cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A255747 Partial sums of A160552.

Original entry on oeis.org

0, 1, 2, 5, 6, 9, 14, 21, 22, 25, 30, 37, 42, 53, 70, 85, 86, 89, 94, 101, 106, 117, 134, 149, 154, 165, 182, 201, 222, 261, 310, 341, 342, 345, 350, 357, 362, 373, 390, 405, 410, 421, 438, 457, 478, 517, 566, 597, 602, 613, 630, 649, 670, 709, 758, 793, 814, 853, 906, 965, 1046, 1173, 1302, 1365, 1366, 1369, 1374
Offset: 0

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Author

Omar E. Pol, Mar 05 2015

Keywords

Comments

It appears that the sums of two successive terms give the positive terms of the toothpick sequence A139250.
It appears that the odd terms (a bisection) give A162795.
It appears that a(n) is also the total number of ON cells at stage n+1 in one of the four wedges of two-dimensional cellular automaton "Rule 750" using the von Neumann neighborhood (see A169707). Therefore a(n) is also the total number of ON cells at stage n+1 in one of the four quadrants of the NW-NE-SE-SW version of that cellular automaton.
See also the formula section.
First differs from A169779 at a(11).

Examples

			Also, written as an irregular triangle in which the row lengths are the terms of A011782 (the number of compositions of n, n >= 0), the sequence begins:
0;
1;
2,   5;
6,   9, 14, 21;
22, 25, 30, 37, 42, 53, 70, 85;
86, 89, 94,101,106,117,134,149,154,165,182,201,222,261,310,341;
...
It appears that the first column gives 0 together with the terms of A047849, hence the right border gives A002450.
It appears that this triangle only shares with A151920 the positive elements of the columns 1, 2, 4, 8, 16, ... (the powers of 2).
		

Crossrefs

Programs

  • Mathematica
    Accumulate[Nest[Join[#, 2 # + Append[Rest@#, 1]] &, {0}, 6]] (* Ivan Neretin, Feb 09 2017 *)

Formula

It appears that a(n) + a(n-1) = A139250(n), n >= 1.
It appears that a(2n-1) = A162795(n), n >= 1.
It appears that a(n) = (A169707(n+1) - 1)/4.