A002315 -id:A002315 - cp's OEIS frontend

A143608 A005319 and A002315 interleaved.

0, 1, 4, 7, 24, 41, 140, 239, 816, 1393, 4756, 8119, 27720, 47321, 161564, 275807, 941664, 1607521, 5488420, 9369319, 31988856, 54608393, 186444716, 318281039, 1086679440, 1855077841, 6333631924, 10812186007, 36915112104, 63018038201, 215157040700

Offset: 0

Originally submitted by Clark Kimberling, Aug 27 2008. Merged with an essentially identical sequence submitted by Kenneth J Ramsey, Jun 01 2012, by N. J. A. Sloane, Aug 02 2012

Also, numerators of the lower principal and intermediate convergents to 2^(1/2). The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496.
Sequence a(n) such that a(2*n) = sqrt(2*A001108(2*n)) and a(2*n+1) = sqrt(A001108(2*n+1)).
For n > 0, a(n) divides A******(k+1,n+1)-A******(k,n+1) where A****** is any one of A182431, A182439, A182440, A182441 and k is any nonnegative integer.
If p is a prime of the form 8*r +/- 3 then a(p+1) == 0 (mod p); if p is a prime of the form 8*r +/- 1 then a(p-1) == 0 (mod p).
Numbers n such that sqrt(floor(n^2/2 + 1)) is an integer. The integer square roots are given by A079496. - Richard R. Forberg, Aug 01 2013
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. Then we have
a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
a(2*n) = sqrt(2)*(1 o 1 o ... o 1) (2*n terms). Cf. A084068.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = sqrt(2)*U(2*n) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ). (End)

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Colin Barker, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Creighton Kenneth Dement, Comments on A143608 and A143609
Vincent Granville, Successive records in mathematical sequences: surprising result, Mathematics Stack Exchange, 2019.
Clark Kimberling, Best lower and upper approximations to irrational numbers, Elem. Math. vol. 52 iss. 3 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A002315, A005319, A182431, A001108, A049629, A084068.

I:=[0,1,4,7]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 27 2018

A143608 := proc(n)
    option remember;
    if n <= 3 then
        op(n+1,[0,1,4,7]) ;
    else
        6*procname(n-2)-procname(n-4) ;
    end if;
end proc: # R. J. Mathar, Jul 22 2012

a = -4; b = -1; Reap[While[b<2000000000, t = 4*b-a; Sow[t]; a=b; b=t; t = 2*b-a; Sow[t]; a=b; b=t]][[2,1]]
CoefficientList[Series[x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 24 2014 *)
LinearRecurrence[{0, 6, 0, -1}, {0, 1, 4, 7}, 31] (* Jean-François Alcover, Sep 21 2017 *)

a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,6,0]^n*[0;1;4;7])[1,1] \\ Charles R Greathouse IV, Jun 11 2015

concat(0, Vec(x*(1+4*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Mar 27 2016

a(2*n) = (a(2*n - 1) + a(2*n + 1))/2.
a(2*n + 1) = (a(2*n) + a(2*n + 2))/4.
a(2*n) = 4*A001109(n).
a(2*n + 1) = 4*A001109(n) + A001541(n).
From Colin Barker, Jun 29 2012: (Start)
a(n) = 6*a(n-2) - a(n-4).
G.f.: x*(1 + 4*x + x^2)/((1 + 2*x - x^2)*(1 - 2*x - x^2)) = x*(1 + 4*x + x^2)/(1 - 6*x^2 + x^4). (End)
2*a(n) = A078057(n) - A123335(n-1). - R. J. Mathar, Jul 04 2012
a(2n) = A005319(n); a(2n+1) = A002315(n). - R. J. Mathar, Jul 17 2009
a(n)*a(n+1) + 1 = A001653(n+1). - Charlie Marion, Dec 11 2012
a(n) = (((-2 - sqrt(2) + (-1)^n * (-2+sqrt(2))) * ((-1+sqrt(2))^n - (1+sqrt(2))^n)))/(4*sqrt(2)). - Colin Barker, Mar 27 2016
a(n) = A084068(n) - A079496(n). - César Aguilera, Feb 14 2023

A113224 a(2n) = A002315(n), a(2n+1) = A082639(n+1).

