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Search: a024612 -id:a024612
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Numbers of the form x^2 + xy + y^2, where x and y are positive integers.
+10
17
3, 7, 12, 13, 19, 21, 27, 28, 31, 37, 39, 43, 48, 49, 52, 57, 61, 63, 67, 73, 75, 76, 79, 84, 91, 93, 97, 103, 108, 109, 111, 112, 117, 124, 127, 129, 133, 139, 147, 148, 151, 156, 157, 163, 169, 171, 172, 175, 181, 183, 189, 192, 193, 196, 199, 201, 208, 211, 217, 219, 223, 228
OFFSET
1,1
COMMENTS
Equivalently, sequence A024612 with duplicates removed; i.e., numbers of the form i^2 - i*j + j^2, where 1 <= i < j.
A subsequence of A135412, which consists of multiples (by squarefree factors) of the numbers listed here. It appears that this lists numbers > 1 which have in their factorization: (a) no even power of 3 unless there is a factor == 1 (mod 6); (b) no odd power of 2 or of a prime == 5 (mod 6) and no even power unless there is a factor 3 or == 1 (mod 6). - M. F. Hasler, Aug 17 2016
If we regroup the entries in a triangle with row lengths A004526
3,
7,
12, 13,
19, 21,
27, 28, 31,
37, 39, 43,
... it seems that the j-th row of the triangle contains the numbers i^2+j^2-i*j in row j>=2 and column i = floor((j+1)/2) .. j-1. - R. J. Mathar, Aug 21 2016
Proof of the above characterization: the sequence is the union of 3*(the squares A000290) and A024606 (numbers x^2+xy+y^2 with x > y > 0). For the latter it is known that these are the numbers with a factor p==1 (mod 6) and any prime factor == 2 (mod 3) occurring to an even power. The former (3*n^2) are the same as (odd power of 3)*(even power of any other prime factor). The union of the two cases yields the earlier characterization. - M. F. Hasler, Mar 04 2018
Least term that can be written in exactly n ways is A300419(n). - Altug Alkan, Mar 04 2018
For the general theory see the Fouvry et al. reference and A296095. Bounds used in the Julia program are based on the theorems in this paper. - Peter Luschny, Mar 10 2018
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Peter Luschny)
Étienne Fouvry, Claude Levesque, and Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.
EXAMPLE
3 = 1^2 + 1^2 + 1*1, 7 = 2^2 + 1^2 + 2*1, ...
MAPLE
isA024614 := proc(n)
local i, j, disc;
# n=i^2+j^2-i*j = (j-i)^2+i*j, 1<=i<j
# so (j-i)>=1 and i*j>=j and i^2+j^2-i*j >= 1+j max search radius
for j from 2 to n-1 do
# i=(j +- sqrt(4n-3j^2))/2
disc := 4*n-3*j^2 ;
if disc >= 0 then
if issqr(disc) then
i := (j+sqrt(disc))/2 ;
if type(i, 'integer') and i >= 1 and i<j then
return true;
end if;
i := (j-sqrt(disc))/2 ;
if type(i, 'integer') and i >= 1 and i<j then
return true;
end if;
end if;
end if;
end do:
return false;
end proc:
n := 1 :
for t from 1 to 228 do
if isA024614(t) then
printf("%d %d\n", n, t) ;
n := n+1 ;
end if;
end do: # R. J. Mathar, Aug 21 2016
# second Maple program:
a:= proc(n) option remember; local k, x;
for k from a(n-1)+1 do for x while x^2<k do
if (y-> x^2+(x+y)*y=k)((isqrt(4*k-3*x^2)-x)/2) then return k fi
od od
end: a(0):=0:
seq(a(n), n=1..200); # Alois P. Heinz, Mar 02 2018
MATHEMATICA
max = 228; T0 = {}; xm = Ceiling[Sqrt[max]]; While[T = T0; T0 = Table[x^2 + x y + y^2, {x, 1, xm}, {y, x, xm}] // Flatten // Union // Select[#, # <= max&]&; T != T0, xm = 2 xm]; T (* Jean-François Alcover, Mar 23 2018 *)
PROG
(Julia)
function isA024614(n)
n % 3 >= 2 && return false
n == 3 && return true
M = Int(round(2*sqrt(n/3)))
for y in 2:M, x in 1:y
n == x^2 + y^2 + x*y && return true
end
return false
end
A024614list(upto) = [n for n in 1:upto if isA024614(n)]
println(A024614list(228)) # Peter Luschny, Mar 02 2018 updated Mar 17 2018
