A007583 -id:A007583 - cp's OEIS frontend

A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2n+1) = A000120((2n+1)*k), A000120 is the Hamming weight.

3, 13, 11, 3, 205, 43, 57, 5, 3277, 171, 35, 3641, 7, 52429, 683, 21, 47, 233017, 19, 838861, 2731, 3, 79, 99, 14913081, 23, 13421773, 10923, 241, 5, 197, 187, 954437177, 37, 214748365, 43691, 7, 61681, 7, 325, 419, 61083979321, 39, 3435973837, 174763

Offset: 1

Thomas Scheuerle, Jan 07 2021

Solutions to related equation A000120(k) = A000120(k*n) are A340351.

			Five initial terms of rows 1-5 are listed below:
   1:  3,   11,     43,       171,       683, ...
   2: 13,  205,   3277,     52429,    838861, ...
   3:  3,    5,      7,        19,        23, ...
   4: 57, 3641, 233017,  14913081, 954437177, ...
   5: 35,   47,      99,      187,       419, ...
T(3,4) = 19 because: (3*2+1) in binary is 111 and (3*2+1)*19 = 133 in binary is 10000101, both have 3 bits set to 1.

Pontus von Brömssen, Antidiagonals n = 1..100, flattened

Cf. A000120, A340351, A340069.

Cf. A263132 (superset of 1st row), A007583 (1st row), A299960 (2nd row).

If 2*n = 2^j, then T(n, m) = (1+2^(j+2*j*m))/(2*n+1) for m > 0. In particular:
  T(1, m) = (1+2^(1+2*m))/3 = A007583(m),
  T(2, m) = (1+2^(2+4*m))/5 = A299960(m),
  T(4, m) = (1+2^(3+6*m))/9.
The third row consists of all numbers of the form (1+2^(1+b*3)+2^(2+c*3))/7, where b and c are natural numbers >= 0 and b+c > 0.
The seventh row consists of all numbers of the form (1+2^(1+b*2)+2^(2+c*2)+2^(3+d*2))/15 where b, c, and d are natural numbers >= 0 and b+c+d > 1.

More terms from Pontus von Brömssen, Jan 08 2021

A362903 Array read by antidiagonals: T(n,k) is the number of nonisomorphic k-tuples of involutions on a (2n)-set that pairwise commute.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 11, 4, 1, 1, 16, 43, 24, 5, 1, 1, 32, 171, 176, 46, 6, 1, 1, 64, 683, 1376, 611, 80, 7, 1, 1, 128, 2731, 10944, 9281, 1864, 130, 8, 1, 1, 256, 10923, 87424, 146445, 54384, 5161, 200, 9, 1, 1, 512, 43691, 699136, 2334181, 1696352, 285939, 13184, 295, 10, 1

Offset: 0

Andrew Howroyd, May 11 2023

Two involutions x,y commute if x*y = y*x. Isomorphism is up to permutation of the elements of the (2n)-set. T(n,k) also gives the values for a (2n+1)-set.

			Array begins:
======================================================
n/k| 0 1   2     3       4          5            6 ...
---+--------------------------------------------------
0  | 1 1   1     1       1          1            1 ...
1  | 1 2   4     8      16         32           64 ...
2  | 1 3  11    43     171        683         2731 ...
3  | 1 4  24   176    1376      10944        87424 ...
4  | 1 5  46   611    9281     146445      2334181 ...
5  | 1 6  80  1864   54384    1696352     53885632 ...
6  | 1 7 130  5161  285939   17562679   1110290303 ...
7  | 1 8 200 13184 1372224  165343616  20774749952 ...
8  | 1 9 295 31532 6101080 1436647664 358238974304 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (first 51 antidiagonals)

Columns k=0..3 are A000012, A000027(n-1), A001752, A362904.
Rows n=1..3 are A000079, A007583, A103334(n+1).
Cf. A022166, A362648, A362824, A362826.

