A008233 -id:A008233 - cp's OEIS frontend

A059403 Quarter-squared applied twice.

0, 0, 0, 1, 4, 9, 20, 36, 64, 100, 156, 225, 324, 441, 600, 784, 1024, 1296, 1640, 2025, 2500, 3025, 3660, 4356, 5184, 6084, 7140, 8281, 9604, 11025, 12656, 14400, 16384, 18496, 20880, 23409, 26244, 29241, 32580, 36100, 40000, 44100, 48620, 53361, 58564, 64009

Offset: 0

Henry Bottomley, Mar 21 2001

			a(9)=100 since the ninth quarter-square is 20 and the twentieth quarter-square is 100.

Harvey P. Dale, Table of n, a(n) for n = 0..1000 (first 501 terms from Harry J. Smith)
Index entries for linear recurrences with constant coefficients, signature (2, 1, -4, 2, 0, -2, 4, -1, -2, 1).

Cf. A008233 for an alternative approach.

Floor[Floor[Range[0,50]^2/4]^2/4] (* or *) LinearRecurrence[{2,1,-4,2,0,-2,4,-1,-2,1},{0,0,0,1,4,9,20,36,64,100},50] (* Harvey P. Dale, Dec 13 2014 *)

a(n) = { (n^2\4)^2\4 } \\ Harry J. Smith, Jun 26 2009

a(n) = floor(floor(n^2/4)^2/4) = A002620(A002620(n)).
a(4*n) = 4n^4; a(4*n+1) = n^2*(2*n+1)^2;
a(4*n+2) = 2*n*(n+1)*(2*n*(n+1)+1); a(4*n+3) = (n+1)^2*(2*n+1)^2.
a(2n) = A060494(2n); a(2n-1) = A060494(2n-1)-A011861(n).
G.f.: x^3*(1 + 2*x + 2*x^3 + x^4)/((1 - x)^5*(1 + x)^3*(1 + x^2)). - R. J. Mathar, Sep 09 2008

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

Thomas Scheuerle, Aug 19 2022

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:
  A000027 <--> A000142; A000079 <--> A000079; A000045 <--> A001654;
  A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;
  A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;
  A098558 <--> A004277; A010551 <--> A087811; A100538 <--> A329227;
  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.

			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.

Cf. A000005.
Cf. A000027, A000045, A000079, A000142, A001405.
Cf. A001654, A004277, A006501, A008233, A019442.
Cf. A010551, A019464, A026549, A031923, A038754.
Cf. A057895, A058295, A062112, A066332, A100071.
Cf. A079352, A082458, A087811, A093968, A098011.
Cf. A098558, A100538, A111286, A166447, A137326.
Cf. A138278, A171647, A205825, A208147, A264557.
Cf. A264635, A308546, A329227, A336496, A349079.
Cf. A349080, A359039.

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A059403 Quarter-squared applied twice.

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