A001652 -id:A001652 - cp's OEIS frontend

A157088 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

0, 21, 312, 4365, 60816, 847077, 11798280, 164328861, 2288805792, 31878952245, 444016525656, 6184352406957, 86136917171760, 1199732487997701, 16710117914796072, 232741918319147325, 3241676738553266496, 45150732421426583637, 628868577161418904440, 8759009347838438078541

Offset: 0

Charlie Marion, Mar 12 2009

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=4, then first(2) = 680 = 18*36 - 0 + 32.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=4 and n=2, first(2) = 680 = 9*36 + 8*44 + 4 and last(2) = 764 = 10*36 + 9*44 + 8.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=4 and n=2, then first(2) = 680 = (4^3((1+sqrt(5/4)^5 + (1-sqrt(5/4))^5)-2*4)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(3) = 4365 = 3*15*97 and a(4) = 60816 = 4*84*181.
In general, if first(n) is the first term of the n-th consecutive integer Pythagorean 2k+1-tuple, then lim_{n->inf} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2)=312 since 312^2 + 313^2 + 314^2 + 315^2 = 361^2 + 361^2 + 363^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..870
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001652, A157084, A157092, A157096.

[Round((3^(n+1)*((1+Sqrt(4/3))^(2*n+1)+(1-Sqrt(4/3))^(2*n+1))-2*3)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017

CoefficientList[Series[3*x*(-7 + x)/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Nov 04 2017 *)

my(x='x+O('x^50)); concat([0], Vec(3*x*(-7+x)/((x-1)*(x^2-14*x+1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 14*a(n-1) - a(n-2) + 18.
For n > 0, a(n) = 7*a(n-1) + 6*A157089(n-1) + 3.
Limit_{n->oo} a(n+1)/a(n) = 3*(1+sqrt(4/3))^2 = 7 + 2*sqrt(12).
a(n) = (3^(n+1)*((1+sqrt(4/3))^(2*n+1) + (1-sqrt(4/3))^(2*n+1)) - 2*3)/4.
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 3*x*(-7+x)/((x-1)*(x^2-14*x+1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) = 3*A001921(n). (End)

A330276 NSW pseudoprimes: odd composite numbers k such that A002315((k-1)/2) == 1 (mod k).

Original entry on oeis.org

169, 385, 961, 1105, 1121, 3827, 4901, 6265, 6441, 6601, 7107, 7801, 8119, 10945, 11285, 13067, 15841, 18241, 19097, 20833, 24727, 27971, 29953, 31417, 34561, 35459, 37345, 37505, 38081, 39059, 42127, 45451, 45961, 47321, 49105, 52633, 53041, 55969, 56953, 58241

Offset: 1

Amiram Eldar, Dec 08 2019

nonn

If p is an odd prime, then A002315((p-1)/2) == 1 (mod p). This sequence consists of the odd composite numbers for which this congruence holds.

Equivalently, odd composite numbers k such that A001652((k-1)/2) is divisible by k.

			169 = 13^2 is a term since it is composite and A002315((169-1)/2) - 1 = A002315(84) - 1 is divisible by 169.

Amiram Eldar, Table of n, a(n) for n = 1..1000
Morris Newman, Daniel Shanks, and H. C. Williams, Simple groups of square order and an interesting sequence of primes, Acta Arithmetica, Vol. 38, No. 2 (1980), pp. 129-140.

Cf. A001652, A002315.

a0 = 1; a1 = 7; k = 5; seq = {}; Do[a = 6 a1 - a0; a0 = a1; a1 = a; If[CompositeQ[k] && Divisible[a - 1, k], AppendTo[seq, k]]; k += 2, {n, 2, 10^4}]; seq

A336624 Triangular numbers that are one-eighth of other triangular numbers; T(t) such that 8*T(t)=T(u) for some u where T(k) is the k-th triangular number.

Original entry on oeis.org

0, 15, 66, 17391, 76245, 20069280, 87986745, 23159931810, 101536627566, 26726541239541, 117173180224500, 30842405430498585, 135217748442445515, 35592109140254127630, 156041164529401899891, 41073263105447832786516, 180071368649181350028780, 47398510031577658781511915

Offset: 0

Vladimir Pletser, Aug 07 2020

The triangular numbers T(t) that are one-eighth of other triangular numbers T(u) : T(t)=T(u)/8. The t's are in A336623, the T(u)'s are in A336626 and the u's are in A336625.

Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

			a(1)= 15 is a term because it is triangular and 8*15 = 120 is also triangular.
a(2) = 1154*a(0) - a(-2) + 81 = 0 - 15 + 81 = 66;
a(3) = 1154*a(1) - a(-1) + 81 = 1154*15 - 0 + 81 = 17391, etc.

Vladimir Pletser, Table of n, a(n) for n = 0..650
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,1154,-1154,-1,1).

Subsequence of A000217.
Cf. A336623, A336626, A336625.
Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400.

f := gfun:-rectoproc({a(n) = 1154*a(n - 2) - a(n - 4) + 81, a(1) = 15, a(0) = 0, a(-1) = 0, a(-2) = 15}, a(n), remember): map(f, [$ (0 .. 40)])[]; #

LinearRecurrence[{1, 1154, -1154, -1, 1}, {0, 15, 66, 17391, 76245}, 18] (* Amiram Eldar, Aug 08 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(4*n + 2)*(11 - 6*(-1)^n*Sqrt[2]) + (Sqrt[2] - 1)^(4*n + 2)*(11 + 6*(-1)^n*Sqrt[2]) - 18)/256, {n, 0, 17}]] (* Vaclav Kotesovec, Sep 08 2020 *)
Select[Accumulate[Range[0, 10^6]]/8, OddQ[Sqrt[8 # + 1]] &] (* The program generates the first 8 terms of the sequence. *) (* Harvey P. Dale, Jan 15 2024 *)

concat(0, Vec(3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)) + O(x^40))) \\ Colin Barker, Aug 08 2020

a(n) = 1154*a(n-2) - a(n-4) + 81, for n>=2 with a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(2)=66, a(1)=15, a(0)=0, a(-1)=0, a(-2)=15.
a(n) = b(n)*(b(n)+1)/2 where b(n) is A336623(n).
G.f.: 3*x*(5 + 17*x + 5*x^2) / ((1 - x)*(1 - 34*x + x^2)*(1 + 34*x + x^2)). - Colin Barker, Aug 08 2020
a(n) = ((sqrt(2) + 1)^(4*n + 2) * (11 - 6*(-1)^n*sqrt(2)) + (sqrt(2) - 1)^(4*n + 2) * (11 + 6*(-1)^n*sqrt(2)) - 18)/256. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((11 - 6*sqrt(2))*(1 + sqrt(2))^(4n + 2) + (11 + 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for even n.
a(n) = ((11 + 6*sqrt(2))*(1 + sqrt(2) )^(4n + 2) + (11 - 6*sqrt(2))*(1 - sqrt(2) )^(4n + 2) - 18) / 256 for odd n. (End)
128*a(n) = -9+33*A077420(n)-24*(-1)^n*A046176(n+1). - R. J. Mathar, May 05 2023

A336625 Indices of triangular numbers that are eight times other triangular numbers.

Original entry on oeis.org

0, 15, 32, 527, 1104, 17919, 37520, 608735, 1274592, 20679087, 43298624, 702480239, 1470878640, 23863649055, 49966575152, 810661587647, 1697392676544, 27538630330959, 57661384427360, 935502769664975, 1958789677853712, 31779555538278207, 66541187662598864, 1079569385531794079, 2260441590850507680

Offset: 1

Vladimir Pletser, Aug 13 2020

Second member of the Diophantine pair (b(n), a(n)) that satisfies a(n)^2 + a(n) = 8*(b(n)^2 + b(n)) or T(a(n)) = 8*T(b(n)) where T(x) is the triangular number of x. The T(a)'s are in A336626, the T(b)'s are in A336624 and the b's are in A336623.

Can be defined for negative n by setting a(-n) = -a(n+1) - 1 for all n in Z.

			a(3) = 34*a(1) - a(-1) + 16 = 0 - (-16) + 16 = 32,
a(4) = 34*a(2) - a(0) + 16 = 34*15 - (-1) + 16 = 527, etc.

Vladimir Pletser, Table of n, a(n) for n = 1..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A336623, A336624, A336626, A166477 (at n=8).