Original entry on oeis.org

1, 2, 7, 16, 41, 98, 239, 576, 1393, 3362, 8119, 19600, 47321, 114242, 275807, 665856, 1607521, 3880898, 9369319, 22619536, 54608393, 131836322, 318281039, 768398400, 1855077841, 4478554082, 10812186007, 26102926096, 63018038201

Offset: 0

Creighton Dement, Oct 18 2005

From Paul D. Hanna, Oct 22 2005: (Start)
The logarithmic derivative of this sequence is twice the g.f. of A113282, where a(2*n) = A113282(2*n), a(4*n+1) = A113282(4*n+1) - 3, a(4*n+3) = A113282(4*n+3) - 1.
Equals the self-convolution of integer sequence A113281. (End)
With an offset of 1, this sequence is the case P1 = 2, P2 = 0, Q = -1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 19 2015
Floretion Algebra Multiplication Program, FAMP Code: -2ibaseiseq[B*C], B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'; C = + .5'i + .5i' + .5'ii' + .5e

Creighton Dement, Floretion Online Multiplier. [broken link]
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).

Cf. A113225, A002315, A082639, A100828, A113281, A113282, A113283, A113284, A129744.

[Floor((1+Sqrt(2))^(n+1)/2): n in [0..30]]; // Vincenzo Librandi, Mar 20 2015

a[n_] := n*Sum[ Sum[ Binomial[i, n-k-i]*Binomial[k+i-1, k-1], {i, Ceiling[(n-k)/2], n-k}]*(1-(-1)^k)/(2*k), {k, 1, n}]; Table[a[n], {n, 1, 29}] (* Jean-François Alcover, Feb 26 2013, after Vladimir Kruchinin *)
CoefficientList[Series[(1 + x^2) / ((x^2 - 1) (x^2 + 2 x - 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)
LinearRecurrence[{2,2,-2,-1},{1,2,7,16},30] (* Harvey P. Dale, Oct 10 2017 *)

a(n):=n*sum(sum(binomial(i,n-k-i)*binomial(k+i-1,k-1),i,ceiling((n-k)/2),n-k)*(1-(-1)^k)/(2*k),k,1,n); /* Vladimir Kruchinin, Apr 11 2011 */

{a(n)=local(x=X+X*O(X^n));polcoeff((1+x^2)/(1-x^2)/(1-2*x-x^2),n,X)} \\ Paul D. Hanna

G.f.: (1+x^2)/((x-1)*(x+1)*(x^2+2*x-1)).
a(n+2) - a(n+1) - a(n) = A100828(n+1).
a(n) = -(u^(n+1)-1)*(v^(n+1)-1)/2 with u = 1+sqrt(2), v = 1-sqrt(2). - Vladeta Jovovic, May 30 2007
a(n) = n * Sum_{k=1..n} Sum_{i=ceiling((n-k)/2)..n-k} binomial(i,n-k-i)*binomial(k+i-1,k-1)*(1-(-1)^k)/(2*k). - Vladimir Kruchinin, Apr 11 2011
a(n) = A001333(n+1) - A000035(n). - R. J. Mathar, Apr 12 2011
a(n) = floor((1+sqrt(2))^(n+1)/2). - Bruno Berselli, Feb 06 2013
From Peter Bala, Mar 19 2015: (Start)
a(n) = (1/2) * A129744(n+1).
exp( Sum_{n >= 1} 2*a(n-1)*x^n/n ) = 1 + 2*Sum_{n >= 1} Pell(n) *x^n. (End)
a(n) = A105635(n-1) + A105635(n+1). - R. J. Mathar, Mar 23 2023

A008843 Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.

Original entry on oeis.org

1, 49, 1681, 57121, 1940449, 65918161, 2239277041, 76069501249, 2584123765441, 87784138523761, 2982076586042449, 101302819786919521, 3441313796169221281, 116903366249966604049, 3971273138702695316401, 134906383349641674153601, 4582845760749114225906049, 155681849482120242006652081

Offset: 0

N. J. A. Sloane

Also indices of triangular numbers (A000217) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 16 2015

a(n)-th triangular number is a square; subsequence of A001108. - Jaroslav Krizek, Aug 05 2016

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.
P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 288.
P. F. Teilhet, Note #2094, L'Intermédiaire des Mathématiciens, 10 (1903), pp. 235-238.

M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
Index entries for sequences related to Bernoulli numbers.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A002315, A016754, A146313, A253826.

