A043569 -id:A043569 - cp's OEIS frontend

A023758 Numbers of the form 2^i - 2^j with i >= j.

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 28, 30, 31, 32, 48, 56, 60, 62, 63, 64, 96, 112, 120, 124, 126, 127, 128, 192, 224, 240, 248, 252, 254, 255, 256, 384, 448, 480, 496, 504, 508, 510, 511, 512, 768, 896, 960, 992, 1008, 1016, 1020, 1022, 1023

Offset: 1

Olivier Gérard

Numbers whose digits in base 2 are in nonincreasing order.
Might be called "nialpdromes".
Subset of A077436. Proof: Since a(n) is of the form (2^i-1)*2^j, i,j >= 0, a(n)^2 = (2^(2i) - 2^(i+1))*2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004
Numbers whose binary representation contains no "01". - Benoit Cloitre, May 23 2004
Every polynomial with coefficients equal to 1 for the leading terms and 0 after that, evaluated at 2. For instance a(13) = x^4 + x^3 + x^2 at 2, a(14) = x^4 + x^3 + x^2 + x at 2. - Ben Paul Thurston, Jan 11 2008
From Gary W. Adamson, Jul 18 2008: (Start)
As a triangle by rows starting:
   1;
   2,  3;
   4,  6,  7;
   8, 12, 14, 15;
  16, 24, 28, 30, 31;
  ...,
equals A000012 * A130123 * A000012, where A130123 = (1, 0,2; 0,0,4; 0,0,0,8; ...). Row sums of this triangle = A000337 starting (1, 5, 17, 49, 129, ...). (End)
First differences are A057728 = 1; 1; 1; 1; 2,1; 1; 4,2,1; 1; 8,4,2,1; 1; ... i.e., decreasing powers of 2, separated by another "1". - M. F. Hasler, May 06 2009
Apart from first term, numbers that are powers of 2 or the sum of some consecutive powers of 2. - Omar E. Pol, Feb 14 2013
From Andres Cicuttin, Apr 29 2016: (Start)
Numbers that can be digitally generated with twisted ring (Johnson) counters. This is, the binary digits of a(n) correspond to those stored in a shift register where the input bit of the first bit storage element is the inverted output of the last storage element. After starting with all 0’s, each new state is obtained by rotating the stored bits but inverting at each state transition the last bit that goes to the first position (see link).
Examples: for a(n) represented by three bits
             Binary
a(5)= 4   ->  100   last bit = 0
a(6)= 6   ->  110   first bit = 1 (inverted last bit of previous number)
a(7)= 7   ->  111
and for a(n) represented by four bits
             Binary
a(8) = 8   -> 1000
a(9) = 12  -> 1100  last bit = 0
a(10)= 14  -> 1110  first bit = 1 (inverted last bit of previous number)
a(11)= 15  -> 1111
(End)
Powers of 2 represented in bases which are terms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i - 2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known non-member of this sequence with this property is 5. - Ely Golden, Sep 05 2017
Numbers k such that k = 2^(1 + A000523(k)) - 2^A007814(k). - Daniel Starodubtsev, Aug 05 2021
A002260(n) = v(a(n)/2^v(a(n))+1) and A002024(n) = A002260(n) + v(a(n)) where v is the dyadic valuation (i.e., A007814). - Lorenzo Sauras Altuzarra, Feb 01 2023

			a(22) = 64 = 32 + 32 = 2^5 + a(16) = 2^A003056(20) + a(22-5-1).
a(23) = 96 = 64 + 32 = 2^6 + a(16) = 2^A003056(21) + a(23-6-1).
a(24) = 112 = 64 + 48 = 2^6 + a(17) = 2^A003056(22) + a(24-6-1).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 5051 terms from T. D. Noe)
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
S. M. Shabab Hossain, Md. Mahmudur Rahman and M. Sohel Rahman, Solving a Generalized Version of the Exact Cover Problem with a Light-Based Device, Optical Supercomputing, Lecture Notes in Computer Science, 2011, Volume 6748/2011, 23-31, DOI: 10.1007/978-3-642-22494-2_4.
Eric Weisstein's World of Mathematics, Digit.
Wikipedia, Ring counter.
Index entries for 2-automatic sequences.
Index entries for sequences related to binary expansion of n.

