A134808 -id:A134808 - cp's OEIS frontend

A160711 Cyclops squares: squares (A000290) that are also cyclops numbers (A134808).

0, 11025, 19044, 21025, 24025, 32041, 38025, 42025, 47089, 51076, 58081, 59049, 65025, 66049, 67081, 75076, 87025, 93025, 1110916, 1140624, 1170724, 1190281, 1240996, 1270129, 1290496, 1340964, 1350244, 1380625, 1420864, 1430416

Offset: 1

Omar E. Pol, Jun 08 2009

			19044 is in the sequence because it is a square (138^2) and is also a cyclops number (odd number of digits, middle digit is the only zero).
11025 is in the sequence because it is a square (105^2) and is also a cyclops number (odd number of digits, middle digit is the only zero). - _Michael B. Porter_, Jul 09 2016

Kenny Lau, Table of n, a(n) for n = 1..20001

Cf. A000290, A134808, A134809, A134148, A138131, A153806, A160712, A160717, A239828.

Select[Range[0, 1200]^2, And[OddQ@ Length@ #, #[[Ceiling[Length[#]/2]]] == 0, Count[#, 0] == 1] &@ IntegerDigits@ # &] (* Michael De Vlieger, Jul 08 2016 *)
cnQ[n_]:=Module[{len=IntegerLength[n]},OddQ[len]&&DigitCount[n,10,0]==1 && IntegerDigits[n][[(len+1)/2]]==0]; Join[{0},Select[Range[1200]^2,cnQ]] (* Harvey P. Dale, Mar 19 2018 *)

A160712 Composite cyclops numbers (A134808).

Original entry on oeis.org

102, 104, 105, 106, 108, 201, 202, 203, 204, 205, 206, 207, 208, 209, 301, 302, 303, 304, 305, 306, 308, 309, 402, 403, 404, 405, 406, 407, 408, 501, 502, 504, 505, 506, 507, 508, 602, 603, 604, 605, 606, 608, 609, 702, 703, 704

Offset: 1

Omar E. Pol, Jun 08 2009

Cf. A134808, A134809, A134148, A138131, A153806, A160711, A160717.

Edited by Omar E. Pol, Jul 04 2009

A285845 Powers (A001597) that are also cyclops numbers (A134808).

Original entry on oeis.org

11025, 19044, 21025, 24025, 32041, 38025, 42025, 47089, 51076, 58081, 59049, 65025, 66049, 67081, 74088, 75076, 87025, 93025, 1110916, 1140624, 1170724, 1190281, 1240996, 1270129, 1290496, 1340964, 1350244, 1380625, 1420864, 1430416, 1490841, 1510441

Offset: 1

Colin Barker, Apr 27 2017

The first term not in A160711 is 74088 = 42^3.

Intersection of A001597 and A134808. - Robert G. Wilson v, Apr 27 2017

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 1156 terms from Robert G. Wilson v)

Cf. A001597, A134808, A160711.

Select[NestList[If[# == 1, 4, Min@ Table[(Floor[#^(1/k)] + 1)^k, {k, 2, 1 + Floor@ Log2@ #}]] &, 1, 1400], Function[n, And[OddQ@ Length@ #, #[[ Ceiling[Length[#]/2] ]] == 0, DigitCount[n, 10, 0] == 1] &@ IntegerDigits@ n]] (* Michael De Vlieger, Apr 27 2017, after Robert G. Wilson v at A001597 *)
cyclopsQ[n_Integer, b_: 10] := Module[{digitList = IntegerDigits[n, b], len, pos0s, flag}, len = Length[digitList]; pos0s = Select[Range[len], digitList[[#]] == 0 &]; flag = OddQ[len] && (Length[pos0s] == 1) && (pos0s == {(len + 1)/2}); Return[flag]]; (* from Alonso del Arte in A134808 *) min = 0; max = 1520000; t = Union@ Flatten@ Table[n^expo, {expo, Prime@ Range@ PrimePi@ Log2@ max}, {n, Floor[1 + min^(1/expo)], max^(1/expo)}]; Select[t, cyclopsQ] (* Robert G. Wilson v, Apr 27 2017 *)

