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Revisions by Wolfdieter Lang (See also Wolfdieter Lang's wiki page
and changes approved by Wolfdieter Lang)

(Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Decimal expansion of golden ratio phi (or tau) = (1 + sqrt(5))/2.
(history; published version)
#560 by Wolfdieter Lang at Thu Feb 20 07:00:27 EST 2025
STATUS

editing

proposed

Discussion
Thu Feb 20
07:04
Michel Marcus: Ma.thematical  : Mathematical ?
#559 by Wolfdieter Lang at Thu Feb 20 07:00:10 EST 2025
COMMENTS

The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025

REFERENCES

Michael Doob, The Canadian Mathematical Ma.thematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993 - Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1975, pages 76-77, 1993.

STATUS

approved

editing

Riordan array (1/(1 - x - x^2), x/(1 - x)).
(history; published version)
#53 by Wolfdieter Lang at Thu Feb 13 18:20:48 EST 2025
STATUS

editing

proposed

#52 by Wolfdieter Lang at Thu Feb 13 18:18:25 EST 2025
FORMULA

From ~~~_Wolfdieter Lang_, Feb 13 2025: (Start)

Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(n-j+1) * binomial(j-1, k-1), 0 <= k <= n.

STATUS

proposed

editing

Discussion
Thu Feb 13
18:20
Wolfdieter Lang: Thanks Alois. Yes, Peter, I made a mistake, it should have been F(n-j+1). Sorry!
#48 by Wolfdieter Lang at Thu Feb 13 16:40:27 EST 2025
STATUS

editing

proposed

Discussion
Thu Feb 13
17:37
Alois P. Heinz: please try it again with one ~ more ...
#47 by Wolfdieter Lang at Thu Feb 13 16:38:48 EST 2025
FORMULA

Number triangle T(n, k) = Sum_{j=0..n} binomial(n-j, k+j); T(n, 0)=A000045(n+1);

From ~~~: (Start)

Array A(k, n) = Sum_{j=0..n} F(j+1)*binomial(k-1+n-j, k-1), k >= 0, n >= 0, with F = A000045, (from Riordan triangle k-th convolution in columns without leading 0s).

A(k, n) = Fn(n+1+2*k) - Sum_{j=0..k-1} F(2*(k-j)-1) * binomial(n+1+j, j), (from iteration of partial sums).

Triangle T(n, k) = A(k, n-k) = Sum_{j=k..n} F(j+1) * binomial(j-1, k-1), 0 <= k <= n.

T(n, k) = F(n+1+k)-Sum_{j=0..k-1} F(2*(k-j)-1) *

binomial(n-(k-1-j)), j). (End)

STATUS

approved

editing

Odd numbers of the form x^2 + 2*y^2 with positive integers x and y.
(history; published version)
#11 by Wolfdieter Lang at Tue Jan 14 16:30:34 EST 2025
STATUS

editing

proposed

#10 by Wolfdieter Lang at Tue Jan 14 16:30:00 EST 2025
COMMENTS

For the numbers with positive proper representations see A225771 without member 1, the subsequence without 75 = 3*5^2, 147 = 3*7^2, 225 = (3*5)^2, 275 = 5^2*11, ... - Wolfdieter Lang, Jan 14 2025

CROSSREFS

Cf. A033200 (primes), A033203, A225771.

STATUS

approved

editing

Irregular triangle read by rows: row n gives the proper positive integer fundamental solutions (x, y) of x^2 - 15*y^2 = - A378710(n), for n >= 1.
(history; published version)
#6 by Wolfdieter Lang at Thu Jan 02 04:32:41 EST 2025
STATUS

editing

approved

#5 by Wolfdieter Lang at Thu Jan 02 04:31:01 EST 2025
COMMENTS

For details on the general proper representations of a negative integer k of by forms with discriminant Disc = 60 = 4*15 see A378710, with references. For the Pell case x^2 - 15*y^2 only a subset of these k values are is permitted, namely those that have representative reduced primitive forms (rpapfs) Fpa(-k, j) (see A378710) equivalent to the principal reduced form CR(1) = [1, 6, -6], which is in turn equivalent to the Pell form FPell = [1, 0, -15].

Some rules for the represented -A378710(n) values are: the negative of the prime factors 2, 3 and 5 of 15 are not represented, they are equivalent to some of the other three 2-cycles forms. These Powers of these three primes can never occur in powers because they cannot be lifted (see the Apostol reference, Theorem 5.20, pp. 121-122). The products -2*3 and -3*5 are represented but not -2*5 (the rpapf [-10, 10, -1] is equivalent to [-1, 6, 6], a member of the 2-cycle called CRhat in A378710). -2*3*5 is also not represented ([-30, 30, -7] is equivalent to [2, 6, -3] from the cycle Chat).

The number of infinite families of proper solutions for k = -A378710(n) with positive y is then determined by 2^P, (n), where P (n) is the number of primes >= 7 in the considered admissible -A378710(n) . These numbers 2^P(there n) are also the solutions with a common sign change in x and y). This number 2^P is given in the table below, and in the first comment.

The proper family of solutions {(x(n,i), y(n,i))}_{i = -infinity ... +infinity} are found from the rpapfsrpapf(-A378710(n), j) = [-A378710(n), 2*j, (15 - j^2)/A378710(n)] with the help of the formula (x(n, i), y(n, i))^T = (+ or - B15)*(-Auto15)^i*Rtvalues(n,j)^(-1)*(1, 0)^T, for the solutions of j^2 - 15 == 0 (mod(A378710(n)), for j from 0, 1,.., A378710(n) - 1, (T for transpose) where B15 = R(0)*R(3) = -Matrix([1, 3], [0, 1]), Auto15 = R(-1)*R(6) = - Matrix([1, 6], [1, 7]). For the R(t)-transformation matrix see A378710(n). Rtvalues(n,j) is the product of R(t) matrices with the t-values leading from the rpapf(-A378710(n), j) to the form CR(1). The sign of B15 are is chosen such that no negative values for y appear.

EXAMPLE

For the representation of -A378710(19) = -231 = -3*7*11 see the linked Figure of the directed and weighted Pell cycle graph with the two pairs of conjugate rpapfs (corresponding to solution of the congruence j^2 - 15 = = 0 (mod 231) with j and 231 - j, for j = 57 and j = 90. There the t-values are given as weights. E.g., the rpapf Fpa4 = [-231. 282, -86] has t-values (1-, 2, 2, 6). The pairs of row n = 19 belong to FPa1, FPa3, Fpa4 and FPa2, with the i exponents in the formula above 0, 0, 1, 1, repsectively, respectively, and the sign of B15 is - in all four cases.

STATUS

approved

editing

Discussion
Thu Jan 02
04:32
Wolfdieter Lang: I revisited this entry in connection with the pink box comment in  A378710.