cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A007299 Number of Hadamard matrices of order 4n.

Original entry on oeis.org

1, 1, 1, 1, 5, 3, 60, 487, 13710027
Offset: 0

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Author

Keywords

Comments

More precisely, number of inequivalent Hadamard matrices of order n if two matrices are considered equivalent if one can be obtained from the other by permuting rows, permuting columns and multiplying rows or columns by -1.
The Hadamard conjecture is that a(n) > 0 for all n >= 0. - Charles R Greathouse IV, Oct 08 2012
From Bernard Schott, Apr 24 2022: (Start)
A brief historical overview based on the article "La conjecture de Hadamard" (see link):
1893 - J. Hadamard proposes his conjecture: a Hadamard matrix of order 4k exists for every positive integer k (see link).
As of 2000, there were five multiples of 4 less than or equal to 1000 for which no Hadamard matrix of that order was known: 428, 668, 716, 764 and 892.
2005 - Hadi Kharaghani and Behruz Tayfeh-Rezaie publish their construction of a Hadamard matrix of order 428 (see link).
2007 - D. Z. Djoković publishes "Hadamard matrices of order 764 exist" and constructs 2 such matrices (see link).
As of today, there remain 12 multiples of 4 less than or equal to 2000 for which no Hadamard matrix of that order is known: 668, 716, 892, 1132, 1244, 1388, 1436, 1676, 1772, 1916, 1948, and 1964. (End)
By private email, Felix A. Pahl informs that a Hadamard matrix of order 1004 was constructed in 2013 (see link Djoković, Golubitsky, Kotsireas); so 1004 is deleted from the last comment. - Bernard Schott, Jan 29 2023

References

  • J. Hadamard, Résolution d'une question relative aux déterminants, Bull. des Sciences Math. 2 (1893), 240-246.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Wolfram, A New Kind of Science. Champaign, IL: Wolfram Media, p. 1073, 2002.

Crossrefs

Extensions

a(8) from the H. Kharaghani and B. Tayfeh-Rezaie paper. - N. J. A. Sloane, Feb 11 2012

A004118 Maximal excess of a Hadamard matrix of order 4n.

Original entry on oeis.org

0, 8, 20, 36, 64, 80, 112, 140, 172, 216, 244, 280, 324, 364, 408
Offset: 0

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Author

Keywords

Comments

This is the maximal value of the sum of the entries of any n X n Hadamard matrix (cf. A019442).

References

  • Brown, Thomas A. and Spencer, Joel H., Minimization of +-1 matrices under line shifts. Colloq. Math. 23 (1971), 165-171, 177 (errata).
  • N. Farmakis and S. Kounias, The excess of Hadamard matrices and optimal designs, Discrete Mathematics, 67 (1987), 165-176. [From William P. Orrick, Mar 26 2009]
  • S. Kounias and N. Farmakis, On the excess of Hadamard matrices, Discrete Mathematics, 68 (1988), 59-69. [From William P. Orrick, Mar 26 2009]
  • Seberry, Jennifer and Yamada, Mieko; Hadamard matrices, sequences and block designs, in Dinitz and Stinson, eds., Contemporary design theory, pp. 431-560, Wiley-Intersci. Ser. Discrete Math. Optim., Wiley, New York, 1992.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A019442.

Formula

n^2*2^(-n)*binomial(n,n/2) <= a(n) <= n*sqrt(n).
Contribution from William P. Orrick, Mar 26 2009: (Start)
a(n/4) <= n(2m+1)+8[n/4(n/4-1)/(2(2m+1))], if 4m^2<=n/4<=4m^2+2m+1 or 4m^2+6m+3<=n/4<=4(m+1)^2,
a(n/4) <= 8[nm/4+1/2[n/4(n/4-1)/(2m)]-(n+4)/8]+n+4, if 4m^2+2m+1
a(n/4)<=8[nm/4+1/2[n/4(n/4-1)/(2(m+1))]+(n-4)/8]+n+4, if 4m^2+4m+1<=n/4<4m^2+6m+3.
[x] denotes the integer part. (See Kounias and Farmakis, 1988.) (End)

Extensions

a(7)-a(14) from William P. Orrick, Mar 26 2009

A179901 Triangle read by rows, antidiagonals of an array generated from (1, r, r, r, ...) convolved with (1, 0, r, r, r, ...).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 4, 3, 1, 4, 6, 8, 4, 1, 5, 8, 15, 12, 5, 1, 6, 10, 24, 24, 16, 6, 1, 7, 12, 35, 40, 33, 20, 7, 1, 8, 14, 48, 60, 56, 42, 24, 8, 1, 9, 16, 63, 84, 85, 70, 51, 28, 9, 1, 10, 18, 80, 112, 120, 110, 88, 60, 32, 10, 1, 11, 20, 99, 144, 161, 156, 135, 104, 69, 36, 11
Offset: 1

Author

Gary W. Adamson, Jul 31 2010

Keywords

Comments

Row sums = A179902: (1, 2, 5, 11, 23, 46, 87, 155, ...).