A000337(r) = sum of row T(r, c) with 0 <= c < r. See also A002024, A003056, A140129, A140130, A221975.
Cf. A007088, A130123, A101082 (complement), A340375 (characteristic function).
This is the base-2 version of A064222. First differences are A057728.
Subsequence of A077436, of A129523, of A277704, and of A333762.
Subsequences: A043569 (nonzero even terms, or equally, nonzero terms doubled), A175332, A272615, A335431, A000396 (its even terms only), A324200.
Positions of zeros in A049502, A265397, A277899, A284264.
Positions of ones in A283983, A283989.
Positions of nonzero terms in A341509 (apart from the initial zero).
Positions of squarefree terms in A260443.
Fixed points of A264977, A277711, A283165, A334666.
Distinct terms in A340632.
Cf. also A002024, A002260, A007814, A133797, A152449, A181666, A211705, A277699, A306441.
Cf. also A309758, A309759, A309761 (for analogous sequences).

import Data.Set (singleton, deleteFindMin, insert)
a023758 n = a023758_list !! (n-1)
a023758_list = 0 : f (singleton 1) where
f s = x : f (if even x then insert z s' else insert z $ insert (z+1) s')
where z = 2*x; (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Sep 24 2014, Dec 19 2012

a:=proc(n) local n2,d: n2:=convert(n,base,2): d:={seq(n2[j]-n2[j-1],j=2..nops(n2))}: if n=0 then 0 elif n=1 then 1 elif d={0,1} or d={0} or d={1} then n else fi end: seq(a(n),n=0..2100); # Emeric Deutsch, Apr 22 2006

Union[Flatten[Table[2^i - 2^j, {i, 0, 100}, {j, 0, i}]]] (* T. D. Noe, Mar 15 2011 *)
Select[Range[0, 2^10], NoneTrue[Differences@ IntegerDigits[#, 2], # > 0 &] &] (* Michael De Vlieger, Sep 05 2017 *)

for(n=0,2500,if(prod(k=1,length(binary(n))-1,component(binary(n),k)+1-component(binary(n),k+1))>0,print1(n,",")))

A023758(n)= my(r=round(sqrt(2*n--))); (1<<(n-r*(r-1)/2)-1)<<(r*(r+1)/2-n)
/* or, to illustrate the "decreasing digit" property and analogy to A064222: */
A023758(n,show=0)={ my(a=0); while(n--, show & print1(a","); a=vecsort(binary(a+1)); a*=vector(#a,j,2^(j-1))~); a} \\ M. F. Hasler, May 06 2009

is(n)=if(n<5,1,n>>=valuation(n,2);n++;n>>valuation(n,2)==1) \\ Charles R Greathouse IV, Jan 04 2016

list(lim)=my(v=List([0]),t); for(i=1,logint(lim\1+1,2), t=2^i-1; while(t<=lim, listput(v,t); t*=2)); Set(v) \\ Charles R Greathouse IV, May 03 2016

def a_next(a_n): return (a_n | (a_n >> 1)) + (a_n & 1)
a_n = 1; a = [0]
for i in range(55): a.append(a_n); a_n = a_next(a_n) # Falk Hüffner, Feb 19 2022

from math import isqrt
def A023758(n): return (1<<(m:=isqrt(n-1<<3)+1>>1))-(1<<(m*(m+1)-(n-1<<1)>>1)) # Chai Wah Wu, Feb 23 2025

