Search: a001219 -id:a001219
|
| |
|
|
A097571
|
|
Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of three consecutive integers.
|
|
+10
4
|
| |
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Replacing "three" by "two" we get A001652.
Replacing "three" by "N" we get: {15} for N = 4 and 5, {2079} for N = 6 and no solutions for N >= 7. - J. B. M. Melissen.
t(a(n)) equals x*(x+1)*(x+2) for x = [0, 1, 4, 5, 9, 56, 636]_n. - Zak Seidov, Jun 21 2013
|
|
|
LINKS
|
|
|
|
MATHEMATICA
|
(Sqrt[8#+1]-1)/2&/@Select[Table[n(n+1)(n+2), {n, 0, 23000}], OddQ[Sqrt[8#+1]]&] (* Harvey P. Dale, Jan 12 2023 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,fini,full
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A214838
|
|
Triangular numbers of the form k^2 + 2.
|
|
+10
4
|
|
|
|
3, 6, 66, 171, 2211, 5778, 75078, 196251, 2550411, 6666726, 86638866, 226472403, 2943171003, 7693394946, 99981175206, 261348955731, 3396416785971, 8878171099878, 115378189547778, 301596468440091, 3919462027838451, 10245401755863186, 133146330756959526, 348042063230908203
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Corresponding k values are in A077241.
Except 3, all terms are in A089982: in fact, a(2) = 3+3 and a(n) = (k-2)*(k-1)/2+(k+1)*(k+2)/2, where k = sqrt(a(n)-2) > 2 for n > 2. [Bruno Berselli, Mar 08 2013]
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: -3*x*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)). - Joerg Arndt, Mar 08 2013
a(n) = A000217(t), where t = ((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4. - Bruno Berselli, Mar 08 2013
|
|
|
EXAMPLE
|
2211 is in the sequence because 2211 = 47^2 + 2.
|
|
|
MATHEMATICA
|
LinearRecurrence[{1, 34, -34, -1, 1}, {3, 6, 66, 171, 2211}, 25] (* Bruno Berselli, Mar 08 2013 *)
|
|
|
PROG
|
(Python)
import math
for i in range(2, 1L<<32):
t = i*(i+1)/2 - 2
sr = int(math.sqrt(t))
if sr*sr == t:
print '%10d' % sr, '%10d' % i, t+2
(PARI) for(n=1, 10^9, t=n*(n+1)/2; if(issquare(t-2), print1(t, ", "))); \\ Joerg Arndt, Mar 08 2013
(Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(-3*(x^4+x^3-14*x^2+x+1)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)))); // Bruno Berselli, Mar 08 2013
(Maxima)
t[n]:=((5-2*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2))+(5+2*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))-2)/4$
makelist(expand(t[n]*(t[n]+1)/2), n, 1, 25); /* Bruno Berselli, Mar 08 2013 */
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A165892
|
|
Triangular numbers of form n(n+2)(n+4).
|
|
+10
3
|
| |
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Values of (m^2-1)/8 corresponding to ordinates of integral points on the elliptic curve: m^2 = 8*n^3 + 48*n^2 + 64*n + 1.
Corresponding values of n are given in A165893.
|
|
|
LINKS
|
|
|
|
MATHEMATICA
|
TNQ[m_]:=IntegerQ[Sqrt[1+8*m]]; Do[If[TNQ[m=n*(n+2)*(n+4)], Print[m]], {n, 2*10^3}]
Select[Table[n(n+2)(n+4), {n, 0, 2000}], OddQ[Sqrt[8#+1]]&] (* Harvey P. Dale, Feb 07 2015 *)
|
|
|
CROSSREFS
|
Cf. A001219 Triangular numbers of form a(a+1)(a+2).
|
|
|
KEYWORD
|
nonn,fini,full
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Initial 0 added by Zak Seidov, Oct 04 2009 at the suggestion of Alexander R. Povolotsky.
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A227027
|
|
Triangular numbers representable as x!/y! with y < x-1.
|
|
+10
3
|
|
|
|
6, 120, 210, 990, 7140, 185136, 242556, 2162160, 8239770, 258474216, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Triangular numbers in A045619, except A045619(1)=0. The sequence is infinite because A029549 is a subsequence. According to Melissen's comment in A097571, y > x-7.
