Displaying 21-30 of 136 results found.
page
1
2
3
4
5
6
7
8
9
10
... 14
Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+199)^2 = y^2.
+10
8
0, 21, 504, 597, 704, 3441, 3980, 4601, 20540, 23681, 27300, 120197, 138504, 159597, 701040, 807741, 930680, 4086441, 4708340, 5424881, 23818004, 27442697, 31619004, 138821981, 159948240, 184289541, 809114280, 932247141, 1074118640
COMMENTS
Also values x of Pythagorean triples (x, x+199, y).
Corresponding values y of solutions (x, y) are in A159548. For the generic case x^2+(x+p)^2 = y^2 with p = 2*m^2-1 a (prime) number in A066436 see A118673 or A129836. Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (201+20*sqrt(2))/199 for n mod 3 = {1, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (91443+58282*sqrt(2))/199^2 for n mod 3 = 0.
FORMULA
a(n) = 6*a(n-3) - a(n-6) + 398 for n > 6; a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441.
G.f.: x*(21+483*x+93*x^2-19*x^3-161*x^4-19*x^5) / ((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 199* A001652(k) for k >= 0. a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441, a(7)=3980, a(n)=a(n-1)+6*a(n-3)-6*a(n-4)-a(n-6)+a(n-7). - Harvey P. Dale, Jun 03 2012
MATHEMATICA
LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {0, 21, 504, 597, 704, 3441, 3980}, 30] (* Harvey P. Dale, Jun 03 2012 *)
PROG
(PARI) {forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+398*n+39601), print1(n, ", ")))};
(Magma) I:=[0, 21, 504, 597, 704, 3441, 3980]; [n le 7 select I[n] else Self(n-1) + 6*Self(n-3) - 6*Self(n-4) - Self(n-6) + Self(n-7): n in [1..50]]; // G. C. Greubel, Mar 31 2018
Decimal expansion of 7 + 2*sqrt(2).
+10
8
9, 8, 2, 8, 4, 2, 7, 1, 2, 4, 7, 4, 6, 1, 9, 0, 0, 9, 7, 6, 0, 3, 3, 7, 7, 4, 4, 8, 4, 1, 9, 3, 9, 6, 1, 5, 7, 1, 3, 9, 3, 4, 3, 7, 5, 0, 7, 5, 3, 8, 9, 6, 1, 4, 6, 3, 5, 3, 3, 5, 9, 4, 7, 5, 9, 8, 1, 4, 6, 4, 9, 5, 6, 9, 2, 4, 2, 1, 4, 0, 7, 7, 7, 0, 0, 7, 7, 5, 0, 6, 8, 6, 5, 5, 2, 8, 3, 1, 4, 5, 4, 7, 0, 0, 2
COMMENTS
lim_{n -> infinity} b(n)/b(n-1) = (7+2*sqrt(2))/(7-2*sqrt(2)) for n mod 3 = {1, 2}, b = A129288. lim_{n -> infinity} b(n)/b(n-1) = (7+2*sqrt(2))/(7-2*sqrt(2)) for n mod 3 = {0, 2}, b = A157257.
EXAMPLE
7 + 2*sqrt(2) = 9.82842712474619009760...
MATHEMATICA
RealDigits[7+2Sqrt[2], 10, 120][[1]] (* Harvey P. Dale, Mar 20 2011 *)
Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.
+10
7
-3, 0, 5, 8, 21, 48, 65, 140, 297, 396, 833, 1748, 2325, 4872, 10205, 13568, 28413, 59496, 79097, 165620, 346785, 461028, 965321, 2021228, 2687085, 5626320, 11780597, 15661496, 32792613, 68662368, 91281905, 191129372, 400193625, 532029948, 1113983633
COMMENTS
First two terms included for consistency with A076293. Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (9+4*sqrt(2))/7 for n mod 3 = {1, 2}.
Lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 3 = 0. (End)
For the generic case x^2 + (x+p)^2 = y^2 with p=2*m^2-1 a prime number in A066436, m >= 2, the x values are given by the sequence defined by: a(n) = 6*a(n-3) - a(n-6) + 2p with a(1)=0, a(2)=2m+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21. Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. - Mohamed Bouhamida, Sep 09 2009 For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a prime number, m>=2, the first three consecutive solutions are: (0;p), (2*m+1; 2*m^2+2*m+1), (6*m^2-10*m+4; 10*m^2-14*m+5) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-3)+2*Y(n-3)+p; 4*X(n-3)+3*Y(n-3)+2*p). - Mohamed Bouhamida, Aug 20 2019
FORMULA
a(n) = 6a(n-3) - a(n-6) + 14 = ( A076293(n) - 7)/2. a(n) = 5*(a(n-3) + a(n-6)) - a(n-9) + 28.
a(n) = 7*(a(n-3) - a(n-6)) + a(n-9). (End)
G.f.: (-3 + 3*x + 5*x^2 + 21*x^3 - 5*x^4 - 3*x^5 - 4*x^6)/((1-x)*(1 - 6*x^3 + x^6)). - Klaus Brockhaus, Feb 18 2009
EXAMPLE
8 is in the sequence as the shorter leg of the (8,15,17) triangle.
MATHEMATICA
LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {-3, 0, 5, 8, 21, 48, 65}, 50] (* T. D. Noe, Feb 07 2012 *)
PROG
(PARI) x='x+O('x^30); Vec((-3+3*x+5*x^2+21*x^3-5*x^4-3*x^5-4*x^6)/((1-x)*(1-6*x^3 +x^6))) \\ G. C. Greubel, May 04 2018 (Magma) I:=[-3, 0, 5, 8, 21, 48, 65]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +Self(n-7): n in [1..30]]; // G. C. Greubel, May 04 2018
Numbers n such that 2*n^2 + 14 is a square.
+10
7
1, 5, 11, 31, 65, 181, 379, 1055, 2209, 6149, 12875, 35839, 75041, 208885, 437371, 1217471, 2549185, 7095941, 14857739, 41358175, 86597249, 241053109, 504725755, 1404960479, 2941757281, 8188709765, 17145817931, 47727298111
COMMENTS
The equation "2*n^2 + 14 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = 14.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = (7+n^2)/2. - Ctibor O. Zizka, Nov 09 2009 This sequence gives all positive solutions x = a(n+1), n >= 0, of the Pell equation x^2 - 2*y^2 = -7. For the corresponding y-solutions see y(n) = 2* A006452(n+2) = A077447(n+1)/2. This implies that X^2 - 2*Y^2 = 14 has the general solutions (X(n),Y(n)) = (2*y(n), x(n)). See the first comment above. For the positive first class solutions see ( A054490(n), 2* A038723(n)) and for the second class solutions ( A255236(n), 2* A038725(n+1)). (End) For n > 0, a(n) is the n-th almost Lucas-balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022
REFERENCES
A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
FORMULA
Lim. n-> Inf. a(n)/a(n-2) = 5.8284271247461... = 3 + 2*sqrt(2) = A156035 = RG (Great Ratio). Lim. k-> Inf. a(2*k+1)/a(2*k) = 2.09383632135605... = (9 + 4*sqrt(2))/7 = A156649 = R1 (Ratio 1). Lim. k -> Inf. a(2*k)/a(2*k-1) = 2.78361162489122432754 = (11 + 6*sqrt(2))/7 = R2 (Ratio 2); RG = R1*R2.
a(2*k-1) = [ 2*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] - [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] + [(3+2*Sqrt(2))^(n-2) - (3-2*Sqrt(2))^(n-2)] ] / (4*Sqrt(2)) a(2*k) = [ 5*[(3+2*Sqrt(2))^n - (3-2*Sqrt(2))^n] + [(3+2*Sqrt(2))^(n-1) - (3-2*Sqrt(2))^(n-1)] ] / (4*Sqrt(2)).
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1+x)*(x^2+4*x+1) / ( (x^2+2*x-1)*(x^2-2*x-1) ). - R. J. Mathar, Jul 03 2011 a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=5, a(3)=11, a(4)=31. - Sture Sjöstedt, Oct 08 2012 Bisection: a(2*k+1) = S(k, 6) + 5*S(k-1, 6), a(2*k) = 5*S(k-1, 6) + S(k-2, 6), with the Chebyshev polynomials S(n, x) ( A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1). See A054490 and A255236, and the given g.f.s. - Wolfdieter Lang, Feb 26 2015 E.g.f.: 1 - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x). - Stefano Spezia, Nov 26 2022
EXAMPLE
n = 3: ( A077447(3))^2 - 2*a(3)^2 = 16^2 - 2*11^2 = 14;
MATHEMATICA
LinearRecurrence[{0, 6, 0, -1}, {1, 5, 11, 31}, 50] (* Sture Sjöstedt, Oct 08 2012 *)
CROSSREFS
Cf. A001109, A006452, A038723, A038725, A049310, A054490, A077447, A155765, A156035, A156649, A255236.
Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+617)^2 = y^2.
+10
7
0, 108, 1407, 1851, 2407, 9768, 12340, 15568, 58435, 73423, 92235, 342076, 429432, 539076, 1995255, 2504403, 3143455, 11630688, 14598220, 18322888, 67790107, 85086151, 106795107, 395111188, 495919920, 622448988, 2302878255, 2890434603
COMMENTS
Also values x of Pythagorean triples (x, x+617, y).
Corresponding values y of solutions (x, y) are in A160176. lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (633+100*sqrt(2))/617 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (755667+461578*sqrt(2))/617^2 for n mod 3 = 0.
FORMULA
a(n) = 6*a(n-3) -a(n-6) +1234 for n > 6; a(1)=0, a(2)=108, a(3)=1407, a(4)=1851, a(5)=2407, a(6)=9768.
G.f.: x*(108 +1299*x +444*x^2 -92*x^3 -433*x^4 -92*x^5)/((1-x)*(1 -6*x^3 +x^6)).
a(3*k+1) = 617* A001652(k) for k >= 0.
MATHEMATICA
LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {0, 108, 1407, 1851, 2407, 9768, 12340}, 50] (* G. C. Greubel, May 04 2018 *)
PROG
(PARI) {forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+1234*n+380689), print1(n, ", ")))}
(PARI) x='x+O('x^30); Vec(x*(108 +1299*x +444*x^2 -92*x^3 -433*x^4 -92*x^5)/((1-x)*(1 -6*x^3 +x^6))) \\ G. C. Greubel, May 04 2018 (Magma) I:=[0, 108, 1407, 1851, 2407, 9768, 12340]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +self(n-7): n in [1..30]]; // G. C. Greubel, May 04 2018
Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+127)^2 = y^2.
+10
7
0, 17, 308, 381, 468, 2117, 2540, 3045, 12648, 15113, 18056, 74025, 88392, 105545, 431756, 515493, 615468, 2516765, 3004820, 3587517, 14669088, 17513681, 20909888, 85498017, 102077520, 121872065, 498319268, 594951693, 710322756, 2904417845, 3467632892
COMMENTS
Also values x of Pythagorean triples (x, x+127, y).
Corresponding values y of solutions (x, y) are in A159466. For the generic case x^2+(x+p)^2 = y^2 with p = 2*m^2-1 a (prime) number in A066436 see A118673 or A129836. Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
Lim_{n -> infinity} a(n)/a(n-1) = (129+16*sqrt(2))/127 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (34947+21922*sqrt(2))/127^2 for n mod 3 = 0.
FORMULA
a(n) = 6*a(n-3) - a(n-6) + 254 for n > 6; a(1)=0, a(2)=17, a(3)=308, a(4)=381, a(5)=468, a(6)=2117.
G.f.: x*(17+291*x+73*x^2-15*x^3-97*x^4-15*x^5) / ((1-x)*(1-6*x^3+x^6)).
a(3*k+1) = 127* A001652(k) for k >= 0.
MATHEMATICA
LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {0, 17, 308, 381, 468, 2117, 2540}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *)
PROG
(PARI) {forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+254*n+16129), print1(n, ", ")))};
(Magma) I:=[0, 17, 308, 381, 468, 2117, 2540]; [n le 7 select I[n] else Self(n-1) + 6*Self(n-3) - 6*Self(n-4) - Self(n-6) + Self(n-7): n in [1..50]]; // G. C. Greubel, Mar 31 2018
Positive numbers y such that y^2 is of the form x^2+(x+17)^2 with integer x.