a(n) = 2^s(n) - 2^((s(n)^2 + s(n) - 2n)/2) where s(n) = ceiling((-1 + sqrt(1+8n))/2). - Sam Alexander, Jan 08 2005
a(n) = 2^k + a(n-k-1) for 1 < n and k = A003056(n-2). The rows of T(r, c) = 2^r-2^c for 0 <= c < r read from right to left produce this sequence: 1; 2, 3; 4, 6, 7; 8, 12, 14, 15; ... - Frank Ellermann, Dec 06 2001
For n > 0, a(n) mod 2 = A010054(n). - Benoit Cloitre, May 23 2004
A140130(a(n)) = 1 and for n > 1: A140129(a(n)) = A002262(n-2). - Reinhard Zumkeller, May 14 2008
a(n+1) = (2^(n - r(r-1)/2) - 1) 2^(r(r+1)/2 - n), where r=round(sqrt(2n)). - M. F. Hasler, May 06 2009
Start with A000225. If k is in the sequence, then so is 2k. - Ralf Stephan, Aug 16 2013
G.f.: (x^2/((2-x)*(1-x)))*(1 + Sum_{k>=0} x^((k^2+k)/2)*(1 + x*(2^k-1))). The sum is related to Jacobi theta functions. - Robert Israel, Feb 24 2015
A049502(a(n)) = 0. - Reinhard Zumkeller, Jun 17 2015
a(n) = a(n-1) + a(n-d)/a(d*(d+1)/2 + 2) if n > 1, d > 0, where d = A002262(n-2). - Yuchun Ji, May 11 2020
A277699(a(n)) = a(n)^2, A306441(a(n)) = a(n+1). - Antti Karttunen, Feb 15 2021 (the latter identity from A306441)
Sum_{n>=2} 1/a(n) = A211705. - Amiram Eldar, Feb 20 2022

Definition changed by N. J. A. Sloane, Jan 05 2008

A062289 Numbers n such that n-th row in Pascal triangle contains an even number, i.e., A048967(n) > 0.

Original entry on oeis.org

2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77

Offset: 1

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 02 2001

nonn

Numbers n such that binary representation contains the bit string "10". Union of A043569 and A101082. - Rick L. Shepherd, Nov 29 2004

The asymptotic density of this sequence is 1 (Burns, 2016). - Amiram Eldar, Jan 26 2021

Reinhard Zumkeller, Table of n, a(n) for n = 1..10001
Rob Burns, Asymptotic density of Catalan numbers modulo 3 and powers of 2, arXiv:1611.03705 [math.NT], 2016.
Oliver Kullmann and Xishun Zhao, On variables with few occurrences in conjunctive normal forms, in: K. A. Sakallah and L. Simon (eds), International Conference on Theory and Applications of Satisfiability Testing, Springer, Berlin, Heidelberg, 2011, pp. 33-46; arXiv preprint, arXiv:1010.5756 [cs.DM], 2010-2011.
Oliver Kullmann and Xishun Zhao, Parameters for minimal unsatisfiability: Smarandache primitive numbers and full clauses, arXiv preprint, arXiv:1505.02318 [cs.DM], 2015.

Complement of A000225, so these might be called non-Mersenne numbers.
A132782 is a subsequence.
Cf. A048967, A057716, A043569, A101082.
Cf. A007461, A036987, A102370, A105027.
Cf. A261461, A261922.

a062289 n = a062289_list !! (n-1)
a062289_list = 2 : g 2 where
   g n = nM n : g (n+1)
   nM k = maximum $ map (\i -> i + min i (a062289 $ k-i+1)) [2..k]
   -- Cf. link [Oliver Kullmann, Xishun Zhao], Def. 3.1, page 3.
-- Reinhard Zumkeller, Feb 21 2012, Dec 31 2010

ok[n_] := MatchQ[ IntegerDigits[n, 2], {_, 1, 0, _}]; Select[ Range[100], ok] (* Jean-François Alcover, Dec 12 2011, after Rick L. Shepherd *)

isok(m) = #select(x->((x%2)==0), vector(m+1, k, binomial(m, k-1))); \\ Michel Marcus, Jan 26 2021

