A004731 -id:A004731 - cp's OEIS frontend

A338718 Define b(1)=1 and for n>1, b(n)=n/b(n-1); then a(n) = floor(b(n)).

1, 2, 1, 2, 1, 3, 2, 3, 2, 4, 2, 4, 2, 4, 3, 5, 3, 5, 3, 5, 3, 5, 3, 6, 4, 6, 4, 6, 4, 6, 4, 7, 4, 7, 4, 7, 4, 7, 5, 7, 5, 8, 5, 8, 5, 8, 5, 8, 5, 8, 5, 9, 5, 9, 5, 9, 6, 9, 6, 9, 6, 9, 6, 10, 6, 10, 6, 10, 6, 10, 6, 10, 6, 10, 6, 10, 7, 11, 7, 11, 7, 11, 7, 11, 7, 11, 7, 11, 7, 11

Offset: 1

N. J. A. Sloane, Nov 29 2020, following a suggestion from Anchar Koops, Nov 24 2020

nonn

			The first few fractions b(n) are 1, 2, 3/2, 8/3, 15/8, 16/5, 35/16, 128/35, 315/128, 256/63, 693/256, 1024/231, 3003/1024, 2048/429, ...

Cf. A338719, A338720.

For the numerators and denominators of b(n) see A004731 and A004730.

a[n_] := Floor[n!!/(n - 1)!!]; Array[a, 90] (* Robert G. Wilson v, Dec 06 2020 *)

A086767 Last coefficient of the last term in the numerator of the simplified expansion of the solutions of FLT for n=2 for FLT n=1,2,3,..

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 7, 1, 1, 1, 9, 1, 5, 1, 11, 1, 3, 1, 13, 1, 7, 1, 15, 1, 1, 1, 17, 1, 9, 1, 19, 1, 5, 1, 21, 1, 11, 1, 23, 1, 3, 1, 25, 1, 13, 1, 27, 1, 7, 1, 29, 1, 15, 1, 31, 1, 1, 1, 33, 1, 17, 1, 35, 1, 9, 1, 37, 1, 19, 1, 39, 1, 5, 1, 41, 1, 21, 1, 43, 1, 11, 1, 45, 1, 23

Offset: 0

Cino Hilliard, Aug 02 2003

Integers a > b form the solution to FLT n = 2 as follows. (2ab)^2 = (a^2-b^2)^2 - (a^2+b^2)^2. The sequence is the coefficient c of the last b term in the numerator for the simplified expansion of the solution for n=2 as verification of FLT for n=1, 2, ...

			b/a
1
(3*a^4 + b^4)/(4*b*a^3)
(a^4 + b^4)/(2*b^2*a^2)
(5*a^8 + 10*b^4*a^4 + b^8)/(16*b^3*a^5)
(3*a^8 + 10*b^4*a^4 + 3*b^8)/(16*b^4*a^4)
(7*a^12 + 35*b^4*a^8 + 21*b^8*a^4 + b^12)/(64*b^5*a^7)
(a^12 + 7*b^4*a^8 + 7*b^8*a^4 + b^12)/(16*b^6*a^6)
(9*a^16 + 84*b^4*a^12 + 126*b^8*a^8 + 36*b^12*a^4 + b^16)/(256*b^7*a^9)
(5*a^16 + 60*b^4*a^12 + 126*b^8*a^8 + 60*b^12*a^4 + 5*b^16)/(256*b^8*a^8)
........
(K + cb^m)/2^m1b^m2c^m3
Seq = c for integers K,b,m1,m2,m3,n = 1,2,3...

Anonymous, Fermat's Theorem for Pythagorean Triples.

Cf. A004731, A004730.

sigma := proc(n) local i; add(i,i=convert(n,base,2)) end:
a := proc(n) if n=0 or type(n,odd) then 1 else if type(iquo(n,2),odd) then n/2 else n/2^(1-sigma(n)+sigma(n-1)) fi fi end: # Peter Luschny, Aug 03 2009

\ verification of general solution in integers \ a>b,x = 2ab,y=a^2-b^2,z=a^2+b^2 \ or FLT n=2 x^n+y^n <> z^n = (2ab)^n + (a^2-b^2)^n <> \(a^2+b^2)^n for n > 2 flt(n,a1,b1) = for(x=0,n,print(f(x,a1,b1))) f(n,a,b) = simplify(((a^2+b^2)^n - (a^2-b^2)^n)/(2*a*b)^n) coeffb(m) = { for(y=1,m, n=y; if(n%2,x=1, while(n%2==0,n=n/2); x=n; ); print1(x",") ) }

a(n) = A004731(n+1)/A004730(n). - Flávio V. Fernandes, Feb 13 2025

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A338718 Define b(1)=1 and for n>1, b(n)=n/b(n-1); then a(n) = floor(b(n)).

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A086767 Last coefficient of the last term in the numerator of the simplified expansion of the solutions of FLT for n=2 for FLT n=1,2,3,..

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