\\ B(n, k) is A022166.
B(n, k)={polcoef(x^k/prod(j=0, k, 1-2^j*x + O(x*x^n)), n)}
C(k,n) = Vec(1/prod(j=0, min(k-1, logint(n, 2)), (1 - x^(2^j) + O(x*x^n))^B(k,j+1), 1 - x + O(x*x^n)))
M(n,m=n) = Mat(vector(m+1, k, C(k-1, n)~))
{ my(A=M(7)); for(i=1, #A, print(A[i,])) }

G.f. of column k: 1/((1 - x)*Product_{j=0..k-1} (1 - x^(2^j))^A022166(k,j+1)).

A364215 The number of 1's in the canonical representation of n as a sum of distinct Jacobsthal numbers (A280049).

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Jul 14 2023

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000120, A003159, A007583, A007953, A280049.

DigitCount[Select[Range[200], EvenQ[IntegerExponent[#, 2]] &], 2, 1]

s(n) = if(n < 2, n > 0, n = s(n-1); until(valuation(n, 2)%2 == 0, n++); n); \\ A003159
a(n) = hammingweight(s(n));

a(n) = A007953(A280049(n)).
a(n) = A000120(A003159(n)).
a(A007583(n)) = 1.

A046636 Number of cubic residues mod 8^n.

Original entry on oeis.org

1, 5, 37, 293, 2341, 18725, 149797, 1198373, 9586981, 76695845, 613566757, 4908534053, 39268272421, 314146179365, 2513169434917, 20105355479333, 160842843834661, 1286742750677285, 10293942005418277, 82351536043346213, 658812288346769701, 5270498306774157605, 42163986454193260837

Offset: 0

David W. Wilson

Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.
E. Wilmer and O. Schirokauer, A note on Stephan's conjecture 25, 2004. [broken link]
E. Wilmer and O. Schirokauer, A note on Stephan's conjecture 25, 2004. [cached copy]
Index entries for linear recurrences with constant coefficients, signature (9,-8).

Cf. A007583, A046530, A046630, A047853, A226308.

LinearRecurrence[{9, -8}, {1, 5}, 20] (* Jean-François Alcover, Jan 19 2019 *)

a(n) = (4*8^n + 3)/7.
a(n) = 8*a(n-1) - 3 (with a(0)=1). - Vincenzo Librandi, Nov 18 2010
From R. J. Mathar, Feb 28 2011: (Start)
a(n) = A046530(8^n) = A046630(3*n).
G.f.: (1-4*x)/((1-8*x)*(1-x)). (End)
a(n+1) = A226308(3*n+2). - Philippe Deléham, Feb 24 2014
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: exp(x)*(4*exp(7*x) + 3)/7.
a(n) = 9*a(n-1) - 8*a(n-2).
a(n) = A047853(n+1)/2. (End)

More terms from Elmo R. Oliveira, Apr 03 2025

A059054 Integers which can be written as (b^k+1)/(b+1) for positive integers b and k.

Original entry on oeis.org

1, 3, 7, 11, 13, 21, 31, 43, 57, 61, 73, 91, 111, 133, 157, 171, 183, 205, 211, 241, 273, 307, 343, 381, 421, 463, 507, 521, 547, 553, 601, 651, 683, 703, 757, 813, 871, 931, 993, 1057, 1111, 1123, 1191, 1261, 1333, 1407, 1483, 1561, 1641, 1723, 1807, 1893

Offset: 1

Henry Bottomley, Dec 21 2000

nonn

It seems that all values are odd. For (b^k+1)/(b+1) to be an integer, it seems k must be odd. 2=(0^0+1)/(0+1) has been excluded since neither b nor k would be positive.
When k is a composite, a(n) is a composite.
These numbers are in the form of 111...1 (k of 1s) base b. - Lei Zhou, Feb 08 2012

			43 is in the sequence since (2^7+1)/(2+1)=129/3=43; indeed also (7^3+1)/(7+1)=344/8=43.

H. Dubner and T. Granlund, Primes of the Form (b^n+1)/(b+1), J. Integer Sequences, 3 (2000), #P00.2.7.

Cf. A007583, A002061, A059055.

max = 44; maxdata = (1 + max^3)/(1 + max); a = {}; Do[i = 1; While[i = i + 2; cc = (1 + m^i)/(1 + m); cc <= maxdata, a = Append[a, cc]], {m, 2, max}]; Union[a] (* Lei Zhou, Feb 08 2012 *)

A081547 Smallest composite number which is 1 more than the product of n (not necessarily distinct) prime numbers.