LinearRecurrence[{35,-35,1},{1,49,1681},17] (* Stefano Spezia, Aug 17 2024 *)

a(n) = 34*a(n-1) - a(n-2) + 16 = A002315(n)^2 = 2*A001653(n)^2 - 1 = 2*A008844(n) - 1 = floor(A046176(n)*sqrt(2)) = 6*A055792(n+1) - a(n-1) + 4 = (6*A055792(n+2) + a(n-1) - 20)/35. - Henry Bottomley, Nov 13 2001
a(n) = A001108(2n+1). - Ira M. Gessel, Nov 05 2014
a(n) = Sum_{k=1..2*n+1} 2^(k-1)*binomial(4*n+2, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 03 2003
O.g.f.: -(1+14*x+x^2)/((-1+x)*(1-34*x+x^2)). - R. J. Mathar, Nov 23 2007
a(n) = -(cosh((2*n - 1)*arctanh(sqrt(2))))^2 = -1 - (sinh((2*n - 1)*arctanh(sqrt(2))))^2. - Artur Jasinski, Oct 30 2008
a(n) = Sum_{k=0..4n+1} A000129(k), see Santana and Diaz-Barrero link at A002315. - Ivan N. Ianakiev, Jul 15 2022

a(14)-a(17) from Stefano Spezia, Aug 17 2024

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

A377016 Semiperimeter of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

1, 28, 861, 28680, 970921, 32963140, 1119662181, 38034888528, 1292062686481, 43892073946540, 1491038320325421, 50651410052600280, 1720656899012149561, 58451683130389395028, 1985636569382856677301, 67453191675004485098400, 2291422880375627492063521, 77840924741066359629967420

Offset: 0

Miguel-Ángel Pérez García-Ortega, Oct 13 2024

a(0) = 1 is included by convention. This corresponds to the Pythagorean triple 1^2 + 0^2 = 1^2.

			For n=2, the short leg is A002315(2) = 41 and the hypotenuse is A008844(n) = 841 so the semiperimeter is then a(2) = (41 + 840 + 841)/2 = 861.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Index entries for linear recurrences with constant coefficients, signature (41,-246,246,-41,1).

Cf. A002315, A078522, A008844.

s[n_]:=s[n]=Module[{a, b},a=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2;b=(a^2-1)/2;{(a+2b+1)/2}];semis={};Do[semis=Join[semis,FullSimplify[s[n]]],{n,0,17}];semis

Vec((1 - 13*x - 41*x^2 + 21*x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 - x)) + O(x^20)) \\ Andrew Howroyd, Oct 14 2024

a(n) = (A002315(n) + 2*A008844(n) - 1)/2.

G.f.: (1 - 13*x - 41*x^2 + 21*x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 - x)). - Andrew Howroyd, Oct 14 2024

A377725 Length of the short leg of the unique primitive Pythagorean triple whose inradius is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

3, 15, 83, 479, 2787, 16239, 94643, 551615, 3215043, 18738639, 109216787, 636562079, 3710155683, 21624372015, 126036076403, 734592086399, 4281516441987, 24954506565519, 145445522951123, 847718631141215, 4940866263896163, 28797478952235759, 167844007449518387

Offset: 1

Miguel-Ángel Pérez García-Ortega, Nov 05 2024

			Triangles begin:
  n=1:      3,         4,         5;
  n=2:     15,       112,       113;
  n=3:     83,      3444,      3445;
  n=4:    479,    114720,    114721;
  ...
This sequence gives the first column.

Miguel-Ángel Pérez García-Ortega, Capitulo 5. Catetos, El Libro de las Ternas Pitagóricas.

Cf. A002315, A377016, A377017, A377726, A385977 (long leg).

a(n) = 2*A002315(n) + 1.

A377017 Area of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

0, 84, 17220, 3412920, 675761016, 133797385260, 26491207202460, 5245125232676784, 1038508304885968560, 205619399242324129860, 40711602541676078766516, 8060691683852625858745320, 1595976241800278270688414120, 315995235184771245126273789084, 62565460590342906257639745449100

Offset: 0

Miguel-Ángel Pérez García-Ortega, Oct 13 2024

nonn

a(0)=0 is included by convention. This corresponds to the Pythagorean triple 1^2 + 0^2 = 1^2.

All terms in this sequence are divisible by 84.

