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

0, 1, 1, -2, 3, -11, 1, -87, -167, -306, -500, -552, 688, -3041, -579, 20854, 37075, 55618, 37108, -222296, -147729, 891994, 602155, -3523022, -2228805, 14811346, 11792251, -47737262, -1136517, 375078994, 741065851, 1445763154, 2746052116, 4910207464, 7492827856

Offset: 0

Author

Alex Ratushnyak, Feb 26 2014

Keywords

• sign

Examples

			a(0) = 1 - 1 = 0.
a(1) = 4 - 3 = 1.
a(2) = 16 - 15 = 1.
a(3) = 64 - 66 = -2.
a(4) = 256 - 253 = 3.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000

Crossrefs

Absolute values give the other bisection of A233327.

Cf. A000079, A000217, A053616.

Programs

Mathematica

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[d1Harvey P. Dale, Jul 02 2019 *)

PARI

a(n) = my(p=4^n, t=sqrtint(2*p)); (-t^2 - t + 2*p)/2; \\ Michel Marcus, Jun 16 2022

Python

def isqrt(a):
    sr = 1 << (int.bit_length(int(a)) >> 1)
    while a < sr*sr:  sr>>=1
    b = sr>>1
    while b:
        s = sr + b
        if a >= s*s:  sr = s
        b>>=1
    return sr
for n in range(77):
    nn = 4**n
    s = isqrt(2*nn)
    if s*(s+1)//2 > nn:  s-=1
    d1 = nn - s*(s+1)//2
    d2 = (s+1)*(s+2)//2 - nn
    if d2 < d1:  d1 = -d2
    print(str(d1), end=',')

Formula

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