Cf. A053141, A001652, A075528, A029549, A061278, A001571, A076139, A076140, A077259, A077262, A077260, A077261, A077288, A077291, A077289, A077290, A077398, A077401, A077399, A077400, A000217.

f := gfun:-rectoproc({a(n) = 34*a(n - 2) - a(n - 4) + 16, a(2) = 15, a(1) = 0, a(0) = -1, a(-1) = -16}, a(n), remember); map(f, [$ (0 .. 1000)]); #

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 15, 32, 527, 1104, 17919}, 29] (* Amiram Eldar, Aug 18 2020 *)
FullSimplify[Table[((Sqrt[2] + 1)^(2*n + 1) * (3 - Sqrt[2]*(-1)^n) - (Sqrt[2] - 1)^(2*n + 1) * (3 + Sqrt[2]*(-1)^n) - 2)/4, {n, 0, 20}]] (* Vaclav Kotesovec, Sep 08 2020 *)

concat(0, Vec(x*(15 + 17*x - 15*x^2 - x^3) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^22))) \\ Colin Barker, Aug 14 2020

a(n) = 34*a(n-2) - a(n-4) + 16, for n>=2 with a(2)=15, a(1)=0, a(0)=-1, a(-1)=-16.
a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5), for n>=3 with a(3)=32, a(2)=15, a(1)=0, a(0)=-1, a(-1)=-16.
a(n) = (-1 + sqrt(8*b(n) + 1))/2, where b(n) is A336626(n).
G.f.: x^2*(15 + 17*x - 15*x^2 - x^3) / ((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)). - Colin Barker, Aug 14 2020
a(n) = ((sqrt(2) + 1)^(2*n+1) * (3 - sqrt(2)*(-1)^n) - (sqrt(2) - 1)^(2*n+1) * (3 + sqrt(2)*(-1)^n) - 2)/4. - Vaclav Kotesovec, Sep 08 2020
From Vladimir Pletser, Feb 21 2021: (Start)
a(n) = ((3 - sqrt(2))*(1 + sqrt(2))^(2*n+1) + (3 + sqrt(2))*(1 - sqrt(2))^(2*n+1))/4 - 1/2 for even n.
a(n) = ((3 + sqrt(2))*(1 + sqrt(2))^(2*n+1) + (3 - sqrt(2))*(1 - sqrt(2))^(2*n+1))/4 - 1/2 for odd n. (End)

A006062 Star-hex numbers.

Original entry on oeis.org

1, 37, 1261, 42841, 1455337, 49438621, 1679457781, 57052125937, 1938092824081, 65838103892821, 2236557439531837, 75977114840189641, 2580985347126915961, 87677524687474953037, 2978454854027021487301, 101179787512231255615201, 3437134320561835669429537

Offset: 1

N. J. A. Sloane, Jul 11 1991

Martin Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Paolo Xausa, Table of n, a(n) for n = 1..650
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A001109, A001652, A001653, A046090.

A006062:=-(z+1)**2/(z-1)/(z**2-34*z+1); [Simon Plouffe in his 1992 dissertation for offset zero.]

LinearRecurrence[{35, -35, 1}, {1, 37, 1261}, 20] (* Paolo Xausa, Sep 02 2024 *)

a(n) = 34*a(n-1) - a(n-2) + 4.
G.f.: -x*(x + 1)^2/(x - 1)/(x^2 - 34*x + 1). [from Simon Plouffe, see Maple code].
From Charlie Marion, Aug 03 2005: (Start)
a(n) = A001109(n-1)^2 + A001109(n)^2; e.g., 1261 = 6^2 + 35^2.
a(n) = A001652(n-1)*A046090(n-1) + A001653(n-1)^2; e.g., 1261 = 20*21 + 29^2. (End)

A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

Original entry on oeis.org

-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

Offset: 0

Henry Bottomley, Oct 05 2002

First two terms included for consistency with A076293.
From Klaus Brockhaus, Feb 18 2009: (Start)
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

			8 is in the sequence as the shorter leg of the (8,15,17) triangle.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A001652, A076293, A076294, A076295.

Cf. A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7). - Klaus Brockhaus, Feb 18 2009

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

CoefficientList[Series[(3-3x-5x^2-21x^3+5x^4+3x^5+4x^6)/(-1+x+6x^3-6x^4-x^6+x^7),{x,0,50}],x] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2012 *)
LinearRecurrence[{1,0,6,-6,0,-1,1}, {-3,0,5,8,21,48,65}, 50] (* T. D. Noe, Feb 07 2012 *)

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

a(n) = 6a(n-3) - a(n-6) + 14 = (A076293(n) - 7)/2.
a(n) = sqrt(A076294(n)^2 - A076295(n)^2) = A076295(n) - 7.
a(3*n+1) = 7*A001652(n).
From Mohamed Bouhamida, Jul 06 2007: (Start)
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

More terms from Klaus Brockhaus, Feb 18 2009

A115135 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+617)^2 = y^2.

