cp's OEIS Frontend

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

%I A309878 #37 Oct 14 2024 11:21:37
%S A309878 0,1,2,1,-4,-13,-22,-23,-8,23,54,53,-12,-141,-270,-271,-16,495,1006,
%T A309878 1005,-20,-2069,-4118,-4119,-24,8167,16358,16357,-28,-32797,-65566,
%U A309878 -65567,-32,131039,262110,262109,-36,-524325,-1048614,-1048615,-40,2097111,4194262,4194261,-44
%N A309878 The real part of b(n) where b(n) = (n + b(n-1)) * (1 + i) with b(-1)=0; i = sqrt(-1).
%C A309878 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.
%C A309878 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
%C A309878 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.
%C A309878 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
%H A309878 Sela Fried, <a href="/A309878/a309878.pdf">Sequence A309878</a>, 2024.
%H A309878 Sela Fried, <a href="https://arxiv.org/abs/2410.07237">Proofs of some Conjectures from the OEIS</a>, arXiv:2410.07237 [math.NT], 2024. See p. 9.
%F A309878 Conjectures from _Colin Barker_, Aug 21 2019: (Start)
%F A309878 G.f.: x*(1 - 2*x) / ((1 - x)^2*(1 - 2*x + 2*x^2)).
%F A309878 a(n) = 4*a(n-1) - 7*a(n-2) + 6*a(n-3) - 2*a(n-4) for n>3.
%F A309878 a(n) = i*((1-i)^n - (1+i)^n + i*n) where i=sqrt(-1).
%F A309878 (End)
%F A309878 E.g.f.: exp(x)*(-x + 2*sin(x)). - Conjectured by _Stefano Spezia_, Aug 21 2019 after _Colin Barker_
%F A309878 Conjecture: a(n) = 2*A009545(n)-n. - _R. J. Mathar_, Mar 06 2022
%F A309878 All conjectures stated above hold true. See links. - _Sela Fried_, Jul 27 2024.
%e A309878 For n = -1; b(n) = 0
%e A309878 For n =  0; b(n) = (0+0)*(1+i) = 0
%e A309878 For n =  1; b(n) = (1+0)*(1+i) = 1+i ; a(1) = Re(1+i) = 1
%e A309878 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
%e A309878 For n =  3; b(n) = 1+9i ; a(3) = 1
%e A309878 For n =  4; b(n) = -4+14i ; a(4) = -4
%e A309878 For n =  5; b(n) = -13+15i ; a(5) = -13
%e A309878 For n =  6; b(n) = -22+8i ; a(6) = -22
%e A309878 For n =  7; b(n) = -23-7i ; a(7) = -23
%e A309878 ...
%e A309878 For n =  31; b(n) = -65567-65503i ; a(31) = -65567
%e A309878 For n =  32; b(n) = (32-65567-65503i)*(1+i) = (-65535-65503i)*(1+i) = -65535-65503i-65535i+65503 = -32-131038i ; a(32) = -32
%e A309878 For n =  33; b(n) = 131039-131037i ; a(33) = 131039
%o A309878 (PARI) b(n) = if (n==0, 0, (n + b(n-1)) * (1 + I));
%o A309878 for (n=0, 50, print1(real(b(n)), ", ")) \\ _Michel Marcus_, Aug 21 2019
%K A309878 sign,easy
%O A309878 0,3
%A A309878 _Jesse Fiedler_, Aug 21 2019