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For n>1, a(n) = 22*a(n-1)-a(n-2)-40; e.g., 30365=22*1385-65-40.
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=6, then
last(2)=2274=26*90-6-60.
For n>0, a(n)=12*A157096 (n-1)+11*a(n-1)+10; e.g., 30365=12*1260+11*1385+10.
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=6 and n=2, then
first(2)=2100=13*78+12*90+6 and last(2)=2274=14*78+13*90+12.
a(n)=5^n*6((1+sqrt(6/5))^(2n+1)-(1-sqrt(6/5))^(2n+1))/(4*sqrt(6/5))+4/2;
e.g., 1385=5^2*6((1+sqrt((6/5))^5-(1-sqrt(6/5))^5)/(4*sqrt(6/5))+4/2;
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=6 and n=2, then
last(2)=2274=6^2*7((1+sqrt((7/6))^5-(1-sqrt(7/6))^5)/(4*sqrt(7/6))+5/2.
In general, if u(n) is the numerator and e(n) is the denominator of the nth
continued fraction convergent 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)=30365=(220^2+5^2*21^2+5*4*220*21)/5 and
a(4)=666605=(5(505^2+241^2+4*505*241))/6.
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+4+5+61+62+63+64+65+1381+1382+1383+1384+1385=7245=15*483.
Lim n->inf a(n+1)/a(n)=5(1+sqrt(6/5))^2=11+2sqrt(30).
In general, if last(n) is the last term of the nth Consecutive Integer
Pythagorean 2k+1-tuple, then
lim n->inf last(n+1)/last(n)= k*(1+sqrt((k+1)/k))^2=2k+1+2sqrt(k^2+k).
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