cp's OEIS Frontend

A238455 Difference between 4^n and the nearest triangular number.

%I A238455 #26 Jun 16 2022 10:25:06
%S A238455 0,1,1,-2,3,-11,1,-87,-167,-306,-500,-552,688,-3041,-579,20854,37075,
%T A238455 55618,37108,-222296,-147729,891994,602155,-3523022,-2228805,14811346,
%U A238455 11792251,-47737262,-1136517,375078994,741065851,1445763154,2746052116,4910207464,7492827856
%N A238455 Difference between 4^n and the nearest triangular number.
%H A238455 Harvey P. Dale, <a href="/A238455/b238455.txt">Table of n, a(n) for n = 0..1000</a>
%F A238455 a(n) = (1/2)*(-t^2 - t + 2*4^n), where t = floor(sqrt(2*4^n)) after formula in A053616. - _Michel Marcus_, Jun 16 2022
%e A238455 a(0) = 1 - 1 = 0.
%e A238455 a(1) = 4 - 3 = 1.
%e A238455 a(2) = 16 - 15 = 1.
%e A238455 a(3) = 64 - 66 = -2.
%e A238455 a(4) = 256 - 253 = 3.
%t A238455 db4n[n_]:=Module[{c=4^n,tr,t1,t2,d1,d2},tr=Floor[(Sqrt[8c+1]-1)/2];t1= (tr (tr+1))/ 2;t2=((tr+1)(tr+2))/2;d1=c-t1;d2=c-t2;If[d1<Abs[ d2], d1,d2]]; Array[ db4n,40,0] (* _Harvey P. Dale_, Jul 02 2019 *)
%o A238455 (Python)
%o A238455 def isqrt(a):
%o A238455     sr = 1 << (int.bit_length(int(a)) >> 1)
%o A238455     while a < sr*sr:  sr>>=1
%o A238455     b = sr>>1
%o A238455     while b:
%o A238455         s = sr + b
%o A238455         if a >= s*s:  sr = s
%o A238455         b>>=1
%o A238455     return sr
%o A238455 for n in range(77):
%o A238455     nn = 4**n
%o A238455     s = isqrt(2*nn)
%o A238455     if s*(s+1)//2 > nn:  s-=1
%o A238455     d1 = nn - s*(s+1)//2
%o A238455     d2 = (s+1)*(s+2)//2 - nn
%o A238455     if d2 < d1:  d1 = -d2
%o A238455     print(str(d1), end=',')
%o A238455 (PARI) a(n) = my(p=4^n, t=sqrtint(2*p)); (-t^2 - t + 2*p)/2; \\ _Michel Marcus_, Jun 16 2022
%Y A238455 Absolute values give the other bisection of A233327.
%Y A238455 Cf. A000079, A000217, A053616.
%K A238455 sign
%O A238455 0,4
%A A238455 _Alex Ratushnyak_, Feb 26 2014