+10
7
13, 17, 25, 53, 85, 137, 305, 493, 797, 1777, 2873, 4645, 10357, 16745, 27073, 60365, 97597, 157793, 351833, 568837, 919685, 2050633, 3315425, 5360317, 11951965, 19323713, 31242217, 69661157, 112626853, 182092985, 406014977, 656437405
COMMENTS
(-5,a(1)) and ( A118120(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+17)^2 = y^2. (Offset 1 is assumed for A118120.) lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (19+6*sqrt(2))/17 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (387+182*sqrt(2))/17^2 for n mod 3 = 1.
For the generic case x^2+(x+p)^2=y^2 with p=2*m^2-1 a prime number in A066436, m>=2, the x values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2p with a(1)=0, a(2)=2m+1, a(3)=6m^2-10m+4, a(4)=3p, a(5)=6m^2+10m+4, a(6)=40m^2-58m+21.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=2*m^2+2m+1, b(3)=10m^2-14m+5, b(4)=5p, b(5)=10m^2+14m+5, b(6)=58m^2-82m+29. [From Mohamed Bouhamida, Sep 09 2009]
FORMULA
a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1) = 13, a(2) = 17, a(3) = 25, a(4) = 53, a(5) = 85, a(6) = 137.
G.f.: x*(1-x)*(13+30*x+55*x^2+30*x^3+13*x^4)/(1-6*x^3+x^6).
EXAMPLE
(-5,a(1)) = (-5,13) is a solution: (-5)^2+(-5+17)^2 = 25+144 = 169 = 13^2;
( A118120(1), a(2)) = (0, 17) is a solution: 0^2+(0+17)^2 = 289 = 17^2; ( A118120(2), a(3)) = (7, 25) is a solution: 7^2+(7+17)^2 = 49+576 = 625 = 25^2.
MATHEMATICA
LinearRecurrence[{0, 0, 6, 0, 0, -1}, {13, 17, 25, 53, 85, 137}, 50] (* Harvey P. Dale, Feb 11 2015 *)
PROG
(PARI) {forstep(n=-5, 660000000, [1, 3], if(issquare(2*n*(n+17)+289, &k), print1(k, ", ")))}
CROSSREFS
Cf. A118120, A156035 (decimal expansion of 3+2*sqrt(2)), A156163 (decimal expansion of (19+6*sqrt(2))/17), A157649 (decimal expansion of (387+182*sqrt(2))/17^2).
EXTENSIONS
G.f. corrected, first and fourth comment and examples edited, cross-reference added by Klaus Brockhaus, Sep 22 2009
Positive numbers y such that y^2 is of the form x^2+(x+23)^2 with integer x.
+10
7
17, 23, 37, 65, 115, 205, 373, 667, 1193, 2173, 3887, 6953, 12665, 22655, 40525, 73817, 132043, 236197, 430237, 769603, 1376657, 2507605, 4485575, 8023745, 14615393, 26143847, 46765813, 85184753, 152377507, 272571133, 496493125
COMMENTS
(-8, a(1)) and( A118337(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+23)^2 = y^2. lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (27+10*sqrt(2))/23 for n mod 3 = {0, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (627+238*sqrt(2))/23^2 for n mod 3 = 1.
For the generic case x^2+(x+p)^2=y^2 with p=m^2-2 a prime number in A028871, m>=2, the x values are given by the sequence defined by: a(n)=6*a(n-3)-a(n-6)+2p with a(1)=0, a(2)=2m+2, a(3)=3m^2-10m+8, a(4)=3p, a(5)=3m^2+10m+8, a(6)=20m^2-58m+42.Y values are given by the sequence defined by: b(n)=6*b(n-3)-b(n-6) with b(1)=p, b(2)=m^2+2m+2, b(3)=5m^2-14m+10, b(4)=5p, b(5)=5m^2+14m+10, b(6)=29m^2-82m+58. [From Mohamed Bouhamida, Sep 09 2009]
FORMULA
a(n) = 6*a(n-3)-a(n-6) for n > 6; a(1)=17, a(2)=23, a(3)=37, a(4)=65, a(5)=115, a(6)=205.
G.f.: x*(1-x)*(17+40*x+77*x^2+40*x^3+17*x^4)/(1-6*x^3+x^6).
EXAMPLE
(-8, a(1)) = (-8, 17) is a solution: (-8)^2+(-8+23)^2 = 64+225 = 289 = 17^2.
