A132417 a(16j+i) = 8(16j+i) + e_i, for j >= 0, 0 <= i <= 15, where e_0, ..., e_15 are 2, -2, -6, -10, -14, -18, -22, -26, -30, -34, -38, -42, -46, -50, -54, 6.

2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 254, 258, 262, 266, 270, 274, 278, 282, 286, 290, 294, 298, 302, 306, 310, 314, 382, 386, 390, 394, 398, 402, 406, 410, 414

Offset: 0

Author

Philippe Deléham, Nov 13 2007, Mar 29 2009

Keywords

• nonn

Comments

Certainly by term n = 8*(2^119 - 1) = 10^36.72..., this sequence and A103747 disagree.

The point of divergence is substantially earlier, described precisely by Charlie Neder in his Feb 2019 comment in A103747. - Peter Munn, Jan 14 2024

Links

• David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).
• Index entries for sequences which agree for a long time but are different

Crossrefs

Cf. A102370 (Sloping binary numbers), A103747 (trajectory of 2).

Formula

a(n) = a(n-1) + a(n-16) - a(n-17). - R. J. Mathar, Jul 21 2013

G.f.: (2 + 4*x + 4*x^2 + 4*x^3 + 4*x^4 + 4*x^5 + 4*x^6 + 4*x^7 + 4*x^8 + 4*x^9 + 4*x^10 + 4*x^11 + 4*x^12 + 4*x^13 + 4*x^14 + 68*x^15 + 2*x^16 ) / ( (1+x) *(x^2+1) *(x^4+1) *(x^8+1) *(x-1)^2 ). - R. J. Mathar, Jul 21 2013