Joshua Oliver - cp's OEIS frontend

A332025 Sum of the lengths of the longest runs of 0, 1, and 2 in the ternary expression of n.

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

Offset: 0

Joshua Oliver, Feb 05 2020

All positive integers appear in this sequence. Given some number k, there will always be some ternary number that has k 1's or k 2's.

The number 0 never appears in this sequence, as every number has at least 1 digit.

			For n = 268, the ternary expansion of 268 is 100221. The length of the run of 0's in the ternary expansion of 268 is 2. The length of the runs of 1's in the ternary expansion of 268 are 1 and 1 respectively. The length of the run of 2's in the ternary expansion of 268 is 2. The sum of 2, 1, and 2 is 5, so a(268) = 5.
   n [ternary n] A330166(n) + A330167(n) + A330168(n) = a(n)
   0 [        0] 1          + 0          + 0          = 1
   1 [        1] 0          + 1          + 0          = 1
   2 [        2] 0          + 0          + 1          = 1
   3 [      1 0] 1          + 1          + 0          = 2
   4 [      1 1] 0          + 2          + 0          = 2
   5 [      1 2] 0          + 1          + 1          = 2
   6 [      2 0] 1          + 0          + 1          = 2
   7 [      2 1] 0          + 1          + 1          = 2
   8 [      2 2] 0          + 0          + 2          = 2
   9 [    1 0 0] 2          + 1          + 0          = 3
  10 [    1 0 1] 1          + 1          + 0          = 2
  11 [    1 0 2] 1          + 1          + 1          = 3
  12 [    1 1 0] 1          + 2          + 0          = 3
  13 [    1 1 1] 0          + 3          + 0          = 3
  14 [    1 1 2] 0          + 2          + 1          = 3
  15 [    1 2 0] 1          + 1          + 1          = 3
  16 [    1 2 1] 0          + 1          + 1          = 2
  17 [    1 2 2] 0          + 1          + 2          = 3
  18 [    2 0 0] 2          + 0          + 1          = 3
  19 [    2 0 1] 1          + 1          + 1          = 3
  20 [    2 0 2] 1          + 0          + 1          = 2

Wikipedia, Ternary numeral system.

Cf. A003462, A007089, A062756, A330036.

Equals A330166 + A330167 + A330168.

Table[Sum[Max@FoldList[If[#2==k,#1+1,0]&,0,IntegerDigits[n,3]],{k,0,2}],{n,1,90}]

a(n) = A330166(n) + A330167(n) + A330168(n).

a(A003462(n)) = a(A024023(n)) = n.

A330168 Length of the longest run of 2's in the ternary expression of n.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 0, 0, 1, 0, 0, 1, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 4, 0, 0, 1, 0, 0, 1, 1, 1, 2, 0

Offset: 0

Joshua Oliver, Dec 04 2019

All numbers appear in this sequence. Numbers of the form 3^n-1 (A024023(n)) have n 2's in their ternary expression.

The longest run of zeros possible in this sequence is 2, as the last digit of the ternary expression of the integers cycles between 0, 1, and 2, meaning that at least one of three consecutive numbers has a 2 in its ternary expression.

			For n = 74, the ternary expression of 74 is 2202. The length of the runs of 2's in the ternary expression of 74 are 2 and 1, respectively. The larger of these two values is 2, so a(74) = 2.
   n [ternary n] a(n)
   0 [        0] 0
   1 [        1] 0
   2 [        2] 1
   3 [      1 0] 0
   4 [      1 1] 0
   5 [      1 2] 1
   6 [      2 0] 1
   7 [      2 1] 1
   8 [      2 2] 2
   9 [    1 0 0] 0
  10 [    1 0 1] 0
  11 [    1 0 2] 1
  12 [    1 1 0] 0
  13 [    1 1 1] 0
  14 [    1 1 2] 1
  15 [    1 2 0] 1
  16 [    1 2 1] 1
  17 [    1 2 2] 2
  18 [    2 0 0] 1
  19 [    2 0 1] 1
  20 [    2 0 2] 1

Wikipedia, Ternary numeral system.

Cf. A007089, A024023, A081603, A330036, A330166, A330167.

Equals zero iff n is in A005836.

Table[Max@FoldList[If[#2==2,#1+1,0]&,0,IntegerDigits[n,3]],{n,0,90}]

a(A024023(n)) = a(3^n-1) = n.

a(n) = 0 iff n is in A005836.