(PARI) is(n)={n>2&&!for(i=1, #n=Set(Col(factor(n)%6))/*consider prime factors mod 6*/, n[i][1]>1||next/*skip factors = 1 mod 6*/; /* odd power: ok only if p=3 */n[i][2]%2&&(n[i][1]!=3 || next) && return; /*even power: ok if there's a p==1, listed first*/ n[1][1]==1 || /*also ok if it's not a 3 and if there's a 3 elsewhere */ (n[i][1]==2 && i<#n && n[i+1][1]==3) || (n[i][1]>3 && for(j=1, i-1, n[j][1]==3 && next(2))||return))} \\ M. F. Hasler, Aug 17 2016, documented & bug fixed (following an observation by Altug Alkan) Mar 04 2018
(PARI) is(n)={(n=factor(n))||return/*n=1*/; /*exponents % 2, primes % 3*/ n[, 2]%=2; n[, 1]%=3; (n=Set(Col(n))) /*odd power of a prime == 2? will be last*/ [#n][2] && n[#n][1]==2 && return; /*factor == 1? will be 1st or after 3*/ n[1+(!n[1][1] && #n>1)][1]==1 || /*thrice a square?*/ (!n[1][1]&&n[1][2]&&!for(i=2, #n, n[i][2]&&return))} \\ Alternate code, 5% slower, maybe a bit less obscure. - M. F. Hasler, Mar 04 2018
(PARI) N=228; v=vector(N);
for(x=1, N, x2=x*x; if(x2>N, break); for(y=x, N, t=x2+y*(x+y); if(t>N, break); v[t]=1; ));
for(n=1, N, if(v[n], print1(n, ", "))); \\ Joerg Arndt, Mar 10 2018
(PARI) list(lim)=my(v=List(), x2); lim\=1; for(x=1, sqrtint(4*lim\3), x2=x^2; for(y=1, min((sqrt(4*lim-3*x2)-x)/2, x), listput(v, y*(x+y)+x2))); Set(v) \\ Charles R Greathouse IV, Mar 23 2018
CROSSREFS
Cf. A003136, A007645 (prime terms), A024612, A135412, A296095, A300419.
KEYWORD
nonn,easy
EXTENSIONS
Edited by M. F. Hasler, Aug 17 2016
b-file for values a(1)..a(10^4) double-checked with PARI code by M. F. Hasler, Mar 04 2018
STATUS
approved
Primitive triples for integer-sided triangles whose angles A < B < C are in arithmetic progression.
+10
10
3, 7, 8, 5, 7, 8, 7, 13, 15, 8, 13, 15, 5, 19, 21, 16, 19, 21, 11, 31, 35, 24, 31, 35, 7, 37, 40, 33, 37, 40, 13, 43, 48, 35, 43, 48, 16, 49, 55, 39, 49, 55, 9, 61, 65, 56, 61, 65, 32, 67, 77, 45, 67, 77, 17, 73, 80, 63, 73, 80, 40, 79, 91, 51, 79, 91, 11, 91, 96
OFFSET
1,1
COMMENTS
The triples are displayed in nondecreasing order of middle side, and if middle sides coincide then by increasing order of the largest side, hence, each triple (a, b, c) is in increasing order.
These three properties below are equivalent:
-> integer-sided triangles whose angles A < B < C are in arithmetic progression,
-> integer-sided triangles such that B = (A+C)/2 with A < C,
-> integer-sided triangles such that A < B < C with B = Pi/3.
When A < B < C are in arithmetic progression with B = A + phi and C = B + phi, then 0 < phi < Pi/3.
The corresponding metric relation between sides is b^2 = a^2 - a*c + c^2.
There exists such primitive triangle iff b^2 is an odd square term of A024612. Hence, the first few middle sides b are 7, 13, 19, 31, 37, 43, 49, 61, 67, ... and b is a term of A004611 \ {1}. Indeed, b cannot be even if the triple is primitive.
As B = Pi/3 and C runs from Pi/3 to 2*Pi/3, sin(C) gets a maximum when C = Pi/2 with sin(C) = 1, hence, from law of sines (see link): b/sin(B) = c/sin(C), and c < b/sin(Pi/3) = b * 2/sqrt(3) < 6*b/5. This bound is used in the PARI and Maple programs below.
When triple (a, b, c) is solution, then triple (c-a, b, c) is another solution. Hence, for each b odd solution, there exist 2 triples with same middle side b and same largest side c.
The common tangent to the nine-point circle and the incircle of a triangle ABC is parallel to the Euler line iff angles A < B < C are in arithmetic progression (see Crux Mathematicorum for Indian team selection). - Bernard Schott, Apr 14 2022
These triples are called (primitive) Eisenstein triples (Wikipedia). - Bernard Schott, Sep 21 2022
REFERENCES
V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.
LINKS
Bill Sands, Indian team selection test 2007, question 5, Crux Mathematicorum, Vol. 36, No. 5 (2010), p. 278.
Eric Weisstein's World of Mathematics, Law of Cosines.