is_cyclops(k) = {
  if(k==0, return(1));
  my(d=digits(k), j);
  if(#d%2==0 || d[#d\2+1]!=0, return(0));
  for(j=1, #d\2, if(d[j]==0, return(0)));
  for(j=#d\2+2, #d, if(d[j]==0, return(0)));
  return(1)}
L=List(); for(n=1, 100000, if(ispower(n) && is_cyclops(n), listput(L, n))); Vec(L)

A286662 Numbers k such that k, k^2 and k^3 are cyclops numbers (A134808).

Original entry on oeis.org

0, 16075, 18039, 1130239, 1130363, 1130668, 1150474, 1220156, 1230423, 1250928, 1290628, 1330162, 1350478, 1390313, 1390989, 1510414, 1510712, 1530314, 1530461, 1530585, 1540896, 1540977, 1560186, 1560324, 1570341, 1580342, 1620244, 1620389, 1630871, 1650288

Offset: 1

Colin Barker, May 12 2017

For k = 1130239, k^4 = 1631853457220539336688641 is also a cyclops number.

			16075 is in the sequence because k^2 = 258405625, k^3 = 4153870421875 and these three numbers are cyclops numbers.

Cf. A000290, A000578, A134808, A160711, A239587, A239588, A239589, A239590, A239591, A239827, A239828.

cycQ[n_]:=DigitCount[n,10,0]==1&&OddQ[IntegerLength[n]]&& IntegerDigits[ n][[(IntegerLength[n]+1)/2]]==0; Join[{0},Table[Select[Range[ 10^n, 10^(n+1)-1],AllTrue[{#,#^2,#^3},cycQ]&],{n,2,6,2}]]//Flatten (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Oct 25 2017 *)

is_cyclops(k) = {
  if(k==0, return(1));
  my(d=digits(k), j);
  if(#d%2==0 || d[#d\2+1]!=0, return(0));
  for(j=1, #d\2, if(d[j]==0, return(0)));
  for(j=#d\2+2, #d, if(d[j]==0, return(0)));
  return(1)}
L=List(); for(n=0, 10000000, if(is_cyclops(n) && is_cyclops(n^2) && is_cyclops(n^3), listput(L, n))); Vec(L)

A357033 a(n) is the smallest number that has exactly n divisors that are cyclops numbers (A134808).

Original entry on oeis.org

1, 101, 202, 404, 606, 1212, 2424, 7272, 21816, 41208, 84048, 123624, 144144, 336336, 288288, 504504, 432432, 865368, 864864, 1009008, 2378376, 1729728, 3459456, 3027024, 4756752, 6054048, 9081072, 11099088, 12108096, 16648632, 23207184, 29405376, 36324288

Offset: 0

Marius A. Burtea, Sep 20 2022

			The divisors of 101 are 1 and 101. Of those, only 101 is a cyclops number; it is the smallest cyclops number, so a(1) = 101.
The divisors of 202 are 1, 2, 101, and 202, the cyclops numbers being 101 and 202, so a(2) = 202.
The divisors of 404 are 1, 2, 4, 101, 202, and 404, the cyclops numbers being 101, 202 and 404, so a(3) = 404.

