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A111138
Let b(n) denote the number of nontriangular numbers less than or equal to n. Then a(n) = b(n-1) + a(b(n-1)), 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
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 p-groups of class two and exponent p, where p is an odd prime.
Partial sums of A002262. - Gionata Neri, Sep 04 2015
LINKS
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).
a(N) = Comb(T,2)+Comb(R,3) where R:=Round(Sqrt(2*N)) and T:=N-Comb(R,2). - 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 *)
PROG
(PARI) a(n) = my(r, m=sqrtint(n<<1, &r)); if(r<m, r+=m, r-=m; m++); binomial(m, 3) + binomial(r>>1, 2); \\ Kevin Ryde, Oct 26 2024
CROSSREFS
Cf. A083920.
Sequence in context: A118001 A182414 A256984 * A349462 A340761 A035625
KEYWORD
nonn,easy
AUTHOR
Arturo Magidin, Oct 17 2005; definition corrected Feb 01 2006
EXTENSIONS
More terms from Robert G. Wilson v, Feb 01 2006
STATUS
approved