def A062289(n): return n+(m:=n.bit_length())-(not n>=(1<Chai Wah Wu, Jun 30 2024

a(n) = A057716(n+1) - 1.
a(n) = 2 if n=1, otherwise max{min{2*i, a(n-i+1) + i}: 1 < i <= n}.
A036987(a(n)) = 0. - Reinhard Zumkeller, Mar 06 2012
A007461(a(n)) mod 2 = 0. - Reinhard Zumkeller, Apr 02 2012
A102370(n) = A105027(a(n)). - Reinhard Zumkeller, Jul 21 2012
A261461(a(n)) = A261922(a(n)). - Reinhard Zumkeller, Sep 17 2015

More terms from Rick L. Shepherd, Nov 29 2004

A101082 Numbers n such that binary representation contains bit strings "10" and "01" (possibly overlapping).

Original entry on oeis.org

5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92

Offset: 1

Rick L. Shepherd, Nov 29 2004

Subsequence of A062289; set difference A062289 minus A043569.
Complement of A023758. Also numbers not the sum of consecutive powers of 2. - Omar E. Pol, Mar 04 2013
Equivalently, numbers not the difference of two powers of two. - Charles R Greathouse IV, Mar 07 2013
The terms >=9 are bases in which a power of 2 exists, which does not contain a digit that is a power of 2. In base 10, 2^16 = 65536 is such a number, as it does not contain any one-digit power of 2, which in base 10 are 1, 2, 4 and 8. - Patrick Wienhöft, Jul 28 2016

			29 = 11101_2 is a term, "10" and "01" are contained (here overlapping).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Patrick Wienhöft, Python program
Index entries for 2-automatic sequences.

Complement: A023758.

Cf. A043569, A062289, A049502.

a101082 n = a101082_list !! (n-1)
a101082_list = filter ((> 0) . a049502) [0..]
-- Reinhard Zumkeller, Jun 17 2015

Select[Range@ 120, Function[d, Times @@ Total@ Map[Map[Function[k, Boole@ MatchQ[#, k]], {{1, 0}, {0, 1}}] &, Partition[d, 2, 1]] > 0]@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Dec 23 2016 *)
Select[Range[100],With[{c=IntegerDigits[#,2]},SequenceCount[c,{1,0}]>0&&SequenceCount[c,{0,1}]>0]&] (* Harvey P. Dale, Jun 08 2025 *)

is(n)=n>>=valuation(n,2);n+1!=1<Charles R Greathouse IV, Mar 07 2013

def A101082(n):
    def f(x): return n+((k:=x.bit_length())*(k-1)>>1)+sum(1 for i in range(k) if (1<Chai Wah Wu, Feb 23 2025

a(n) ~ n. In particular a(n) = n + (log_2 n)^2/2 + O(log n). - Charles R Greathouse IV, Mar 07 2013

A049502(a(n)) > 0. - Reinhard Zumkeller, Jun 17 2015

A276349 Numbers consisting of a nonempty string of 1's followed by a nonempty string of 0's.

Original entry on oeis.org

10, 100, 110, 1000, 1100, 1110, 10000, 11000, 11100, 11110, 100000, 110000, 111000, 111100, 111110, 1000000, 1100000, 1110000, 1111000, 1111100, 1111110, 10000000, 11000000, 11100000, 11110000, 11111000, 11111100, 11111110, 100000000, 110000000, 111000000

Offset: 1

Jaroslav Krizek, Aug 30 2016

Intersection of A037415 and A009996 except for 1 [Corrected by David A. Corneth, Aug 30 2016].
Set of terms from sequence A052983.
a(n) is the binary expansion of A043569(n). - Michel Marcus, Sep 04 2016

			60 is of the form binomial(a, 2) + b where 0 < b <= a and a = 11, b = 5. So a(60) has (11 + 1) digits and 5 leading ones. The other digits are 0. Giving a(60) = 111110000000. It has 7 (more than 1) trailing zeros so the next one, a(61) is a(60) + 10^(7 - 1). - _David A. Corneth_, Aug 30 2016

Robert Israel, Table of n, a(n) for n = 1..10000
Luboš Pick, Dirichletovy šuplíčky, Pokroky matematiky, fyziky a astronomie, Vol. 61, No. 2 (2016), pp. 106-118. (In Czech; The Dirichlet pigeonhole principle)

Cf. A002024, A009996, A037415, A043569, A052983, A073668, A227362, A276348, A309761.