( A118337(1), a(2)) = (0, 23) is a solution: 0^2+(0+23)^2 = 529 = 23^2. ( A118337(3), a(4)) = (33, 65) is a solution: 33^2+(33+23)^2 = 1089+3136 = 4225 = 65^2.
PROG
(PARI) {forstep(n=-8, 360000000, [1, 3], if(issquare(2*n*(n+23)+529, &k), print1(k, ", ")))}
CROSSREFS
Cf. A118337, A156035 (decimal expansion of 3+2*sqrt(2)), A156571 (decimal expansion of (27+10*sqrt(2))/23), A157472 (decimal expansion of (627+238*sqrt(2))/23^2).
EXTENSIONS
G.f. corrected, third and fourth comment edited, cross-reference added by Klaus Brockhaus, Sep 18 2009
Decimal expansion of 2*(sqrt(2) - 1).
+10
7
8, 2, 8, 4, 2, 7, 1, 2, 4, 7, 4, 6, 1, 9, 0, 0, 9, 7, 6, 0, 3, 3, 7, 7, 4, 4, 8, 4, 1, 9, 3, 9, 6, 1, 5, 7, 1, 3, 9, 3, 4, 3, 7, 5, 0, 7, 5, 3, 8, 9, 6, 1, 4, 6, 3, 5, 3, 3, 5, 9, 4, 7, 5, 9, 8, 1, 4, 6, 4, 9, 5, 6, 9, 2, 4, 2, 1, 4, 0, 7, 7, 7, 0, 0, 7, 7, 5, 0, 6, 8, 6, 5, 5, 2, 8, 3, 1, 4, 5
COMMENTS
Decimal expansion of shortest length, (B), of segment from side BC through incenter to side BA in right triangle ABC with sidelengths (a,b,c)=(1,1,sqrt(2)). (See A195284.) - Clark Kimberling, Sep 14 2011
REFERENCES
J. M. Steele, Probability Theory and Combinatorial Optimization, SIAM, 1997, p. 3.
FORMULA
Equals Sum_{k>=0} (-1)^k * binomial(2*k,k)/((k+1) * 4^k). - Amiram Eldar, May 06 2022
EXAMPLE
0.82842712474619009760337744841939615713934375075389614635335...
MATHEMATICA
RealDigits[2(Sqrt[2]-1), 10, 120][[1]] (* Harvey P. Dale, May 27 2016 *)
Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+569)^2 = y^2.
+10
6
0, 111, 1260, 1707, 2280, 8791, 11380, 14707, 52624, 67711, 87100, 308091, 396024, 509031, 1797060, 2309571, 2968224, 10475407, 13462540, 17301451, 61056520, 78466807, 100841620, 355864851, 457339440, 587749407, 2074133724, 2665570971
COMMENTS
Also values x of Pythagorean triples (x, x+569, y).
Corresponding values y of solutions (x, y) are in A160090. lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (587+102*sqrt(2))/569 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (617139+371510*sqrt(2))/569^2 for n mod 3 = 0.
FORMULA
a(n) = 6*a(n-3) - a(n-6) + 1138 for n > 6; a(1)=0, a(2)=111, a(3)=1260, a(4)=1707, a(5)=2280, a(6)=8791.
G.f.: x*(111 +1149*x +447*x^2 -93*x^3 -383*x^4 -93*x^5)/((1-x)*(1-6*x^3 +x^6)).
a(3*k+1) = 569* A001652(k) for k >= 0.
MATHEMATICA
LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {0, 111, 1260, 1707, 2280, 8791, 11380}, 50] (* G. C. Greubel, Apr 21 2018 *)
PROG
(PARI) {forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+1138*n+323761), print1(n, ", ")))}
(PARI) x='x+O('x^30); concat([0], Vec(x*(111 +1149*x +447*x^2 -93*x^3 -383*x^4 -93*x^5)/((1-x)*(1-6*x^3 +x^6)))) \\ G. C. Greubel, Apr 21 2018 (Magma) I:=[0, 111, 1260, 1707, 2280, 8791, 11380]; [n le 7 select I[n] else Self(n-1) +6*Self(n-3) -6*Self(n-4) -Self(n-6) +Self(n-7): n in [1..30]]; // G. C. Greubel, Apr 21 2018
Search completed in 0.072 seconds
|