Original entry on oeis.org

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

Offset: 1

Mohamed Bouhamida, Jun 03 2007

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.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).

Cf. A160176, A001652, A111258, A156035 (decimal expansion of 3+2*sqrt(2)), A160177 (decimal expansion of (633+100*sqrt(2))/617), A160178 (decimal expansion of (755667+461578*sqrt(2))/617^2).

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

LinearRecurrence[{1,0,6,-6,0,-1,1}, {0,108,1407,1851,2407,9768,12340}, 50] (* G. C. Greubel, May 04 2018 *)

{forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+1234*n+380689), print1(n, ",")))}

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

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.

Edited and two terms added by Klaus Brockhaus, May 18 2009

A123737 Partial sums of (-1)^floor(n*sqrt(2)).

Original entry on oeis.org

-1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 1, 0, 1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, -2, -1, -2, -3, -2, -1, -2, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 1, 0, 1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, 1, 0

Offset: 1

T. D. Noe, Oct 11 2006

Conjecture: A001652(n) is the index of the first occurrence of n in sequence A123737, A001108(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015

T. D. Noe, Table of n, a(n) for n = 1..10000
Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See p. 2.
Kevin O'Bryant, Bruce Reznick and Monika Serbinowska, Almost alternating sums, arXiv:math/0308087 [math.NT], 2003-2205; Amer. Math. Monthly, Vol. 113 (October 2006), pp. 673-688.

Cf. A123724 (sum for 2^(1/3)), A123738 (sum for Pi), A123739 (sum for e).

Cf. A001652, A001951, A228639.

[&+[(-1)^Floor(j*Sqrt(2)): j in [1..n]]: n in [1..130]]; // G. C. Greubel, Sep 05 2019

ListTools:-PartialSums([seq((-1)^floor(n*sqrt(2)),n=1..100)]); # Robert Israel, Jun 02 2015

Rest[FoldList[Plus,0,(-1)^Floor[Sqrt[2]*Range[120]]]]
Accumulate[(-1)^Floor[Range[120]Sqrt[2]]] (* Harvey P. Dale, Jan 16 2012 *)
(* The positions of the first occurrences of n and -n in this sequence: *) stab = Rest[FoldList[Plus,0,(-1)^Floor[Sqrt[2]*Range[1000000]]]]; Print[Table[FirstPosition[stab,n][[1]],{n,1,8}]]; Print[Table[FirstPosition[stab,-n][[1]],{n,1,8}]]; (* Vaclav Kotesovec, Jun 02 2015 *)

a(n)=sum(i=1,n,(-1)^sqrtint(2*i^2)) \\ Charles R Greathouse IV, Feb 07 2013

[sum((-1)^floor(j*sqrt(2)) for j in (1..n)) for n in (1..130)] # G. C. Greubel, Sep 05 2019

O'Bryant, Reznick, & Serbinowska show that |a(n)| <= k log n + 1, with k = 1/(2 log (1 + sqrt(2))), and further -a(n) > k log n + 0.78 infinitely often. - Charles R Greathouse IV, Feb 07 2013

A129992 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+127)^2 = y^2.

Original entry on oeis.org

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

Offset: 1

Mohamed Bouhamida, Jun 14 2007

nonn

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.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1, 0, 6, -6, 0, -1, 1).

Cf. A159466, A066436, A118673, A118674, A129836, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159467 (decimal expansion of (129+16*sqrt(2))/127), A159468 (decimal expansion of (34947+21922*sqrt(2))/127^2).

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

LinearRecurrence[{1,0,6,-6,0,-1,1},{0,17,308,381,468,2117,2540},80] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2012 *)

{forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+254*n+16129), print1(n, ",")))};

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.