Cf. A134808.

ints:=func; cyc:=func; a:=[]; for n in [0..32]  do k:=1; while #[s:s in Divisors(k)| cyc(s)] ne n do k:=k+1; end while; Append(~a,k); end for; a;

L:= Vector(10^8):
C:= [0]:
for d from 3 to 7 by 2 do
  C:= [seq(seq(seq(a*10^(d-1)+10*b+c,c=1..9),b=C),a=1..9)];
  for x in C do
    Mx:= [seq(i,i=x..10^8,x)];
    L[Mx]:= map(`+`,L[Mx],1)
  od;
od:
V:= Array(0..max(L)):
for n from 1 to 10^8 do
  if V[L[n]] = 0 then V[L[n]]:= n; fi
od:
if member(0,V,'k') then convert(V[0..k-1],list)
else convert(V,list)
fi; # Robert Israel, Sep 20 2022

cyclopQ[n_] := Module[{d = IntegerDigits[n], len}, OddQ[len = Length[d]] && Position[d, 0] == {{(len + 1)/2}}]; f[n_] := DivisorSum[n, 1 &, cyclopQ[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[10, 10^5] (* Amiram Eldar, Sep 26 2022 *)

A138148 Cyclops numbers with binary digits only.

Original entry on oeis.org

0, 101, 11011, 1110111, 111101111, 11111011111, 1111110111111, 111111101111111, 11111111011111111, 1111111110111111111, 111111111101111111111, 11111111111011111111111, 1111111111110111111111111, 111111111111101111111111111, 11111111111111011111111111111

Offset: 0

Omar E. Pol, Mar 18 2008

All members are palindromes A002113. The first five members are mentioned in A129868.
Also, binary representation of A129868.
a(A090748(n)) is equal to A138831(n), the n-th perfect number minus 1, written in base 2.
Except for the first term (replace 0 with 1) the binary representation of the n-th iteration of the elementary cellular automaton, Rule 219 starting with a single ON (black) cell. - Robert Price, Feb 21 2016
a(1) = 101 is only prime number in this sequence since a(n) = (10^(n+1)+1)*(10^n-1)/9. - Altug Alkan, May 11 2016

			n ........ a(n) .... A129868(n): value of a(n) read in base 2.
0 ......... 0 ......... 0
1 ........ 101 ........ 5
2 ....... 11011 ....... 27
3 ...... 1110111 ...... 119
4 ..... 111101111 ..... 495
5 .... 11111011111 .... 2015
6 ... 1111110111111 ... 8127

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Gennady Eremin, Partitioning the set of natural numbers into Mersenne trees and into arithmetic progressions; Natural Matrix and Linnik's constant, arXiv:2405.16143 [math.CO], 2024. See pp. 9, 14.
Brady Haran and Simon Pampena, Glitch Primes and Cyclops Numbers, Numberphile video, video (2015).
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Index to Elementary Cellular Automata
Index entries for sequences related to cellular automata
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cyclops numbers: A134808. Cf. A002113, A129868.
Cf. A000396, A090748, A135627, A138831.
Cf. A002275 (repunits R_n = (10^n-1)/9), A011557 (10^n).

[(-1 - 9*10^n + 10^(1 + 2*n))/9 : n in [0..15]]; // Wesley Ivan Hurt, Dec 08 2015

A138148:=n->(-1-9*10^n+10^(1+2*n))/9: seq(A138148(n), n=0..15); # Wesley Ivan Hurt, Dec 08 2015

Table[(-1 - 9*10^n + 10^(1 + 2*n))/9, {n, 0, 15}] (* Wesley Ivan Hurt, Dec 08 2015 *)

for(n=1, 20, if(n%2==1, c=((10^n-1)/9)-10^((n-1)/2); print1(c, ", "))) \\ Felix Fröhlich, Jul 07 2014

apply( {A138148(n)=10^(n*2+1)\9-10^n}, [0..15]) \\ M. F. Hasler, Feb 08 2020

From Colin Barker, Feb 21 2013: (Start)
a(n) = (-1-9*10^n+10^(1+2*n))/9.
G.f.: x*(200*x-101) / ((x-1)*(10*x-1)*(100*x-1)). (End)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2. - Wesley Ivan Hurt, Dec 08 2015
a(n) = A000533(n+1)*A002275(n). - Altug Alkan, May 12 2016
E.g.f.: (-1 - 9*exp(9*x) + 10*exp(99*x))*exp(x)/9. - Ilya Gutkovskiy, May 12 2016
a(n) = A002275(2n+1) - A011557(n). - M. F. Hasler, Feb 08 2020

More terms from Omar E. Pol, Feb 09 2020

A134809 Cyclops primes.