The sequence of x's producing a(n): A227026.
a(2) and a(3) have two representations:
a(2) = 120 = 5*4*3*2 = 6*5*4.
a(3) = 210 = 7*6*5 = 15*14.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
990 is in the sequence since 990 = 11!/8! = 11*10*9 is a ratio of factorials and 990 = (44)(44 + 1)/2 is a triangular number.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A165893
|
|
Numbers n with property that n(n+2)(n+4) is a triangular number.
|
|
+10
2
|
| |
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
Abscissas of integral points on the elliptic curve: m^2 = 8*n^3 + 48*n^2 + 64*n + 1.
|
|
|
LINKS
|
|
|
|
MATHEMATICA
|
TNQ[n_]:=IntegerQ[Sqrt[1+8n]]; Select[Range[7500], TNQ[ #(#+2)(#+4)]&]
|
|
|
CROSSREFS
|
Cf. A001219 Triangular numbers of form a(a+1)(a+2), A165892 Triangular numbers of form a(a+2)(a+4).
|
|
|
KEYWORD
|
nonn,fini,full
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
Initial 0 added by Zak Seidov, Oct 04 2009 at the suggestion of Alexander R. Povolotsky.
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A165519
|
|
Integers k for which k(k+1)(k+2) is a triangular number.
|
|
+10
1
|
| |
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
This sequence is complete; there are no other integers k for which k(k+1)(k+2) is a triangular number.
Integers k such that 8*k*(k+1)*(k+2)+1 is a square. - Robert Israel, Nov 07 2014
|
|
|
REFERENCES
|
Guy, R. K.; "Figurate Numbers", D3 in Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, 1994, p. 148.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
The third triangular number which is a product of three consecutive integers is 4*5*6=120=T(15), but 4 is the fifth integer k for which k(k+1)(k+2) is a triangular number, so a(5)=4.
|
|
|
MAPLE
|
select(x -> issqr(8*x^3 + 24*x^2 + 16*x+1), [$-2..1000]); # Robert Israel, Nov 07 2014
|
|
|
MATHEMATICA
|
TriangularNumberQ[k_]:=If[IntegerQ[1/2 (Sqrt[1+8k]-1)], True, False]; Select[Range[750], TriangularNumberQ[ # (#+1)(#+2)] &]
With[{nos=Partition[Range[0, 1000], 3, 1]}, Transpose[Select[nos, IntegerQ[ (Sqrt[1+8Times@@#]-1)/2]&]][[1]]] (* Harvey P. Dale, Dec 25 2011 *)
|
|
|
PROG
|
(PARI) isok(k) = ispolygonal(k*(k+1)*(k+2), 3); \\ Michel Marcus, Oct 31 2014
(Magma) [-2, -1] cat [n: n in [0..1000] | IsSquare(8*n^3+24*n^2 +16*n+1)]; // Vincenzo Librandi, Nov 10 2014
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
sign,fini,full
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A226500
|
|
Triangular numbers representable as 3 * x^2.
|
|
+10
1
|
|
|
|
0, 3, 300, 29403, 2881200, 282328203, 27665282700, 2710915376403, 265642041604800, 26030209161894003, 2550694855824007500, 249942065661590841003, 24491771739980078410800, 2399943688452386093417403, 235169989696593857076494700, 23044259046577745607403063203
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3), for n > 3. a(n) = floor((49 + 20*sqrt(6))^(n-1)/32). - Giovanni Resta, Jun 09 2013
a(n) = (49+20*sqrt(6))^(-n)*(49+20*sqrt(6)-2*(49+20*sqrt(6))^n+(49-20*sqrt(6))*(49+20*sqrt(6))^(2*n))/32. - Colin Barker, Mar 03 2016
|
|
|
MATHEMATICA
|
a[1]=0; a[2]=3; a[3]=300; a[n_] := a[n] = 99*(a[n-1] - a[n-2]) + a[n-3]; Array[a, 10] (* Giovanni Resta, Jun 09 2013 *)
Rest@ CoefficientList[Series[3 x^2 (1 + x)/((1 - x) (1 - 98 x + x^2)), {x, 0, 16}], x] (* or *)
|
|
|
PROG
|
(C)
#include <stdio.h>
#include <math.h>
typedef unsigned long long U64;
U64 isTriangular(U64 a) { // input must be < 1ULL<<63
U64 r = sqrt(a*2);
return (r*(r+1) == a*2);
}
int main() {
for (U64 j, i = 0; (j=i*i*3) < (1ULL<<63); i++)
if (isTriangular(j)) printf("%llu, ", j);
return 0;
}
|
|
|
CROSSREFS
|
Cf. A029549 (triangular numbers representable as x^2 + x).
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
Search completed in 0.013 seconds
|
