A309423 - OEIS

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A309423

Numbers k with property that there is an m = m(k) such that m(m+1)/2 divides k(k+1)/2 and m(k) > m(i) for all i < k.

2, 3, 8, 9, 14, 20, 35, 77, 84, 99, 119, 231, 260, 351, 494, 696, 1665, 1785, 1845, 3479, 4059, 6887, 16302, 17919, 23183, 23660, 56979, 58800, 81968, 83880, 95930, 137903, 340955, 358017, 574925, 803760, 1336139, 3375111, 4684659, 10316619, 14935095, 18610022

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

gcd(a(n),m(n)) < m(n), n > 1.

Let k = r * m, and r = (A000217(k) * (m+1))/((A000217(m) * (k+1)). For known terms 1 < r < 4.

LINKS

Table of n, a(n) for n=1..42.

EXAMPLE

2 is a term because A000217(1) divides A000217(2).

3 is a term because A000217(2) divides A000217(3).

8 is a term because A000217(3) divides A000217(8).

9 is a term because A000217(5) divides A000217(9).

14 is a term because A000217(6) divides A000217(14).

MATHEMATICA

m=1; L={2}; k=2; While[k < 10000, k++; tr = k (k + 1)/2; t = SelectFirst[ Reverse@ Divisors[2 tr], # != k && Mod[tr, # (# + 1)/2] == 0 &]; If[t > m, AppendTo[L, k]; m = t]]; L (* Giovanni Resta, Sep 05 2019 *)

PROG

(PARI) T(n) = {return((n * (n+1)) / 2)}

Tk(n, k) = {for (i = k, n - 1, if ((T(n)%T(i))==0, return(i+1)))}

Tn(n) = {phwm = 1; for (i = 2, n, nhwm = Tk(i, phwm); if(nhwm > phwm, phwm = nhwm; print1(i, ", ")))}

Tn(5000000)

CROSSREFS

Cf. A000217.

Sequence in context: A047243 A277094 A099148 * A029787 A246396 A054462

Adjacent sequences: A309420 A309421 A309422 * A309424 A309425 A309426

KEYWORD

nonn

AUTHOR

Torlach Rush, Jul 30 2019

EXTENSIONS

More terms from Giovanni Resta, Sep 05 2019

STATUS

approved