Original entry on oeis.org

4, 10, 9, 25, 33, 65, 129, 385, 513, 1025, 2049, 4097, 8193, 16385, 32769, 98305, 131073, 262145, 524289, 1048577, 2097153, 4194305, 8388609, 16777217, 33554433, 67108865, 134217729, 268435457, 536870913, 1073741825, 2147483649

Offset: 1

Amarnath Murthy, Apr 01 2003

nonn

a(2*n+1) = 2^(2*n+1)+1, n>0. - Vladeta Jovovic, Apr 02 2003

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

Cf. A081545, A081546, A081548, A073919, A007583.

cno[n_]:=Module[{a=2^n+1},If[PrimeQ[a],2^(n-1)*3+1,a]]; Join[{4,10}, Array[cno,30,3]] (* Harvey P. Dale, Mar 24 2012 *)

from sympy import isprime
def A081547(n): return 10 if n==2 else ((3<Chai Wah Wu, Sep 02 2024

For n>2, a(n) = 2^n+1 unless this is a Fermat prime (A019434), in which case a(n) = 2^(n-1)*3+1 (which is divisible by 5). - Dean Hickerson, Apr 05 2003

More terms from Vladeta Jovovic, Apr 02 2003

A092468 a(n) = Sum_{i+j+k=n, 0<=i<=j<=k<=n} (n+2k)!/(i! * j! * (3*k)!).

Original entry on oeis.org

1, 1, 3, 24, 71, 366, 2142, 8128, 40063, 223182, 919600, 4477496, 24423998, 104469379, 508145588, 2726623094, 11936904927, 58101760966, 307770506832, 1369729219470, 6671075408724, 34985537652681, 157621462924548

Offset: 0

Benoit Cloitre, Mar 25 2004

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000 (terms 0..200 from Seiichi Manyama)
Vaclav Kotesovec, Plot a(n+1)/a(n) for n = 1..999

Cf. A007583, A092469.

a(n)=sum(i=0,n,sum(j=0,i,sum(k=0,j,if(i+j+k-n,0,(n+2*k)!/i!/j!/(3*k)!))))

Conjecture: a(n+1)/a(n) tends to (73 + 6*sqrt(87))^(1/3)/3 + 13/(3*(73 + 6*sqrt(87))^(1/3)) + 7/3 = 4.875129794... - Vaclav Kotesovec, Oct 31 2021

A114192 Riordan array (1/(1-2x),x/(1-2x)^2).

Original entry on oeis.org

1, 2, 1, 4, 6, 1, 8, 24, 10, 1, 16, 80, 60, 14, 1, 32, 240, 280, 112, 18, 1, 64, 672, 1120, 672, 180, 22, 1, 128, 1792, 4032, 3360, 1320, 264, 26, 1, 256, 4608, 13440, 14784, 7920, 2288, 364, 30, 1, 512, 11520, 42240, 59136, 41184, 16016, 3640, 480, 34, 1

Offset: 0

Paul Barry, Nov 16 2005

Factors as (1/(1-x),x/(1-x))*(1/(1-x),x*(1+x)/(1-x)^2) or A007318 times A114188. Also (1/(1-2*x),x/(1-2*x))*(1,x*(1+2*x)). Inverse is A114193. Row sums are A007583. Diagonal sums are A007051.

			Triangle begins
1;
2, 1;
4, 6, 1;
8, 24, 10, 1;
16, 80, 60, 14, 1;
32, 240, 280, 112, 18, 1;

T(n,k) = sum{j=0..n, C(k, j)*C(n, k+j)}*2^(n-k).
T(n,k) = 2^(n-k)*binomial(n+k,2k) = 2^(n-k)*A085478(n,k). - Philippe Deléham, May 05 2006
T(n,k) = A013609(n+k, n-k). - Johannes W. Meijer, Sep 05 2013
T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 4*T(n-2,k), T(0,0) = T(1,1) = 1, T(1,0) = 2, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014

A234038 Smallest positive integer solution x of 9x - 2^ny = 1.