Eric Weisstein's World of Mathematics, Law of Sines.
Wikipedia, Eisenstein triple.
EXAMPLE
(3, 7, 8) is a triple for this sequence because from law of cosines (see link), cos(A) = (7^2 + 8^2 - 3^2)/(2*7*8) = 13/14, cos(B) = (8^2 + 3^2 - 7^2)/(2*8*3) = 1/2 and cos(C) = (3^2 + 7^2 - 8^2)/(2*3*7) = -1/7; then, (A+C)/2 = ( arccos(13/14) + arccos(-1/7) )/2 = Pi/3 = B.
Also, arccos(13/14) ~ 21.787 degrees, arccos(1/2) = 60 degrees, arccos(-1/7) ~ 98.213 degrees, so B-A = C-B ~ 38.213 degrees, hence (A, B, C) are in arithmetic progression.
5^2 - 5*8 + 8^2 = 7^2, hence (5, 7, 8) is another triple for triangle whose angles A < B < C are in arithmetic progression.
MAPLE
for b from 3 to 250 by 2 do
for c from b+1 to 6*b/5 do
a := (c - sqrt(4*b^2-3*c^2))/2;
if gcd(a, b)=1 and issqr(4*b^2-3*c^2) then print(a, b, c, c-a, b, c); end if;
end do;
end do;
PROG
(PARI) lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print(a, ", ", b, ", ", c, ", "); print(c-a, ", ", b, ", ", c, ", "); ); ); ); ); } \\ Michel Marcus, Jul 15 2020
CROSSREFS
Cf. A335894 (smallest side), A335895 (middle side), A335896 (largest side), A335897 (perimeter).
Cf. A103606 (primitive Pythagorean triples), A335034 (primitive triples for triangles with two perpendicular medians).
KEYWORD
nonn,tabf
AUTHOR
Bernard Schott, Jun 29 2020
STATUS
approved
Triangle read by rows, sums of 2 distinct nonzero cubes: T(n,k) = (n+1)^3+k^3, 1 <= k <= n.
+10
1
9, 28, 35, 65, 72, 91, 126, 133, 152, 189, 217, 224, 243, 280, 341, 344, 351, 370, 407, 468, 559, 513, 520, 539, 576, 637, 728, 855, 730, 737, 756, 793, 854, 945, 1072, 1241, 1001, 1008, 1027, 1064, 1125, 1216, 1343, 1512, 1729
OFFSET
1,1
COMMENTS
When n=k: T(n,k) = (2n+1)(n^2+n+1). Therefore, T(n,k)/(2n+1) = A002061(n+1).
A002383 is the sequence of all primes of the form T(n,k)/(2n+1), n=k.
When starting at T(n,k) n=k, diagonal sums are n^2*(2n+1)^2. For example, starting at T(4,4) = 189: 189+243+351+513 = 4^2*9^2 = 1296.
Coefficients in T(n,k) are multiples of n+k+1; therefore, coefficients in all diagonals starting at T(n,1) are multiples of n+2.
Let T"(n,k) = T(n,k)/(n+k+1). Then reading T"(n,k) by rows:
i. Row sums are A162147(n). For example, T"(3,k) = [65/5, 72/6, 91/7] = [13,12,13]. 13+12+13 = 38; A162147(3) = 38.
ii. Extend the triangle in A215631 to a symmetric array by reflection about the main diagonal, and let that array be T"215631(n,k). Then the diagonal starting with T"215631(n,1) is row n in T"(n,k). For example, the diagonal starting at T"215631(4,1) = [21,19,19,21]; T"(4,k) = [126/6, 133/7, 152/8, 189/9] = [21,19,19,21].
iii. Coefficients in T"(n,k) are a permutation of A024612.
FORMULA
T(n,k) = (n+1)^3+k^3.
T(n,k) = (2k+1)(k^2+k+1) + Sum_{j=k+1..n} A003215(j), n>=k+1. For example, T(8,4) = 9*21 + 91 + 127 + 169 + 217 = 793.
EXAMPLE
Triangle starts:
n\k 1 2 3 4 5 6 7 8 9 10 ...
1: 9
2: 28 35
3: 65 72 91
4: 126 133 152 189
5: 217 224 243 280 341
6: 344 351 370 407 468 559
7: 513 520 539 576 637 728 855
8: 730 737 756 793 854 945 1072 1241
9: 1001 1008 1027 1064 1125 1216 1343 1512 1729
10: 1332 1339 1358 1395 1456 1547 1674 1843 2060 2331
...
The successive terms are (2^3+1^3), (3^3+1^3), (3^3+2^3), (4^3+1^3), (4^3+2^3), (4^3+3^3), ...
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Bob Selcoe, Mar 31 2015
STATUS
approved

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