			For n=2, the short leg is A002315(2) = 41 and the long leg is A008844(2)-1 = 840 so the area is then a(2) = 41*840/2 = 17220.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A377016, A002315, A078522, A008844.

s[n_]:=s[n]=Module[{a, b}, a=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2; b=(a^2-1)/2; {(a*b)/2}]; areas={}; Do[areas=Join[areas, FullSimplify[s[n]]], {n, 0, 17}]; areas

a(n) = A002315(n)*(A008844(n)-1)/2.

G.f.: 84*x*(1 + x)/((1 - 198*x + x^2)*(1 - 6*x + x^2)). - Andrew Howroyd, Oct 14 2024

A377726 Lengths of the long leg of the unique primitive Pythagorean triple (x,y,z) such that (x-y+z)/2 is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

84, 3280, 113764, 3878112, 131820084, 4478459440, 152138450884, 5168244315840, 175568258308884, 5964153062868112, 202605638937276964, 6882627588628286880, 233806732478308836084, 7942546277279354556400, 269812766698548756220804, 9165691521493946935370112

Offset: 1

Miguel-Ángel Pérez García-Ortega, Nov 05 2024

			Triangles begins:
  n=1:     13,        84,        85;
  n=2:     81,      3280,      3281;
  n=3:    477,    113764,    113765;
  ...
This sequence gives the middle column.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Cf. A002315, A377016, A377017, A377725, A362545 (short legs).

ra[n_]:=ra[n]=Module[{ra},ra=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2;{2ra-1,2ra^2-2ra,2ra^2-2ra+1}];exradio={};Do[exradio=Join[exradio,FullSimplify[ra[n]]],{n,0,10}];exradio

a(n) = 2 * A002315(n) * (A002315(n) - 1).

A378965 Semiperimeter of the unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

1, 91, 3321, 114003, 3879505, 131828203, 4478506761, 152138726691, 5168245923361, 175568267678203, 5964153117476505, 202605639255558003, 6882627590483364721, 233806732489121022091, 7942546277342372594601, 269812766698916052264003, 9165691521496087693591105, 311363698964228006760021403

Offset: 0

Miguel-Ángel Pérez García-Ortega, Dec 12 2024

			For n=2, the short leg is A377726(2,1) = 13, the long leg is A377725(2,2) = 842 and the hypotenuse is A377725(2,3) = 85 so the semiperimeter is then a(2) = (13 + 84 + 85)/2 = 91.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Cf. A002315, A377016, A377017, A377726, A378965.

s[n_]:=s[n]=Module[{ra},ra=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2;{ra(2ra-1)}];semis={};Do[semis=Join[semis,FullSimplify[s[n]]],{n,0,17}];semis

a(n) = (A377726(n,1) + A377726(n,2) + A377726(n,3))/2.

A111647 a(n) = A001541(n)A001653(n+1)A002315(n).

Original entry on oeis.org

1, 105, 20213, 3998709, 791704585, 156753394977, 31036379835581, 6145046450172525, 1216688160731724433, 240898110778299543129, 47696609245941810082565, 9443687732585695622131557

Offset: 0

Charlie Marion, Aug 24 2005

			a(1) = 105 = 3*5*7.

G. C. Greubel, Table of n, a(n) for n = 0..434
Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A001541, A001653, A002315, A111648, A111649.

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)))); // G. C. Greubel, Jul 15 2018

CoefficientList[Series[(1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)), {x, 0, 30}], x] (* G. C. Greubel, Jul 15 2018 *)

x='x+O('x^30); Vec((1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1))) \\ G. C. Greubel, Jul 15 2018

2*a(n) = A001109(3*n+1) + A001109(n+1).
a(n) = sqrt(A011900(2*n)*A046090(2*n)*A001109(2*n+1)).
a(n) = A001541(3*n) + 2*A001109(n)*A001541(n-1)*A001541(n).
For n>0, a(n) = A001652(3*n) - A053141(2*n)*A002315(n-1) - A001652(n-1).
G.f.: (1+3*x^3-17*x^2-99*x)/((x^2-6*x+1)*(x^2-198*x+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
2*a(n) = A001109(n+1) + A097731(n) + 6*A097731(n-1). - R. J. Mathar, Jan 31 2024

cp's OEIS Frontend

A143608 A005319 and A002315 interleaved.

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A113224 a(2n) = A002315(n), a(2n+1) = A082639(n+1).

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A008843 Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.

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

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A377016 Semiperimeter of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

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A377725 Length of the short leg of the unique primitive Pythagorean triple whose inradius is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

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A377017 Area of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

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A111647 a(n) = A001541(n)A001653(n+1)A002315(n).