

A111138


Let b(n) denote the number of nontriangular numbers less than or equal to n. Then a(n) = b(n1) + a(b(n1)), with a(1) = a(2) = 0, a(3) = 1.


3



0, 0, 1, 1, 2, 4, 4, 5, 7, 10, 10, 11, 13, 16, 20, 20, 21, 23, 26, 30, 35, 35, 36, 38, 41, 45, 50, 56, 56, 57, 59, 62, 66, 71, 77, 84, 84, 85, 87, 90, 94, 99, 105, 112, 120, 120, 121, 123, 126, 130, 135, 141, 148, 156, 165, 165, 166, 168, 171, 175, 180, 186, 193, 201
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OFFSET

1,5


COMMENTS

For a subgroup H of order p^n (p an odd prime) of the subgroup generated by all commutators [x_j,x_i] in the relatively free group F of class three and exponent p, freely generated by x_1, x_2,..., x_k, (k sufficiently large) the minimum size of the subgroup of [H,F] of F_3 is p^{kn  a(n)}.
The sequence arises when finding a purely numerical sufficient condition for the capability of pgroups of class two and exponent p, where p is an odd prime.
Partial sums of A002262.  Gionata Neri, Sep 04 2015


LINKS

Table of n, a(n) for n=1..64.
Arturo Magidin, Capable groups of prime exponent and class two II, arXiv:math/0506578 [math.GR], 2005.


FORMULA

If we write n = (m choose 2) + s, 0<= s <= m, then a(n)=(m choose 3) + (s choose 2).
Set R:=Round(Sqrt(2*N)) & T:=NComb(R,2) then
a(N)=Comb(T,2)+Comb(R,3).  Gerald Hillier, Nov 18 2017


EXAMPLE

a(31) = b(30) + a(b(30)) = 23 + a(23) = 23 + b(22) + a(b(22)) = 23 + 16 + a(16 = 39 + b(15) + a(b(15)) = 39 + 10 + a(10) = 49 + b(9) + a(b(9)) = 49 + 6 + a(6) = 55 + b(5) + a(b(5)) = 55 + 3 + a(3) = 58 + 1 = 59.


MATHEMATICA

a[1] = a[2] = 0; a[3] = 1; a[n_] := a[n] = b[n  1] + a[b[n  1]]; b[n_] := n  Floor[(Sqrt[8n + 1]  1)/2]; Array[a, 64] (* Robert G. Wilson v, Feb 01 2006 *)


CROSSREFS

Cf. A083920.
Sequence in context: A118001 A182414 A256984 * A035625 A219875 A132128
Adjacent sequences: A111135 A111136 A111137 * A111139 A111140 A111141


KEYWORD

nonn


AUTHOR

Arturo Magidin, Oct 17 2005; definition corrected Feb 01 2006


EXTENSIONS

More terms from Robert G. Wilson v, Feb 01 2006


STATUS

approved