[n: n in [1..10^7] | Seqint(Setseq(Set(Sort(Intseq(n))))) eq 10 and Seqint(Sort((Intseq(n)))) eq n];

seq(seq(10^(m+1)*(1-10^(-j))/9,j=1..m),m=1..20); # Robert Israel, Sep 02 2016

Table[FromDigits@ Join[ConstantArray[1, #1], ConstantArray[0, #2]] & @@@ Transpose@ {#, n - #} &@ Range[n - 1], {n, 2, 9}] // Flatten (* Michael De Vlieger, Aug 30 2016 *)
Flatten[Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},k,0]]],{n,8},{k,8}]]//Sort (* Harvey P. Dale, Jan 09 2019 *)

is(n) = vecmin(digits(n))==0 && vecmax(digits(n))==1 && digits(n)==vecsort(digits(n), , 4) \\ Felix Fröhlich, Aug 30 2016

a(n) = my(r =  ceil((sqrt(1+8*n)+1)/2), k = n - binomial(r-1, 2));10^(r-k)*(10^(k)-1)/9
\\ given an element n, computes the next element of the sequence.
nxt(n) = my(d = digits(n), qd=#d, s = vecsum(d)); if(qd-s>1, n+10^(qd-s-1), 10^qd)
\\ given an element n of the sequence, computes its place in the sequence.
inv(n) = my(d = digits(n)); binomial(#d-1,2) + vecsum(d) \\ David A. Corneth, Aug 31 2016

from math import isqrt, comb
def A276349(n): return 10*(10**(m:=isqrt(n<<3)+1>>1)-10**(comb(m+1,2)-n))//9 # Chai Wah Wu, Jun 16 2025

A227362(a(n)) = 10.
From Robert Israel, Sep 02 2016: (Start)
a((m^2-m)/2+j) = 10^(m+1)*(1-10^(-j))/9 for m>=1, 1<=j<=m.
a(n) = 10*(10^m - 10^(-n+m*(m+1)/2))/9 where m = A002024(n). (End)
A002275(A002260(n)) * 10^A004736(n) - Peter Kagey, Sep 02 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Feb 20 2022
a(n) = 10*A309761(n). - Chai Wah Wu, Jun 16 2025

A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 2, 5, 1, 3, 2, 1, 4, 2, 8, 1, 3, 3, 3, 3, 7, 2, 13, 1, 1, 1, 3, 5, 5, 11, 2, 21, 1, 1, 2, 4, 3, 8, 9, 18, 2, 34, 1, 2, 1, 1, 5, 3, 13, 17, 29, 2, 55, 1, 2, 1, 1, 4, 9, 3, 21, 31, 47, 2, 89, 1

Offset: 1

Rémy Sigrist, Feb 17 2021

This table contains all Fibonacci sequences of order m > 0 with positive terms:
- order 1 corresponds to constant sequences (n in A126646),
- order 2 corresponds to Fibonacci-like sequences (n in A043569),
- order 3 corresponds to tribonacci-like sequences (n in A043570),
- order 4 corresponds to tetranacci-like sequences (n in A043571).
For any n > 0, the row A341746(n) corresponds to the n-th row from which the first term has been removed.