A330167 Length of the longest run of 1's in the ternary expression of n.

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 3, 2, 1, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 3, 4, 3, 2, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 3, 2, 1, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1

Offset: 0

Joshua Oliver, Dec 04 2019

All numbers appear in this sequence. Numbers of the form (3^n-1)/2 (A003462(n)) have n 1's in their ternary expression.

The longest run of zeros possible in this sequence is 2, as the last digit of the ternary expression of the integers cycles between 0, 1, and 2, meaning that at least one of three consecutive numbers has a 1 in its ternary expression.

			For n = 43, the ternary expression of 43 is 1121. The length of the runs of 1's in the ternary expression of 43 are 2 and 1, respectively. The larger of these two values is 2, so a(43) = 2.
   n [ternary n] a(n)
   0 [        0] 0
   1 [        1] 1
   2 [        2] 0
   3 [      1 0] 1
   4 [      1 1] 2
   5 [      1 2] 1
   6 [      2 0] 0
   7 [      2 1] 1
   8 [      2 2] 0
   9 [    1 0 0] 1
  10 [    1 0 1] 1
  11 [    1 0 2] 1
  12 [    1 1 0] 2
  13 [    1 1 1] 3
  14 [    1 1 2] 2
  15 [    1 2 0] 1
  16 [    1 2 1] 1
  17 [    1 2 2] 1
  18 [    2 0 0] 0
  19 [    2 0 1] 1
  20 [    2 0 2] 0

Wikipedia, Ternary numeral system.

Cf. A003462, A007089, A062756, A330036, A330166, A330168.

Equals zero iff n is in A005823.

Table[Max@FoldList[If[#2==1,#1+1,0]&,0,IntegerDigits[n,3]],{n,0,90}]
Table[Max[Length/@Select[Split[IntegerDigits[n,3]],MemberQ[#,1]&]],{n,0,100}]/.(-\[Infinity]->0) (* Harvey P. Dale, Jan 07 2023 *)

a(A003462(n)) = a((3^n-1)/2) = n.

a(n) = 0 iff n is in A005823.

A330166 Length of the longest run of 0's in the ternary expression of n.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 3, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 0, 0, 1, 0, 0, 2, 1, 1, 1, 0, 0, 1, 0, 0, 4, 3, 3, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Joshua Oliver, Dec 04 2019

All numbers appear in this sequence. The n-th power of 3 (A000244(n)) has n 0's in its ternary expression.

The longest run of zeros possible in this sequence is 2, as the last digit of the ternary expression of the integers cycles between 0, 1, and 2, meaning that at least one of three consecutive numbers has a 0 in its ternary expression.

			For n = 87, the ternary expression of 87 is 10020. The length of the runs of 0's in the ternary expression of 87 are 2 and 1, respectively. The larger of these two values is 2, so a(87) = 2.
   n [ternary n] a(n)
   0 [        0] 1
   1 [        1] 0
   2 [        2] 0
   3 [      1 0] 1
   4 [      1 1] 0
   5 [      1 2] 0
   6 [      2 0] 1
   7 [      2 1] 0
   8 [      2 2] 0
   9 [    1 0 0] 2
  10 [    1 0 1] 1
  11 [    1 0 2] 1
  12 [    1 1 0] 1
  13 [    1 1 1] 0
  14 [    1 1 2] 0
  15 [    1 2 0] 1
  16 [    1 2 1] 0
  17 [    1 2 2] 0
  18 [    2 0 0] 2
  19 [    2 0 1] 1
  20 [    2 0 2] 1

Wikipedia, Ternary numeral system.

Cf. A000244, A007089, A077267, A087117, A330036, A330167, A330168.

Equals zero iff n is in A032924.

Table[Max@FoldList[If[#2==0,#1+1,0]&,0,IntegerDigits[n,3]],{n,0,90}]

a(A000244(n)) = a(3^n) = n.

a(n) = 0 iff n is in A032924.

