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.

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A377290 For each row n in array A374602(n, k), the period size, as a count of terms, that divides the row into congruent subsequences.

Original entry on oeis.org

1, 2, 2, 6, 4, 4, 14, 10, 8, 10, 6, 8, 14, 16, 34, 10, 8, 34, 8, 12, 22, 22, 32, 18, 18, 30, 14, 18, 16, 12, 38, 22, 28, 26, 42, 20, 74, 36, 14, 54, 12, 16, 34, 38, 54, 26, 58, 50, 24, 36, 102, 46, 32, 78, 14, 22, 38, 46, 118, 22, 30, 68, 36, 32, 130, 74, 34
Offset: 1

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Author

Charles L. Hohn, Oct 23 2024

Keywords

Comments

Here "congruent" means: 1) In the defining formula of A374602: sqrt((d-c)*b^2 + c*(b+1)^2), A374602(n, k) and A374602(n, k+a(n)) have equal c values (see Example), and also 2) A374602(n, k+a(n))/A374602(n, k) converges to a limit as k->oo, shown in A377291.

Examples

			Given formula sqrt((d-c)*b^2 + c*(b+1)^2) from A374602, for n=5, the first few terms of A374602(5, k) are:
sqrt((7-3)*1^2 + 3*(1+1)^2) = 4,
sqrt((7-6)*1^2 + 6*(1+1)^2) = 5,
sqrt((7-1)*4^2 + 1*(4+1)^2) = 11,
sqrt((7-4)*10^2 + 4*(10+1)^2) = 28,
sqrt((7-3)*23^2 + 3*(23+1)^2) = 62,
sqrt((7-6)*29^2 + 6*(29+1)^2) = 79,
sqrt((7-1)*66^2 + 1*(66+1)^2) = 175,
sqrt((7-4)*168^2 + 4*(168+1)^2) = 446,
producing the repeating pattern of c values {3, 6, 1, 4}, of length 4 -> a(5).
		

Crossrefs

A377291 For each row n in array A374602(n, k), the asymptotic geometric growth factor of every A377290(n) terms, represented by its nearest integer.

Original entry on oeis.org

6, 14, 7, 98, 16, 34, 1442, 398, 194, 119, 30, 62, 4354, 1154, 115598, 322, 23, 155234, 48, 98, 10402, 2702, 64514, 727, 482, 3040, 1154, 2114, 70, 142, 21314, 5474, 2498, 1442, 16793602, 674, 48497294, 158402, 47, 48670, 96, 194, 39202, 9998, 1684802, 2599
Offset: 1

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Author

Charles L. Hohn, Oct 23 2024

Keywords

Comments

(a(n)^2-4)/A000037(n) is a square, and as such, a(n) is a member of row x of A298675(x, k), where x is the smallest value >= 3 such that (x^2-4)/A000037(n) is a square. E.g. for n=38: A000037(38)=44, x=20 ((20^2-4)/44 = 3^2), and a(38) = 158402 = A298675(20, 4).
The same row x of A298675(x, k) also results as integer solutions to g+(1/g) where g=(w*sqrt(d) + ceiling(w*sqrt(d)))/2 and d=A000037(n) for integers w >= 1. As such, it follows that g(n) can be expressed as a simple integer arithmetic transformation of sqrt(A000037(n)), e.g. g(1) = 2*sqrt(2)+3 (A156035), g(2) = 4*sqrt(3)+7 (A354129), g(3) = (3*sqrt(5)+7)/2 (A374883), g(4) = 20*sqrt(6)+49, and g(5) = 3*sqrt(7)+8 (A010516+8).

Examples

			For n = 5, the first few terms of A374602(5, k) are {4, 5, 11, 28, 62, 79, 175, 446, 988} and the period size is A377290(5) = 4, giving A374602(5, 1+4)/A374602(5, 1) = 62/4 = 15.5, 79/5 = 15.8, 175/11 = 15.909..., 446/28 = 15.928..., 988/62 = 15.935..., ..., to limit 15.937... -> g(5), from which g(5)+(1/g(5)) = 16 -> a(5).
		

Crossrefs

Formula

Growth factor g(n) = Lim_{k->oo}(A374602(n, k+A377290(n))/A374602(n, k)).
a(n) = round(g(n)) = ceiling(g(n)) = g(n)+(1/g(n)).
Inverse: g(n) = (sqrt(a(n)^2-4)+a(n))/2.
For d = A000037(n) and x in {1, 2, 4}, when d+x is a square (unless x==4 and d+x is even): a(n) = 4/x*d+2.
For d = A000037(n) and x in {-4, 1, 2, 4}, when n > 3 and d-x is a square (unless x==-4 and d-x is odd): a(n) = (4/abs(x))^2*d^2-16/x*d+2.

A375336 For n>=4, irregular triangular array of successive integer solutions to sqrt((d-c)*b^2 + c*(b+1)^2) for square integers d = n^2, where b and c are positive integers and c < d, read by rows.

Original entry on oeis.org

5, 7, 7, 8, 9, 13, 17, 27, 8, 10, 11, 13, 16, 19, 10, 11, 13, 14, 19, 21, 25, 31, 59, 61, 12, 15, 22, 23, 29, 34, 39, 42, 11, 13, 14, 16, 17, 19, 25, 33, 37, 41, 49, 103, 107, 125, 13, 14, 16, 17, 19, 20, 23, 27, 28, 32, 37, 40, 46, 53, 82, 83, 15, 18, 21, 26
Offset: 4

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Author

Charles L. Hohn, Aug 12 2024

Keywords

Comments

Provable that every row n has a finite number of terms, with n < 4 producing no solutions, and T(n, k) never exceeding (n/2)^3.
This sequence excludes cases where c == 0, where all b produce integer solutions d*b.
Nonsquare d produce solutions following a different pattern, shown as A374602.

Examples

			4: {5, 7}
5: {7, 8}
6: {9, 13, 17, 27}
7: {8, 10, 11, 13, 16, 19}
8: {10, 11, 13, 14, 19, 21, 25, 31, 59, 61}
9: {12, 15, 22, 23, 29, 34, 39, 42}
10: {11, 13, 14, 16, 17, 19, 25, 33, 37, 41, 49, 103, 107, 125}
11: {13, 14, 16, 17, 19, 20, 23, 27, 28, 32, 37, 40, 46, 53, 82, 83}
12: {15, 18, 21, 26, 29, 31, 34, 41, 43, 51, 54, 57, 61, 71, 159, 165, 209, 211}
...
sqrt((2^2-1)*1^2 + 1*(1+1)^2) = sqrt(7) -> not an integer so not included.
sqrt((4^2-1)*1^2 + 1*(1+1)^2) = sqrt(19) -> not an integer so not included.
sqrt((4^2-3)*1^2 + 3*(1+1)^2) = 5 -> T(4,1).
sqrt((4^2-11)*1^2 + 11*(1+1)^2) = 7 -> T(4,2).
sqrt((5^2-8)*1^2 + 8*(1+1)^2) = 7 -> T(5,1).
sqrt((6^2-5)*2^2 + 5*(2+1)^2) = 13 -> T(6,2).
		

Crossrefs

Programs

  • PARI
    row(n)=my(d=n^2, t=n, v=List()); while(t
    				

Formula

T(n, 1) = A080782(n+2).
Showing 1-3 of 3 results.