			Array T(n, k) begins:
  n\k|  1  2  3  4  5   6   7   8   9   10   11   12   13    14
  ---+---------------------------------------------------------
    1|  1  1  1  1  1   1   1   1   1    1    1    1    1     1 --> A000012
    2|  1  1  2  3  5   8  13  21  34   55   89  144  233   377 --> A000045
    3|  2  2  2  2  2   2   2   2   2    2    2    2    2     2 --> A007395
    4|  2  1  3  4  7  11  18  29  47   76  123  199  322   521 --> A000032
    5|  1  1  1  3  5   9  17  31  57  105  193  355  653  1201 --> A000213
    6|  1  2  3  5  8  13  21  34  55   89  144  233  377   610 --> A000045
    7|  3  3  3  3  3   3   3   3   3    3    3    3    3     3 --> A010701
    8|  3  1  4  5  9  14  23  37  60   97  157  254  411   665 --> A104449
    9|  1  2  1  4  7  12  23  42  77  142  261  480  883  1624 --> A275778
   10|  1  1  1  1  4   7  13  25  49   94  181  349  673  1297 --> A000288

Rémy Sigrist, Table of n, a(n) for n = 1..11325
Rémy Sigrist, Colored representation of the table for n, k = 1..1024
Rémy Sigrist, PARI program for A341694
Wikipedia, Fibonacci numbers of higher order
Index entries for sequences related to Fibonacci numbers

Cf. A005811, A043569, A043570, A043571, A126646, A136480, A227736, A341699, A341746.

Cf. A000012, A000032, A000045, A000213, A000288, A007395, A010701, A104449, A275778.

PARI
```
See Links section.
```

T(A341746(n), k) = T(n, k+1).

T(n, 1) = A136480(n).

A118586 Numbers whose binary expansion contains group of at least two 1's followed by a nonempty group of 0's.

Original entry on oeis.org

6, 12, 14, 24, 28, 30, 48, 56, 60, 62, 96, 112, 120, 124, 126, 192, 224, 240, 248, 252, 254, 384, 448, 480, 496, 504, 508, 510, 768, 896, 960, 992, 1008, 1016, 1020, 1022, 1536, 1792, 1920, 1984, 2016, 2032, 2040, 2044, 2046, 3072, 3584, 3840, 3968, 4032

Offset: 1

Zak Seidov, May 07 2006

All terms are even.

			4080_10 = 111111110000_2.

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Cf. A043569, A065442.

Sort[Flatten[Table[Sum[2^k,{k,k1,k2}],{k1,1,10},{k2,k1+1,11}]]]

from itertools import count, islice
def agen(): # generator of terms
    for d in count(3):
        for i in range(2, d):
            yield int("1"*i + "0"*(d-i), 2)
print(list(islice(agen(), 50))) # Michael S. Branicky, Feb 20 2022

Sum_{n>=1} 1/a(n) = -1 + A065442. - Amiram Eldar, Feb 20 2022

A231387 a(n) is a prime number that cannot be the center term of a length 3 arithmetic progression prime group with a common difference whose number of runs in binary expansion is 2.

Original entry on oeis.org

2, 3, 199, 1319, 2371, 2437, 3253, 6871, 6991, 7937, 7951, 9041, 9973, 10631, 11941, 12379, 12671, 13009, 13147, 13187, 13267, 13799, 13859, 14479, 16889, 17519, 20051, 21089, 26003, 27281, 27529, 29581, 31477, 32009, 34439, 38561, 41611, 42719, 43543, 44839

Offset: 1

Lei Zhou, Nov 08 2013

Fewer than 0.7% of the first three million primes have this property.
AP-3 is defined as 3 prime groups in arithmetic progression.
Hypothesized property: for n>=3, there exists at least one number k whose number of runs in binary expansion (A005811) equals 4, such that {a(n)-k, a(n), a(n)+k} forms an AP-3 group.