A330036 The length of the largest run of 0's in the binary expansion of n + the length of the largest run of 1's in the binary expansion of n.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 3, 3, 4, 3, 2, 3, 4, 3, 4, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 3, 3, 5, 4, 5, 5, 6, 5, 4, 5, 3, 3, 4, 5, 4, 3, 2, 3, 4, 3, 4, 5, 6, 5, 4, 4, 4, 3, 3, 4, 6, 5, 4, 4, 6, 5, 6, 6, 7, 6, 5, 6, 4, 4, 5, 6, 4, 3, 3, 4, 4, 4, 5, 6, 5, 4, 3

Offset: 0

Joshua Oliver, Nov 27 2019

All numbers appear in this sequence. The number of 1's in the n-th Mersenne number (A000225) is n and the number of 0's in the n-th Mersenne number is 0. 0+n=n. See formula.

			   n [binary n ]  A087117(n) + A038374(n) = a(n)
   0 [       0 ]  1          + 0          = 1
   1 [       1 ]  0          + 1          = 1
   2 [     1 0 ]  1          + 1          = 2
   3 [     1 1 ]  0          + 2          = 2
   4 [   1 0 0 ]  2          + 1          = 3
   5 [   1 0 1 ]  1          + 1          = 2
   6 [   1 1 0 ]  1          + 2          = 3
   7 [   1 1 1 ]  0          + 3          = 3
   8 [ 1 0 0 0 ]  3          + 1          = 4
   9 [ 1 0 0 1 ]  2          + 1          = 3
  10 [ 1 0 1 0 ]  1          + 1          = 2
  11 [ 1 0 1 1 ]  1          + 2          = 3
  12 [ 1 1 0 0 ]  2          + 2          = 4
  13 [ 1 1 0 1 ]  1          + 2          = 3
  14 [ 1 1 1 0 ]  1          + 3          = 4
  15 [ 1 1 1 1 ]  0          + 4          = 4

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000120, A000225, A007088, A023416, A038374, A087117.

f:= proc(n) local L;
  L:= convert(n,base,2);
  max(map(nops,[ListTools:-Split(`=`,L,1)]))+max(map(nops,[ListTools:-Split(`=`,L,0)]))
end proc:
map(f, [$0..100]); # Robert Israel, Apr 06 2020

Table[Sum[Max[Differences[Position[Flatten@{k,IntegerDigits[n,2],k},k]]],{k,0,1}]-2,{n,0,82}]

a(n) = A087117(n) + A038374(n).

a(A000225(n)) = n for n > 0.

A329204 Number of integers less than n having at least as many totatives as n.

Original entry on oeis.org

0, 1, 0, 1, 0, 3, 0, 2, 1, 4, 0, 6, 0, 4, 2, 3, 0, 8, 0, 6, 3, 5, 0, 10, 1, 6, 3, 8, 0, 16, 0, 7, 4, 9, 2, 15, 0, 9, 4, 14, 0, 21, 0, 10, 7, 9, 0, 21, 2, 15, 5, 11, 0, 22, 5, 14, 7, 11, 0, 33, 0, 12, 10, 12, 3, 29, 0, 15, 6, 26, 0, 28, 0, 16, 13, 18, 4, 34, 0, 24, 7, 17

Offset: 1

Joshua Oliver, Nov 22 2019

			a(6) = 3, because 6 has 2 totatives and there are 3 integers less than 6 with 2 or more totatives: 3 with 2 totatives, 4 with 2 totatives, and 5 with 4 totatives.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 2000 terms from Joshua Oliver)
Wikipedia, Totative.

Cf. A000010.

Table[Length[Select[Range[n-1],EulerPhi[#]>=EulerPhi[n]&]],{n,1,100}]

a(n) = sum(k=1, n-1, eulerphi(k) >= eulerphi(n)); \\ Michel Marcus, Nov 22 2019

first(n)=my(u=vectorsmall(n),v=vector(n)); forfactored(f=1,n,u[f[1]]=eulerphi(f)); for(i=1,n, v[i]=sum(j=1,i-1,u[j]>=u[i])); v \\ Charles R Greathouse IV, Dec 11 2019

A329725 a(1)=0, a(n) = n - (product of nonzero digits of n) - a(n-1).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 17, 2, 16, 1, 15, 0, 14, -1, 13, -2, 29, -1, 27, -3, 25, -5, 23, -7, 21, -9, 45, -8, 42, -11, 39, -14, 36, -17, 33, -20, 65, -19, 61, -23, 57, -27, 53, -31, 49, -35, 89, -34, 84, -39, 79, -44, 74, -49, 69, -54, 117

Offset: 1

Joshua Oliver, Nov 19 2019

a(10n+1)-a(10n-1)=1 for all positive integer n (conjectured).

			a(22) = 22 - 2*2 - 2 = 16.

Joshua Oliver, Table of n, a(n) for n = 1..5000

Cf. A063108, A063543.