Original entry on oeis.org

1, 1, 1, 1, 9, 25, 57, 57, 57, 57, 569, 1593, 3641, 3641, 3641, 3641, 36409, 101945, 233017, 233017, 233017, 233017, 2330169, 6524473, 14913081, 14913081, 14913081, 14913081, 149130809, 417566265, 954437177, 954437177, 954437177

Offset: 0

Wolfdieter Lang, Feb 17 2014

The solution of the linear Diophantine equation 9*x - 2^n*y = 1 with smallest positive x is x=a(n), n>= 0, and the corresponding y is given by y(n) = 5^(n+3) (mod 9) = A070366(n+3) with o.g.f. (8-4*x-2*x^2+7*x^3)/((1-x+x^2)*(1-x)*(1+x)) (derived from the one given in A070366). This is the periodic sequence with period [8, 4, 2, 1, 5, 7].

			n = 0:  9*1 - 1*8 = 1; n = 3:  9*1 - 8*1  = 1.
a(4) = (1 + 2^4*5)/9 = 9.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv preprint arXiv:1404.2710, 2014 and J. Int. Seq. 17 (2014) # 14.11.7.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-8,24,-16)

Cf. A070366, A007583 (see the Feb 15 2013 comment).

A234038[n_] := Ceiling[(1 + 2^n*Mod[5^(n - 3), 9])/9]; Array[A234038, 50, 0] (* or *)
LinearRecurrence[{3, -2, -8, 24, -16}, {1, 1, 1, 1, 9}, 50] (* Paolo Xausa, Nov 05 2024 *)

a(n) = (1 + 2^n*(5^(n-3)(mod 9)))/3^2, n >= 3.
O.g.f.: (1-2*x+8*x^3-8*x^4)/((1-x)*(1-4*x^2)*(1-2*x+4*x^2)) (derived from the one for y(n) given above in a comment).
a(n) = 2*(a(n-1) - 4*a(n-3) + 8*a(n-4)) - 1, n >= 4, a(0)=a(1)=a(2)=a(3) = 1 (from the y(n) recurrence given in A070366).

A239125 Smallest positive integer solution x of (3^3)x - 2^ny = 1 for n >= 0.

Original entry on oeis.org

1, 1, 3, 3, 3, 19, 19, 19, 19, 19, 531, 531, 2579, 6675, 6675, 23059, 55827, 121363, 252435, 252435, 776723, 776723, 776723, 4971027, 4971027, 4971027, 4971027, 4971027, 139188755, 139188755, 676059667, 1749801491, 1749801491, 6044768787, 14634703379