			2 and 3 cannot be the second term of an AP-3 prime group, so a(1)=2 and a(2)=3;
For any prime numbers between 5 and 197, there exists at least one number k in A043569 such that {a-k, a, a+k} forms an AP-3 prime group.  For example, when p=197, there are the following groups {5,197,389}, {101,197,293}, {137,197,257}, and {167,197,227} with corresponding k = 192 = 11000000 base 2 = A043569(23), 96 = 1100000 base 2 = A043569(17), 60 = 111100 base 2 = A043569(14), and 30 = 11110 base 2 = A043569(10).
However, when p = 199, among all six AP-3 groups, {19,199,379}, {31,199,367}, {61,199,337}, {67,199,331}, {127,199,271}, and {157,199,241}, none of k value (180 = 10110100 base 2, 168 = 10101000 base 2, 138 = 10001010 base 2, 132 = 10000100 base 2, 72 = 1001000, and 42 = 101010 respectively) is a term of A043569. None of them is in the form of 1..10..0 base 2 thus not an element of A043569.
So a(3)=199.

Lei Zhou, Table of n, a(n) for n = 1..10000

Cf. A005811, A043569.

seed = 1; Table[While[seed = NextPrime[seed]; sum = seed*2; lowbond = sum; cp1 = seed; While[cp1 = NextPrime[cp1]; (lowbond > 2) && (cp1 < sum), cp2 = sum - cp1; If[PrimeQ[cp2], test = cp2 - cp1; rank = Length[Length /@ Split[IntegerDigits[test, 2]]]; lowbond = Min[rank, lowbond]]]; lowbond == 2]; seed, {i, 1, 41}]

A370698 Rectangular array, read by antidiagonals: row n consists of the numbers m whose binary representation has exactly n runs.

Original entry on oeis.org

1, 3, 2, 7, 4, 5, 15, 6, 9, 10, 31, 8, 11, 18, 21, 63, 12, 13, 20, 37, 42, 127, 14, 17, 22, 41, 74, 85, 255, 16, 19, 26, 43, 82, 149, 170, 511, 24, 23, 34, 45, 84, 165, 298, 341, 1023, 28, 25, 36, 53, 86, 169, 330, 597, 682, 2047, 30, 27, 38, 69, 90, 171

Offset: 1

Clark Kimberling, Mar 11 2024

Every positive integer occurs exactly once, and for every n, the numbers in row n have the parity of n.

			Corner:
    1    3    7   15   31   63  127  255
    2    4    6    8   12   14   16   24
    5    9   11   13   17   19   23   25
   10   18   20   22   26   34   36   38
   21   37   41   43   45   53   69   73
   42   74   82   84   86   90  106  138
   85  149  165  169  171  173  181  213
  170  298  330  338  340  342  346  362
  341  597  661  677  681  683  685  693
The binary representation of 22 is 10110, which has 4 runs: 1, 0, 11, 0.

Cf. A007089, A005811 (# runs in binary n), A000225 (row 1), A043569 (row 2), A043570 (row 3), A000975 (column 1), A370893 (ternary).

a[n_] := a[n] = Select[Range[8000], Length[Split[IntegerDigits[#, 2]]] == n &];
t[n_, k_] := a[n][[k]];
Grid[Table[t[n, k], {n, 1, 12}, {k, 1, 12}]] (* array *)
Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* sequence *)

A383038 Positive integers which can be expressed in the form b^r - b^s where b, r, and s are positive integers.

Original entry on oeis.org

2, 4, 6, 8, 12, 14, 16, 18, 20, 24, 28, 30, 32, 42, 48, 54, 56, 60, 62, 64, 72, 78, 90, 96, 100, 110, 112, 120, 124, 126, 128, 132, 156, 162, 180, 182, 192, 210, 216, 224, 234, 240, 248, 252, 254, 256, 272, 294, 306, 336, 342, 380, 384, 420, 448, 462, 480, 486, 496, 500

