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Revision History for A077446 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Numbers n such that 2*n^2 + 14 is a square.
(history; published version)
#56 by Wolfdieter Lang at Tue Aug 15 18:03:19 EDT 2023
STATUS

editing

approved

#55 by Wolfdieter Lang at Tue Aug 15 18:02:30 EDT 2023
FORMULA

Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-+1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015

STATUS

approved

editing

Discussion
Tue Aug 15
18:03
Wolfdieter Lang: I corrected an index: n-1 -> n+1.
#54 by Wolfdieter Lang at Tue Aug 15 14:05:19 EDT 2023
STATUS

editing

approved

#53 by Wolfdieter Lang at Tue Aug 15 14:04:36 EDT 2023
FORMULA

Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(nk-1, 6) + S(nk-2, 6), with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015

STATUS

approved

editing

Discussion
Tue Aug 15
14:05
Wolfdieter Lang: I corrected the indx n -> k in my old formula.
#52 by Michael De Vlieger at Sat Nov 26 08:57:40 EST 2022
STATUS

reviewed

approved

#51 by Michel Marcus at Sat Nov 26 08:52:35 EST 2022
STATUS

proposed

reviewed

#50 by Stefano Spezia at Sat Nov 26 08:46:16 EST 2022
STATUS

editing

proposed

#49 by Stefano Spezia at Sat Nov 26 03:52:34 EST 2022
#48 by Stefano Spezia at Sat Nov 26 03:50:04 EST 2022
#47 by Stefano Spezia at Sat Nov 26 03:48:39 EST 2022
FORMULA

2*(a(n))^2 + 14 = (A077447(n))^2.

CROSSREFS

Cf. 2*(a(n))^2 + 14 = (A077447(n))^2. A006452, A155765, A054490, A255236.