| For n>1, a(n)=22*a(n-1)-a(n-2)+50; e.g., 1260=22*55-0+50.
In general, the first terms of Consecutive Integer Pythagorean 2k+1-tuples
may be found as follows: let first(0)=0, first(1)=k*(2k+1) and,
for n>1, first(n)=(4k+2)*first(n-1)-first(n-2)+2*k^2; e.g., if k=6, then
first(2)=2100=26*78-0+72.
For n>0, a(n)=11*a(n-1)+10*A157097(n-1)+5; e.g., 1260=11*55+10*65+5.
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+1)((1+sqrt(6/5))^(2n+1)+(1-sqrt(6/5))^(2n+1))-2*5)/4;
e.g., 1260=(5^3((1+sqrt(6/5))^5+(1-sqrt(6/5))^5)-2*5)/4.
In general, the first terms of Consecutive Integer Pythagorean 2k+1-tuples
may be found as follows:
first(n)=(k^(n+1)((1+sqrt((k+1)/k))^(2n+1)+(1-sqrt((k+1)/k))^(2n+1))-2*k)/4;
e.g., if k=6 and n=2, then
first(2)=2100=(6^3((1+sqrt((7/6))^5+(1-sqrt(7/6))^5)-2*6)/4.
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 first terms of
Consecutive Integer Pythagorean 2k+1-tuples may be found as follows:
first(2n+1)=k*u(2n)*u(2n+1) and, for n>0, first(2n)=(k+1)*e(2n-1)*e(2n);
e.g., a(1)=55=5*1*11 and a(2)=1260=6*10*21.
Lim n->inf a(n+1)/a(n)=5(1+sqrt(6/5))^2=11+2sqrt(30).
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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