Offset: 1

Boas Bakker, Apr 13 2025

nonn

Terms are all even because b^r - b^s == b - b == 0 (mod 2).
From David A. Corneth, Apr 26 2025: (Start)
By definition, b, s, r and b^r - b^s is positive. Therefore, r > s. If b = 1 then b^r - b^s = 1^r - 1^s = 0 which is excluded by default so b > 1. Suppose we look for terms <= U. Then b is maximal when (s, r) = (1, 2) i.e. b^2 - b <= U. Solving for b gives (sqrt(4*U + 1) + 1) / 2.
As r > s >= 1, r >= 2.
Once b is fixed, b^r - b^s is minimal when s = r-1, enabling the largest r.
So we'd have b^r - b^(r-1) = (b - 1)*(b^(r-1)) <= U. Solving for r gives r <= log(U / (b - 1)) / log(b) + 1.
Once b and r are fixed we have
b^r - b^s <= U so b^r - U <= b^s. Solving for s gives log(b^r - U) / log(b) <= s.
Summarizing we have
2 <= b <= (sqrt(4*U + 1) + 1) / 2,
2 <= r <= log(U / (b - 1)) / log(b) + 1,
log(b^r - U) / log(b) <= s <= r-1. (End)

			a(6) = 14 = 16 - 2 = 2^4 - 2^1.
a(9) = 20 = 25 - 5 = 5^2 - 5^1.
As _David A. Corneth_ said, we know r > s > 0 and b > 1, so r > 1, and the smallest value of b is 2. So b^r - b^s >= b^(r-1) >= 2^(r-1). So to prove 10 is not in the sequence, we only need to check up up to r=4, because for r=5, 2^(s-1) = 2^4 = 16 > 10. This means there are 6 combinations of (r, s) we need to check. We also know b^r - b^s == 0 (mod b) because s > 0, so we only need to check divisors of 10. So with b = 2 and s < r < 5, we get the terms {2, 4, 6, 8, 12, 14}. With b=5 the smallest term is 5^2 - 5^1 = 20, which is bigger than 10. For b>5, b^r - b^s >= b^2 - b > 5^2 - 5, so we don't need to check those values of b.

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000

The terms of the case b=2 are in A043569.

function A383038List(limit::Int)
    res = Set{Int}()
    for b in 2:floor(Int, sqrt(limit)) + 2
        for r in 2:limit
            valr = b^(r-1) * (b-1)
            valr > limit && break
            br = b^r
            for s in 1:r-1
                val = br - b^s
                val > 0 && val <= limit && push!(res, val)
    end end end
    sort(collect(res)) end
println(A383038List(500))  # Peter Luschny, Aug 17 2025

upto(n) = {n++; my(res = List());
    maxb = ceil((sqrtint(4*n + 1) + 1) / 2);
    for(b = 2, maxb,
        maxr = logint(n\(b-1), b) + 1;
        for(r = 2, maxr,
            mins = max(1, ceil(log(max(b^r - n, 1)) / log(b)));
            mins = min(mins, r-1); \\ min() to fix messed up rounding when b^r - n is a power of b. Also solved by increasing n by 1 initially (n++).
            forstep(s = r-1, mins, -1,
                c = b^r - b^s;
                listput(res, c);
            );
        );
    );
    Set(res)
} \\ David A. Corneth, Apr 26 2025

from sympy import integer_nthroot
aupto = 500
b_max, A383038 = (i := integer_nthroot(aupto,2))[0] + 2 - i[1], set()
for b in range(2, b_max):
    r, s = 2, 1
    while (br:=b**r) - b**s <= aupto:
        while s > 0 and (res := br - b**s) <= aupto: A383038.add(res); s -= 1
        r += 1
        s = r - 1
print(sorted(A383038)) # Karl-Heinz Hofmann, Apr 29 2025

cp's OEIS Frontend

A023758 Numbers of the form 2^i - 2^j with i >= j.

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A062289 Numbers n such that n-th row in Pascal triangle contains an even number, i.e., A048967(n) > 0.

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A101082 Numbers n such that binary representation contains bit strings "10" and "01" (possibly overlapping).

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A276349 Numbers consisting of a nonempty string of 1's followed by a nonempty string of 0's.

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A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

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A118586 Numbers whose binary expansion contains group of at least two 1's followed by a nonempty group of 0's.

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