Original entry on oeis.org

101, 103, 107, 109, 307, 401, 409, 503, 509, 601, 607, 701, 709, 809, 907, 11027, 11047, 11057, 11059, 11069, 11071, 11083, 11087, 11093, 12011, 12037, 12041, 12043, 12049, 12071, 12073, 12097, 13033, 13037, 13043, 13049, 13063

Offset: 1

Omar E. Pol, Nov 25 2007

Cyclops numbers that are prime numbers: primes with an odd number of digits with middle digit 0 that have only one digit 0.
The only known Fibonacci number in this sequence is 99194853094755497 (see A005478 and A182809).
The only known Lucas number in this sequence is 688846502588399 (see A005479 and A182811).

Harvey P. Dale, Table of n, a(n) for n = 1..10000
G. L. Honaker, Jr. and Chris K. Caldwell, Prime Curios! 688846502588399
G. L. Honaker, Jr. and Chris K. Caldwell, Prime Curios! 99194853094755497

Intersection of prime numbers A000040 and cyclops numbers A134808.

Cf. A005478, A005479, A056709, A069675, A182809, A182811.

(* First run the program given for A134808 *) Select[Prime[Range[2000]], cyclopsQ] (* Alonso del Arte, Dec 16 2010 *)
cycQ[n_]:=Module[{idn=IntegerDigits[n],len},len=Length[idn];OddQ[len] && Count[idn,0] == 1 && idn[[(len+1)/2]]==0]; Select[Flatten[Table[Prime[ Range[ PrimePi[10^(2n)+1],PrimePi[10^(2n+1)]]],{n,2}]],cycQ] (* Harvey P. Dale, Jun 20 2014 *)

# cyclops() in A134808
from sympy import isprime
print([c for c in cyclops(upto=13063) if isprime(c)]) # Michael S. Branicky, Jan 05 2021

Links added by Omar E. Pol, Mar 25 2011

A129868 Binary palindromic numbers with only one 0 bit.

Original entry on oeis.org

0, 5, 27, 119, 495, 2015, 8127, 32639, 130815, 523775, 2096127, 8386559, 33550335, 134209535, 536854527, 2147450879, 8589869055, 34359607295, 137438691327, 549755289599, 2199022206975, 8796090925055, 35184367894527, 140737479966719, 562949936644095

Offset: 0

Zak Seidov, May 24 2007

Binary expansion is 0, 101, 11011, 1110111, 111101111, ... (see A138148).
9 + 8a(n) = s^2 is a perfect square with s = 2^(n + 2) -1 = 3, 7, 15, 31, 63, ...
Numbers with middle bit 0, that have only one bit 0, and the total number of bits is odd.
The fractional part of the base 2 logarithm of a(n) approaches 1 as n approaches infinity.
Also called binary cyclops numbers.
Last digit of the decimal representation follows the pattern 5, 7, 9, 5, 5, 7, 9, 5, ... . - Alex Ratushnyak, Dec 08 2012

Robert Israel, Table of n, a(n) for n = 0..1630
Gennady Eremin, Partitioning the set of natural numbers into Mersenne trees and into arithmetic progressions; Natural Matrix and Linnik's constant, arXiv:2405.16143 [math.CO], 2024. See pp. 9, 14.
Brady Haran and Simon Pampena, Glitch Primes and Cyclops Numbers, Numberphile video, video (2015)
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Base 10 analog is A134808.
Binary palindromic numbers, including repunits (or Mersenne numbers A000225) are in A006995. The sequence of binary pandigital (having both 0's and 1's) palindromic numbers begins 5, 9, 17, 21, 27, 33, 45, 51, 65, 73, ...
Cf. A006516, A138148.

