] (* Jean-François Alcover, Feb 08 2023 *)

isok(n) = issquare(numdiv(n)) && issquare(n-numdiv(n)); \\ Michel Marcus, Jul 15 2014

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User: Reinhard Muehlfeld

A245599 Numbers m with A030101(m) XOR A030109(m) = m for the binary representation of m.

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A245471 If n is odd, then a(n) = A065621(n+1). If n is even, then a(n) = n/2.

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A245202 Numbers k such that tau(k) + phi(k) is a perfect square.

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A245199 Numbers n where phi(n) and tau(n) are perfect squares.

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A245047 Numbers n where phi(n)|n or tau(n)|n.

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A245197 Numbers n where tau(n) and n-tau(n) are perfect squares, with tau(n) the number of divisors of n (A000005).

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