Edited and two more terms added by Klaus Brockhaus, Apr 13 2009

A157092 Consider all consecutive integer Pythagorean 9-tuples (X, X+1, X+2, X+3, X+4, Z-3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Original entry on oeis.org

0, 36, 680, 12236, 219600, 3940596, 70711160, 1268860316, 22768774560, 408569081796, 7331474697800, 131557975478636, 2360712083917680, 42361259535039636, 760141959546795800, 13640194012307284796, 244763350261984330560, 4392100110703410665316, 78813038642399407645160

Offset: 0

Charlie Marion, Mar 12 2009

nonn

In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0, first(1) = k*(2k+1) and, for n > 1, first(n) = (4k+2)*first(n-1) - first(n-2) + 2*k^2; e.g., if k=5, then first(2) = 1260 = 22*55 - 0 + 50.
In general, the first and last terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=5 and n=2, then first(2) = 1260 = 11*55 + 10*65 + 5 and last(2) = 1385 = 12*55 + 11*65 + 10.
In general, the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(n) = (k^(n+1)((1+sqrt((k+1)/k))^(2n+1) + (1-sqrt((k+1)/k))^(2n+1)) - 2*k)/4; e.g., if k=5 and n=2, then first(2) = 1260 = (5^3((1+sqrt((6/5))^5 + (1-sqrt(6/5))^5) - 2*5)/4.
In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the first terms of consecutive integer Pythagorean 2k+1-tuples may be found as follows: first(2n+1) = k*u(2n)*u(2n+1) and, for n > 0, first(2n) = (k+1)*e(2n-1)*e(2n); e.g., a(1) = 36 = 4*1*9 and a(2) = 680 = 5*8*17.
In general, if first(n) is the first term of the n-th consecutive integer Pythagorean 2k+1-tuple, then lim_{n->inf} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(2)=680 since 680^2 + 681^2 + 682^2 + 683^2 + 684^2 = 761^2 + 762^2 + 763^2 + 764^2.

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1964, pp. 122-125.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. Dover Publications, Inc., Mineola, NY, 2005, pp. 181-183.
W. Sierpinski, Pythagorean Triangles. Dover Publications, Mineola NY, 2003, pp. 16-22.

G. C. Greubel, Table of n, a(n) for n = 0..790
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A001652, A157084, A157088, A157096.

[Round((4^(n+1)*((1+Sqrt(5/4))^(2*n+1) + (1-Sqrt(5/4))^(2*n+1)) - 2*4)/4): n in [0..50]]; // G. C. Greubel, Nov 04 2017

RecurrenceTable[{a[0]==0,a[1]==36,a[n]==18a[n-1]-a[n-2]+32},a,{n,20}] (* or *) LinearRecurrence[{19,-19,1},{0,36,680},20] (* Harvey P. Dale, Oct 09 2012 *)

x='x+O('x^50); concat([0], Vec(4*x*(-9+x)/((x-1)*(x^2-18*x+1)))) \\ G. C. Greubel, Nov 04 2017

For n > 1, a(n) = 18*a(n-1) - a(n-2) + 32.
For n > 0, a(n) = 9*a(n-1) + 8*A157093(n-1) + 4.
a(n) = (4^(n+1)((1+sqrt(5/4))^(2n+1) + (1-sqrt(5/4))^(2n+1)) - 2*4)/4.
Lim_{n->inf} a(n+1)/a(n) = 4*(1+sqrt(5/4))^2 = 9 + 2*sqrt(20).
From R. J. Mathar, Mar 19 2009: (Start)
G.f.: 4*x*(-9+x)/((x-1)*(x^2-18*x+1)).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3).
a(n) = 4*A119032(n+1). (End)
For n > 0, 1/a(n) = Sum_{k>=1} F(3*k)/phi^(6*k*n + 3*k), where F(n) = A000045(n) and phi = A001622 = (sqrt(5)+1)/2. - Diego Rattaggi, Dec 28 2019
E.g.f.: (1/2)*((2 + sqrt(5))*exp((9+4*sqrt(5))*x) + (2 - sqrt(5))*exp((9-4*sqrt(5))*x) - 4*exp(x)). - Stefano Spezia, Dec 29 2019

Terms a(15) onward added by G. C. Greubel, Nov 06 2017

cp's OEIS Frontend

A157088 Consider all consecutive integer Pythagorean septuples (X, X+1, X+2, X+3, Z-2, Z-1, Z) ordered by increasing Z; sequence gives X values.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A330276 NSW pseudoprimes: odd composite numbers k such that A002315((k-1)/2) == 1 (mod k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A336624 Triangular numbers that are one-eighth of other triangular numbers; T(t) such that 8*T(t)=T(u) for some u where T(k) is the k-th triangular number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A336625 Indices of triangular numbers that are eight times other triangular numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A006062 Star-hex numbers.

Views

Author

Keywords

References

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A076296 Consider all Pythagorean triples (X,X+7,Z); sequence gives X values.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A115135 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+617)^2 = y^2.

Views

Author

Keywords