Examples

			First few rows of the array:
.
1,.1,..2,...3,...4,...5,...6,...7,....8,...
1,.2,..4,...8,..12,..16...20,..24,...28,... = A019442
1,.3,..6,..15,..24,..33,..42,..51,...60,... = A179805
1,.4,..8,..24,..40,..56,..70,..88,..104,...
.
Example: row 4 = (1, 4, 8, 24, ...) = (1, 4, 4, 4, ...) * (1, 0, 4, 4, 4, ...) = (1, r, 2*r, (2*r + r^2), ...).
.
First few rows of the triangle:
.
1,
1, 1;
1, 2, 2;
1, 3, 4, 3;
1, 4, 6, 8, 4;
1, 5, 8, 15, 12, 5;
1, 6, 10, 24, 24, 16, 6;
1, 7, 12, 35, 40, 33, 20, 7;
1, 8, 14, 48, 60, 56, 42, 24, 8;
1, 9, 16, 63, 84, 85, 70, 51, 28, 9;
1, 10, 18, 80, 112, 120, 110, 88, 60, 32, 10;
1, 11, 20, 99, 144, 161, 156, 135, 104, 69, 36, 11;
1, 12, 22, 120, 180, 208, 210, 192, 160, 120, 78, 40, 12;
1, 13, 24, 143, 220, 261, 272, 259, 228, 185, 136, 87, 44, 13;
...
		

Crossrefs

Formula

Triangle read by rows, antidiagonals of an array generated from (1, r, r, r, ...) convolved with (1, 0, r, r, r, ...), such that the r-th row of the array = (1, r, 2*r, ...) then for n > 3, a(n) = r^2 + a(n-1).

A356639 Number of integer sequences b with b(1) = 1, b(m) > 0 and b(m+1) - b(m) > 0, of length n which transform under the map S into a nonnegative integer sequence. The transform c = S(b) is defined by c(m) = Product_{k=1..m} b(k) / Product_{k=2..m} (b(k) - b(k-1)).

Original entry on oeis.org

1, 1, 3, 17, 155, 2677, 73327, 3578339, 329652351
Offset: 1

Author

Thomas Scheuerle, Aug 19 2022

Keywords

Comments

This sequence can be calculated by a recursive algorithm:
Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').
The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).
There exist some transformation pairs of infinite sequences in the database:
A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;
A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;
A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;
A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;
A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;
A279312 <--> A------; A031923 <--> A------.
These transformation pairs are conjectured:
A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;
A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;
A349079 <--> A------.
("A------" means not yet in the database.)
Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.
If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.

Examples

			a(4) = 17. The 17 transformation pairs of length 4 are:
  {1, 2, 3, 4}  = S({1, 2, 6, 24}).
  {1, 2, 3, 5}  = S({1, 2, 6, 15}).
  {1, 2, 3, 6}  = S({1, 2, 6, 12}).
  {1, 2, 3, 9}  = S({1, 2, 6, 9}).
  {1, 2, 3, 12} = S({1, 2, 6, 8}).
  {1, 2, 3, 21} = S({1, 2, 6, 7}).
  {1, 2, 4, 5}  = S({1, 2, 4, 20}).
  {1, 2, 4, 6}  = S({1, 2, 4, 12}).
  {1, 2, 4, 8}  = S({1, 2, 4, 8}).
  {1, 2, 4, 12} = S({1, 2, 4, 6}).
  {1, 2, 4, 20} = S({1, 2, 4, 5}).
  {1, 2, 6, 7}  = S({1, 2, 3, 21}).
  {1, 2, 6, 8}  = S({1, 2, 3, 12}).
  {1, 2, 6, 9}  = S({1, 2, 3, 9}).
  {1, 2, 6, 12} = S({1, 2, 3, 6}).
  {1, 2, 6, 15} = S({1, 2, 3, 5}).
  {1, 2, 6, 24} = S({1, 2, 3, 4}).
b(1) = 1 by definition, b(2) = 1+1 as 1 has only 1 as divisor.
a(3) = A000005(b(2)*b(2)) = 3.
The divisors of b(2) are 1,2,4. So b(3) can be b(2)+1, b(2)+2 and b(2)+4.
a(4) = A000005((b(2)+1)*(b(2)+4)) + A000005((b(2)+2)*(b(2)+2)) + A000005((b(2)+4)*(b(2)+1)) = 17.
		
Showing 1-4 of 4 results.