[2^(2*n+1)-2^n-1: n in [0..25]]; // Vincenzo Librandi, Dec 08 2015

A129868:=n->2^(2*n + 1) - 2^n - 1: seq(A129868(n), n=0..30); # Wesley Ivan Hurt, Dec 08 2015

(* 1st *) FromDigits[ #,2]&/@NestList[Append[Prepend[ #, 1], 1]&, {0}, 25] (* 2nd *) NestList[(1/2)(7 + 8# + Sqrt[9 + 8# ])&, 0, 22] (* both of these are from Zak Seidov *)
f[n_] := 2^(2n + 1) - 2^n - 1; Table[f@n, {n, 0, 22}] (* Robert G. Wilson v, Aug 24 2007 *)
Table[EulerE[2, 2^n], {n, 1, 60}]/2 - 1 (* Vladimir Joseph Stephan Orlovsky, Nov 03 2009 *)
(* After running the program in A134808 *) Select[Range[0, 2^16 - 1], cyclopsQ[#, 2] &] (* Alonso del Arte, Dec 17 2010 *)
LinearRecurrence[{7, -14, 8}, {0, 5, 27}, 30] (* Vincenzo Librandi, Dec 08 2015 *)

concat(0, Vec(x*(5-8*x)/(1-7*x+14*x^2-8*x^3) + O(x^100))) \\ Altug Alkan, Dec 08 2015

def A129868(n): return ((m:=1<Chai Wah Wu, Mar 19 2024

a(n) = 2^(2n + 1) - 2^n - 1 = 2*4^n - 2^n - 1 = (2^n - 1)(2*2^n + 1).
G.f.: x*(8*x-5)/((x-1)*(2*x-1)*(4*x-1)).
Recurrences:
a(n) = (1/2)*(7 + 8*a(n - 1) + sqrt(9 + 8*a(n - 1))), a(0) = 0;
a(n) = 6*a(n - 1) - 8*a(n - 2) - 3, a(0) = 0, a(1) = 5;
a(n) = 7*a(n - 1) - 14*a(n - 2) + 8*a(n - 3), a(0) = 0, a(1) = 5, a(2) = 27.
a(n) = A006516(n+1) - 1.

A262259 Numbers k such that the symmetric representation of sigma(k) has only two parts and they meet at the center of the Dyck path.

Original entry on oeis.org

3, 10, 78, 136, 666, 820, 1830, 2628, 4656, 5886, 6328, 16290, 18528, 28920, 32896, 39340, 48828, 56616, 62128, 78606, 80200, 83436, 88410, 93528, 100576, 104196, 135460, 146070, 166176, 180300, 187578, 190036

Offset: 1

Hartmut F. W. Hoft, Sep 16 2015

For a proof of the formula see the link and also the links in A239929 and A071561. This formula allows for a fast computation of numbers in the sequence that does not require computations of Dyck paths.
Subsequence of A239929.
A191363 is a subsequence.
All terms are triangular numbers.
More precisely, all terms are second hexagonal numbers (A014105). There are no terms with middle divisors. - Omar E. Pol, Oct 31 2018
Numbers k such that the concatenation of the widths of the symmetric representation of sigma(k) is a cyclops numbers (A134808). - Omar E. Pol, Aug 29 2021