R:= ListTools:-PartialSums(map(n -> (-1)^n*(n - convert(subs(0=NULL,convert(n,base,10)),`*`)), [$1..100])):
seq((-1)^n*R[n],n=1..100); # Robert Israel, Nov 20 2019

Nest[Append[#1, #2 - Last[#1] - Times @@ DeleteCases[IntegerDigits[#2], 0]] & @@ {#, Length@ # + 1} &, {0}, 69] (* Michael De Vlieger, Nov 19 2019 *)

for (n=1, 70, print1 (v=if (n==1, 0, n - vecprod(select(sign, digits(n))) - v)", ")) \\ Rémy Sigrist, Nov 28 2019

a(n) = Sum_{k=2..n} (-1)^(n-k)*A063543(k). - Robert Israel, Nov 20 2019

A329934 a(1)=1, a(2)=1, a(n) = (number of times a(n-1) has appeared before) + (number of times a(n-2) has appeared before).

Original entry on oeis.org

1, 1, 4, 3, 2, 2, 4, 4, 6, 4, 5, 5, 4, 7, 6, 3, 4, 8, 7, 3, 5, 6, 6, 8, 6, 7, 8, 6, 9, 7, 5, 8, 8, 10, 6, 8, 13, 7, 6, 13, 10, 4, 9, 9, 6, 12, 10, 4, 11, 9, 5, 9, 10, 9, 10, 11, 7, 8, 13, 10, 9, 13, 11, 7, 10, 14, 8, 9, 16, 9, 10, 17, 9, 11, 14, 6, 12, 12, 6, 14, 14

Offset: 1

Joshua Oliver, Nov 24 2019

nonn

Conjecture: This sequence grows logarithmically.

			a(n)=4 where n=3 because 1 (a(n-1)) has appeared twice before, and 1 (a(n-2)) has appeared twice before as well. 2+2 = 4.

Joshua Oliver, Table of n, a(n) for n = 1..4000
Rémy Sigrist, Density plot of the first 10000000 terms

Cf. A158416, A306246, A316774.

b:= proc(n) option remember; `if`(n=0, 0, b(n-1)+x^a(n)) end:
a:= proc(n) option remember; `if`(n<3, 1, (p->
      coeff(p, x, a(n-1))+coeff(p, x, a(n-2)))(b(n-1)))
    end:
seq(a(n), n=1..120);  # Alois P. Heinz, Nov 24 2019

A={1,1};For[n=3,n<=81,n++,A=Append[A,Sum[Count[Table[Part[A,i],{i,1,n-1}],Part[A,n-k]],{k,2}]]];A

o=vector(17); for (n=1, 81, print1 (v=if (n<3, 1, o[pp]+o[p]) ", "); o[v]++; [pp,p]=[p,v]) \\ Rémy Sigrist, Nov 27 2019

A319604 Largest Brady number with n digits.

Original entry on oeis.org

6569, 73857, 818996, 9082813, 62254526, 690413273, 7656800529, 84915219092, 941724210541, 6454673763134, 71583428964617, 793872392373921, 8804179745077748, 97639849588229149, 669233485047004862, 7421913081219006521, 82310277378456076593, 912834964244235849044

Offset: 4

Joshua Oliver, Sep 24 2018

If a(n) = A247698(k), then a(n+1) = A247698(k+4) or a(n+1) = A247698(k+5). - Charles R Greathouse IV, Dec 10 2019

Brady Haran and Matt Parker, Brady Numbers, Numberphile video (2014).

Cf. A247698, A247839.

b[1] = 2308; b[2] = 4261; b[n_] := b[n] = b[n-1] + b[n-2]; a[n_] := Block[ {k=1}, While[ IntegerLength[b@k] <= n, k++]; b[k-1]]; a /@ Range[4, 21] (* Giovanni Resta, Sep 24 2018 *)

cp's OEIS Frontend

User: Joshua Oliver

A332025 Sum of the lengths of the longest runs of 0, 1, and 2 in the ternary expression of n.

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A330168 Length of the longest run of 2's in the ternary expression of n.

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A330167 Length of the longest run of 1's in the ternary expression of n.

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A330166 Length of the longest run of 0's in the ternary expression of n.

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A330036 The length of the largest run of 0's in the binary expansion of n + the length of the largest run of 1's in the binary expansion of n.

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A329204 Number of integers less than n having at least as many totatives as n.

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A329725 a(1)=0, a(n) = n - (product of nonzero digits of n) - a(n-1).

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