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For n>1, a(n) = 14*a(n-1)-a(n-2)-12; e.g., 5043=14*363-27-12.
In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples
may be found as follows: let last(0)=0, last(1)=k*(2k+3) and,
for n>1, last(n)=(4k+2)*last(n-1)-last(n-2)-2*k*(k-1); e.g., if k=4, then
last(2)=764=18*44-4-24
For n>0, a(n)=8*a(n-1)+7*A157088(n-1)+6; e.g., 5043=8*312+7*363+6.
In general, the first and last terms of Consecutive Integer Pythagorean
2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n>0,
let first(n)=(2k+1)*first(n-1)+2k*last(n-1)+k and
last(n)=(2k+2)*first(n-1)+(2k+1)*last(n-1)+2k; e.g., if k=4 and n=2, then
first(2)=680=9*36+8*44+4 and last(2)=764=10*36+9*44+8.
a(n)=3^n*4((1+sqrt(4/3))^(2n+1)-(1-sqrt(4/3))^(2n+1))/(4*sqrt(4/3))+2/2;
e.g., 363=3^2*4((1+sqrt((4/3))^5-(1-sqrt(4/3))^5)/(4*sqrt(4/3))+2/2;
In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples
may be found as follows: if q=(k+1)/k, then last(n)=
k^n*(k+1)*((1+sqrt(q))^(2*n+1)-(1-sqrt(q))^(2*n+1))/(4*sqrt(q))+(k-1)/2;
e.g., if k=4 and n=2, then
last(2)=764=4^2*5((1+sqrt((5/4))^5-(1-sqrt(5/4))^5)/(4*sqrt(5/4))+3/2;
In general, if u(n) is the numerator and e(n) is the denominator of the nth
continued fraction fraction to sqrt((k+1)/k), then the last terms
of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows:
last(2n+1)=(e(2n+1)^2+k^2*e(2n)^2+k*(k-1)*e(2n+1)*e(n))/k and, for n>0, last(2n)=(k*(u(2n)^2+u(2n-1)^2+(k-1)*u(2n)*u(2n-1)))/(k+1) ;
e.g., a(3)=5043=(84^2+3^2*13^2+3*2*84*13)/3 and
a(4)=70227=(3(209^2+97^2+2*209*97))/4.
In general, if b(0)=1, b(1)=4k+2 and, for n>1, b(n)=(4k+2)*b(n-1)-b(n-2),
and last(n) is the last term of the n-th Consecutive Integer
Pythagorean 2k+1-tuple as defined above, then
sum_{i=0...n}(k*last(i)-k(k-1)/2)=k(k+1)/2*b(n); e.g., if n=3, then
1+2+3+25+26+27+361+362+363=1170=6*195.
Lim n->inf a(n+1)/a(n)=3(1+sqrt(4/3))^2=7+2sqrt(12).
In general, if first(n) is the first term of the n-th Consecutive Integer
Pythagorean 2k+1-tuple, then
lim n->inf first(n+1)/first(n)= k*(1+sqrt((k+1)/k))^2=2k+1+2sqrt(k^2+k).
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