Offset: 0

Wolfdieter Lang, Mar 13 2014

a(n) is the smallest positive integer solution of the linear Diophantine equation 27*a(n) - 2^n*b(n) = 1, n >= 0, with b(n) the period length 18 = phi(27) sequence repeat(26, 13, 20, 10, 5, 16, 8, 4, 2, 1, 14, 7, 17, 22, 11, 19, 23, 25). Here phi(n) = A000010(n) (Euler's totient). These 18 members are a permutation of the smallest nonnegative numbers of the reduced residue system modulo 27.
This is the instance m = 3 of an m-family of sequence pairs [x0(m, n), y0(m, n)], n >= 0, providing a special solution of the linear Diophantine equation 3^m*x - 2^n*y = 1; in fact the one with smallest positive x. The general formula is y0(m, n) = ((3^m+1)/2)^(n+3^(m-1)) (mod 3^m) and x0(m, n) = (1 + 2^n*y0(m, n))/3^m. For m = 0 this is x0(0, n) = 1 + 2^n with y0(0, n) = 1, n >= 0. Obviously, y0(m, n) is a positive integer (y0 = 0 is out). The proof that x0(m, n) is also a positive integer is done by showing that 1 + 2*y0(m, n) == 0 (mod 3^m). Because (3^m+1)/2 == 1/2 (mod 3^m) one shows that ((3^m+1)/2)^(3^(m-1)) + 1 == 0 (mod 3^m). This can be done writing (3^m+1)/2 = 3*q - 1, with q = (3^(m-1) + 1)/2, a natural number for m >= 1. Then the binomial theorem is used. Finally one has to show that binomial(3^(m-1) - 1, n -1)/n is a (positive) integer. Here the triangle A107711 helps (for a nice proof that A107711 is a positive integer triangle see the history with the remark by Peter Bala from Fri Feb 28, after #13).
The general family of positive solutions of 3^m*x - 2^n*y = c (c an integer) is then x(m, n; k) = x0(m, n) + 2^n*tmin(m, n) + 2^n*k and y(m, n; k) = y0(m, n) + 3^m*tmin(m, n) + 3^m*k for k>=0, with tmin(m, n) = ceiling(-c*y0(m, n)/3^m) if c>=0 and tmin(m, n) = ceiling(|c|*x0(m, n)/2^n) if c < 0.
See the Niven-Zuckerman-Montgomery reference, pp. 212-214, for integer solutions of a*x + b*y = c provided gcd(a,b)|c. Note that in their treatment of positive solutions a and b are assumed to be positive, but here we use b < 0.
For this instance m=3 one can prove directly that the a(n) formula given below in terms of b(n) produces (positive) integers. One uses 1/2 (mod 27) = 14 and 14^9 + 1 == 0 (mod 27).

			a(0) = 1 because 27*1 - 1*b(0) = 27 - 26 = 1.
a(1) = 1 because 27*1 - 2*b(1) = 27 - 2*13 = 1.
a(5) = 19, because 27*19 - 32*b(5) = 27*19 - 32*16 = 1.

I. Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv:1404.2710 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.11.7.
Index entries for linear recurrences with constant coefficients, signature (3,-2,0,0,0,0,0,0,-512,1536,-1024).

Cf. A007583 (comment Feb 15 2014).

LinearRecurrence[{3, -2, 0, 0, 0, 0, 0, 0, -512, 1536, -1024}, {1, 1, 3, 3, 3, 19, 19, 19, 19, 19, 531}, 40] (* Bruno Berselli, Mar 15 2014 *)

a(n) = (1 + 2^n*b(n))/27 with b(n) = 14^(n+9) (mod 27), n >= 0. The sequence b(n) has period length 18, and it is given in a comment above.
a(n) = 3*a(n-1) -2*a(n-2) -512*a(n-9) +1536*a(n-10) -1024*a(n-11) for n>10, with initial values as shown. - Bruno Berselli, Mar 15 2014
G.f.: -(512*x^10-512*x^9+32*x^6-16*x^5+4*x^3-2*x^2+2*x-1) / ((x-1)*(2*x-1)*(2*x+1)*(4*x^2-2*x+1)*(64*x^6-8*x^3+1)). - Colin Barker, Mar 20 2014

cp's OEIS Frontend

A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2*n+1) = A000120((2*n+1)*k), A000120 is the Hamming weight.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

Extensions

A362903 Array read by antidiagonals: T(n,k) is the number of nonisomorphic k-tuples of involutions on a (2n)-set that pairwise commute.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A364215 The number of 1's in the canonical representation of n as a sum of distinct Jacobsthal numbers (A280049).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A046636 Number of cubic residues mod 8^n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A059054 Integers which can be written as (b^k+1)/(b+1) for positive integers b and k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A081547 Smallest composite number which is 1 more than the product of n (not necessarily distinct) prime numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A092468 a(n) = Sum_{i+j+k=n, 0<=i<=j<=k<=n} (n+2k)!/(i! * j! * (3*k)!).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A114192 Riordan array (1/(1-2x),x/(1-2x)^2).

Views

Author

Keywords

Comments

Examples

Formula

A340441 Square array, read by ascending antidiagonals, where row n gives all odd solutions k > 1 and n > 0 to A000120(2n+1) = A000120((2n+1)*k), A000120 is the Hamming weight.

A234038 Smallest positive integer solution x of 9x - 2^ny = 1.

A239125 Smallest positive integer solution x of (3^3)x - 2^ny = 1 for n >= 0.