| Displaying 1-4 of 4 results found.
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0, 0, 0, 0, 1, 1, 0, 1, 2, 2, 1, 1, 3, 3, 2, 0, 4, 4, 2, 7, 5, 5, 1, 7, 6, 6, 10, 3, 7, 7, 0, 5, 8, 8, 10, 4, 9, 9, 7, 2, 10, 10, 5, 16, 11, 11, 1, 10, 17, 12, 19, 6, 13, 22, 17, 10, 14, 14, 7, 7, 15, 15, 10, 6, 27, 16, 8, 25, 17, 17, 4, 4, 18, 18, 28, 9
Numbers k such that A332547(k) = 3.
+20
1
5, 6, 8, 11, 12, 23, 47, 96, 191, 192, 383, 768, 6143, 12288, 786431, 786432, 3221225472, 51539607551, 206158430208, 824633720831, 6597069766656, 26388279066623, 108086391056891903, 55340232221128654847, 221360928884514619392, 226673591177742970257407
COMMENTS
The numbers k such that A332547(k) = 1 are given by A068194, a sequence of interest to Mersenne and Fermat, so this sequence may also be interesting. The factors of the initial terms are 5, 2*3, 2^3, 11, 2^2*3, 23, 47, 2^5*3, 191, 2^6*3, 383, 2^8*3, 6143, 2^12*3, 786431, 2^18*3, ...
There are essentially two cases. Firstly n can be an odd prime and n+1 of the form 3*2^k. These are the terms of A007505 with 2 excluded. Otherwise n can be of the form 3*2^k and n+1 a prime. These are 1 less than the terms of A039687. In addition, 8 is a term which is a special case. - Andrew Howroyd, Feb 21 2020
PROG
(PARI) upto(n)={Set(concat([if(n<8, [], [8]), select(isprime, [3*2^k-1 |k<-[1..logint((n+1)\3, 2)]]), select(p->isprime(p+1), [3*2^k |k<-[1..logint(n\3, 2)]])]))} \\ Andrew Howroyd, Feb 21 2020
a(n) = (n*(n+1)/2)/Q(n) - (Q(n)+1)/2, where Q(n) = A332547(n).
+20
1
2, 5, 9, 3, 5, 27, 10, 6, 8, 20, 24, 9, 11, 21, 135, 12, 14, 35, 6, 15, 17, 90, 12, 18, 20, 7, 54, 21, 23, 495, 42, 24, 26, 19, 69, 27, 29, 44, 161, 30, 32, 80, 13, 33, 35, 374, 45, 17, 38, 14, 99, 39, 10, 26, 65, 42, 44, 110, 114, 45, 47, 85, 153, 11
Smallest k > 0 such that T(n) + T(k) = T(m), for some m, T(i) being the triangular numbers, n > 1.
+10
17
2, 5, 9, 3, 5, 27, 10, 4, 8, 14, 17, 9, 5, 21, 135, 12, 14, 35, 6, 9, 17, 30, 12, 18, 10, 7, 54, 21, 23, 495, 42, 14, 26, 8, 49, 27, 15, 20, 98, 30, 32, 80, 9, 19, 35, 62, 45, 17, 20, 14, 99, 39, 10, 18, 54, 24, 44, 78, 81, 45, 25, 85, 153, 11, 50, 125, 20, 29, 53, 94, 97
COMMENTS
For 16 years this entry stood with no upper bound, and indeed with no proof that a(n) always existed. In February 2020 the following three bounds and formulas arrived. They are listed in chronological order. Here k = k(n) denotes the smallest number such that T(n)+T(k) is a triangular number T(m) for some m = m(n). - N. J. A. Sloane, Feb 22 2020 k = T(n) - 1 is an upper bound on k(n) = a(n). For T(k) makes a huge triangle; all the elements of the T(n) triangle can be thinly plated onto the side of the big one as a single additional row, producing T(k+1) with m = k+1. - Allan C. Wechsler, Feb 19 2020 Let Q be the largest odd number < n dividing T(n). Then T(n) is the sum of Q consecutive integers, the last Q rows of the triangle T(m) with m = T(n)/Q + (Q-1)/2, giving the upper bound k <= T(n)/Q - (Q+1)/2. [This bound is now A332554, the values of Q are in A332547.] This bound is not tight: for n=9 it gives a(9) <= 6 when in fact a(9) = 4. - Michael J. Collins, Feb 19 2020 Comments from Richard C. Schroeppel, Feb 19 2020: (Start)
2T(n) = 2T(m) - 2T(k) = m^2 + m - k^2 - k = (m-k) (m + k + 1). Now (m-k) and (m+k+1) are of opposite parity. Factor 2T(n) into the product of an odd number times an even number. We can take one of these to be m-k, and the other to be m+k+1.
The factorization 2T(n) = n^2 + n gives two obvious solutions, n * (n+1) and 1 * (n^2+n). Equating these to (m-k) * (m+k+1) gives the two "trivial" solutions k=0, m=n and k=T(n)-1, m=T(n).
Unless n is a Mersenne prime, or n+1 is a Fermat prime [these are the n such that Q=1, see A068194] there will be a nontrivial odd divisor of n(n+1) other than 1, n, or n+1. Select the odd divisor d logarithmicly closest to n + 1/2 that isn't n or n+1. Let q be the quotient n(n+1)/q. Then m-k = min(d,q) and m+k+1 = max(d,q). Solve for k, which is the required minimum k(n) = a(n).
Example: n=5, T(n)=15, 2T(n)=30 = 3*10, d=3, q=10, k=3, m=6, 15+6 = 21. (End)
MAPLE
f:= proc(n) local e, t, te;
t:= n*(n+1);
e:= padic:-ordp(t, 2);
te:= 2^e;
min(map(d -> (abs(te*d-t/(te*d))-1)/2, numtheory:-divisors(t/te)) minus {0}):
MATHEMATICA
Table[SelectFirst[Range[10^3], Function[m, PolygonalNumber@ Floor@ Sqrt[2 m] == m][PolygonalNumber[n] + PolygonalNumber[#]] &], {n, 2, 72}] (* Michael De Vlieger, Sep 19 2017, after Maple by Robert Israel *)
PROG
(PARI) for(n=2, 100, t=n*(n+1)/2; for(k=1, 10^9, u=t+k*(k+1)/2; v=floor(sqrt(2*u)); if(v*(v+1)/2==u, print1(k", "); break)))
(Python)
from __future__ import division
from sympy import divisors
t = n*(n+1)
ds = divisors(t)
for i in range(len(ds)//2-2, -1, -1):
x = ds[i]
y = t//x
a, b = divmod(y-x, 2)
if b:
return a
CROSSREFS
See A055527 for a very similar sequence involving Pythagorean triples. - Bradley Klee, Feb 20 2020 See also A309332 (number of ways to write a triangular number as a sum of two triangular numbers), A309507 (... as a difference ...).
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