A309878 The real part of b(n) where b(n) = (n + b(n-1)) * (1 + i) with b(-1)=0; i = sqrt(-1).

0, 1, 2, 1, -4, -13, -22, -23, -8, 23, 54, 53, -12, -141, -270, -271, -16, 495, 1006, 1005, -20, -2069, -4118, -4119, -24, 8167, 16358, 16357, -28, -32797, -65566, -65567, -32, 131039, 262110, 262109, -36, -524325, -1048614, -1048615, -40, 2097111, 4194262, 4194261, -44

Offset: 0

Author

Jesse Fiedler, Aug 21 2019

Keywords

• sign
• easy

Comments

Observe that (starting with n=1) the sequence has a pattern of a cluster of 3 positive numbers followed by a cluster of 5 negative numbers.

Observe also if the clusters of 3 positive numbers are represented by x, y, z; then y = (x * 2) + (8 * k) where k a positive integer ; when this happens, k = (n - 1) / 8 ; therefore y = x * 2 + n - 1; z = y - 1

Observe also that within each cluster of 5 negative numbers, the first and last are orders of magnitude less than the middle 3 values. The first and last always differ by 4 and are always equal to -n.

Observe also if the clusters of 5 negative numbers are represented by c, d, e, f, g ; then d - c = e - d; f = e - 1; g = c - 4

Examples

			For n = -1; b(n) = 0
For n =  0; b(n) = (0+0)*(1+i) = 0
For n =  1; b(n) = (1+0)*(1+i) = 1+i ; a(1) = Re(1+i) = 1
For n =  2; b(n) = (2+1+i)*(1+i) = (3+i)*(1+i) = 3+i+3i-1 = 2+4i ; a(2) = Re(2+4i) = 2
For n =  3; b(n) = 1+9i ; a(3) = 1
For n =  4; b(n) = -4+14i ; a(4) = -4
For n =  5; b(n) = -13+15i ; a(5) = -13
For n =  6; b(n) = -22+8i ; a(6) = -22
For n =  7; b(n) = -23-7i ; a(7) = -23
...
For n =  31; b(n) = -65567-65503i ; a(31) = -65567
For n =  32; b(n) = (32-65567-65503i)*(1+i) = (-65535-65503i)*(1+i) = -65535-65503i-65535i+65503 = -32-131038i ; a(32) = -32
For n =  33; b(n) = 131039-131037i ; a(33) = 131039
		

Links

• Sela Fried, Sequence A309878, 2024.
• Sela Fried, Proofs of some Conjectures from the OEIS, arXiv:2410.07237 [math.NT], 2024. See p. 9.

Programs

PARI

b(n) = if (n==0, 0, (n + b(n-1)) * (1 + I));
for (n=0, 50, print1(real(b(n)), ", ")) \\ Michel Marcus, Aug 21 2019

Formula

Conjectures from Colin Barker, Aug 21 2019: (Start)

G.f.: x*(1 - 2*x) / ((1 - x)^2*(1 - 2*x + 2*x^2)).

a(n) = 4*a(n-1) - 7*a(n-2) + 6*a(n-3) - 2*a(n-4) for n>3.

a(n) = i*((1-i)^n - (1+i)^n + i*n) where i=sqrt(-1).

(End)

E.g.f.: exp(x)*(-x + 2*sin(x)). - Conjectured by Stefano Spezia, Aug 21 2019 after Colin Barker

Conjecture: a(n) = 2*A009545(n)-n. - R. J. Mathar, Mar 06 2022

All conjectures stated above hold true. See links. - Sela Fried, Jul 27 2024.