			q = 128 = 2^7 is the 15th term in A174973 for which 2*n+1 = 2^8 + 1 is prime so that a(15) = 2^7 * (2^8 + 1) = 32896. The two parts in the symmetric representation of sigma of a(15) have width 1 and sigma(a(15)) = 2 * a(15) - 2.
q = 308 is the 32nd term in A174973 for which 2*n+1 is prime so that a(32) = 308 * 617 = 190036. The maximum width of the two regions is 2 and sigma(a(32)) = 415296.
For n = 21, the symmetric representation of sigma(21) has two parts that meet at the center of the Dyck path, but 21 is not in the sequence because the symmetric representation of sigma(21) has more than two parts. - _Omar E. Pol_, Sep 18 2015
From _Omar E. Pol_, Oct 05 2015: (Start)
Illustration of initial terms (n = 1, 2):
. y
.  |
.  |_ _ _ _ _ _
.  |_ _ _ _ _  |
.  |         | |_
.  |         |_ _|_
.  |             | |_ _
.  |             |_ _  |
.  |                 | |
.  |_ _              | |
.  |_ _|_            | |
.  |   | |           | |
.  |_ _|_|_ _ _ _ _ _|_|_ _ x
.       3             10
.
The symmetric representation of sigma(3) = 2 + 2 = 4 has two parts and they meet at the point (2, 2), so a(1) = 3.
The symmetric representation of sigma(10) = 9 + 9 = 18 has two parts and they meet at the point (7, 7), so a(2) = 10.
(End)
Also 10 is in the sequence because the concatenation of the widths of the symmetric representation of sigma(10) is 1111111110111111111 and it is a cyclops number (A134808). - _Omar E. Pol_, Aug 29 2021

Hartmut F. W. Hoft, proof of formula

Cf. A000203, A000217, A014105, A071561, A134808, A174973, A191363, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A239929, A245092, A249351 (widths), A262045, A262048, A262626.

(* test for membership in A174973 *)
a174973Q[n_]:=Module[{d=Divisors[n]}, Select[Rest[d] - 2 Most[d], #>0&]=={}]
a174973[n_]:=Select[Range[n], a174973Q]
(* compute numbers in the sequence *)
a262259[n_]:=Map[#(2#+1)&, Select[a174973[1, n], PrimeQ[2#+1]&]]
a262259[308] (* data *)

Terms are equal to q*(2*q + 1) where q is in A174973 and 2*q + 1 is prime.

A138131 Palindromic cyclops numbers.

Original entry on oeis.org

0, 101, 202, 303, 404, 505, 606, 707, 808, 909, 11011, 12021, 13031, 14041, 15051, 16061, 17071, 18081, 19091, 21012, 22022, 23032, 24042, 25052, 26062, 27072, 28082, 29092, 31013, 32023, 33033, 34043, 35053, 36063, 37073, 38083

Offset: 1

Omar E. Pol, Mar 09 2008

For prime entries in the sequence see A136098. - Lekraj Beedassy, Mar 15 2008, May 21 2008

			101 is a member because 101 is a palindromic number A002113 and also a cyclops number A134808.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A002113, A002385, A134808.

f:= proc(n,d) local L,m,k;
  L:= convert(9^d+n,base,9);
  add((1+L[d+1-i])*(10^(i-1)+10^(2*d+1-i)),i=1..d)
end proc:
seq(seq(f(n,d),n=0..9^d-1),d=0..2); # Robert Israel, Feb 18 2018

Join[{0},Flatten[Table[Select[Range[10^(2n),10^(2n+1)-1],PalindromeQ[ #] && DigitCount[ #,10,0]==1&&IntegerDigits[#][[(IntegerLength[#]+1)/2]]==0&],{n,2}]]] (* Harvey P. Dale, Dec 03 2022 *)

cp's OEIS Frontend

A160711 Cyclops squares: squares (A000290) that are also cyclops numbers (A134808).

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A160712 Composite cyclops numbers (A134808).

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A285845 Powers (A001597) that are also cyclops numbers (A134808).

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A286662 Numbers k such that k, k^2 and k^3 are cyclops numbers (A134808).

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A357033 a(n) is the smallest number that has exactly n divisors that are cyclops numbers (A134808).

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A138148 Cyclops numbers with binary digits only.

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A134809 Cyclops primes.

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A129868 Binary palindromic numbers with only one 0 bit.

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