A253175 -id:A253175 - cp's OEIS frontend

A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

2, 2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Essentially the same as A157085. - R. J. Mathar, Feb 07 2022

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A031138, A054318, A097784, A253175, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350922, A350924, A350925, A350926.

LinearRecurrence[{11, -11, 1}, {2, 2, 14}, 25] (* Paolo Xausa, May 30 2025 *)

G.f.: 2*(1 - 10*x + 7*x^2)/((1 - x)*(1 - 10*x + x^2)). - Stefano Spezia, Jan 22 2022
From Hugo Pfoertner, Jan 22 2022: (Start)
a(n) = A031138(n) + 1.
a(n) = 3*A054318(n) - 1.
a(n) = 12*A097784(n-2) + 2 for n >= 2. (End)
a(n) = 2 * A253175(n) for n>=1. - Alois P. Heinz, Jan 22 2022

A157085 Consider all Consecutive Integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives Z values.

Original entry on oeis.org

2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402

Offset: 0

Charlie Marion, Mar 12 2009

"Consecutive Integer Pythagorean quintuple" means that X^2 + (X+1)^2 + (X+2)^2 = (Z-1)^2 + Z^2. - M. F. Hasler, Oct 04 2014
The last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: let last(0)=0, last(1)=k*(2k+3) and, for n > 1, last(n) = (4k+2)*last(n-1) - last(n-2) - 2*k*(k-1); e.g., if k=3, then last(2) = 363 = 14*27 - 3 - 12.
For n > 0, a(n) = 6*a(n-1) + 5*A157084(n-1) + 4; e.g., 1322 = 6*108 + 5*134 + 4.
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=3 and n=2, then first(2) = 312 = 7*21 + 6*27+3 and last(2) = 363 = 8*21 + 7*27 + 6.
a(n) = 2^n*3*((1+sqrt(3/2))^(2*n+1) - (1-sqrt(3/2))^(2*n+1))/(4*sqrt(3/2)) + 1/2; e.g., 134 = 2^2*3*((1+sqrt(3/2))^5 - (1-sqrt(3/2))^5)/(4*sqrt(3/2)) + 1/2.
The last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: if q=(k+1)/k, then last(n) = k^n*(k+1)*((1+sqrt(q))^(2*n+1) - (1-sqrt(q))^(2*n+1))/(4*sqrt(q)) + (k-1)/2; e.g., if k=3 and n=2, then last(2) = 363 = 3^2*4*((1+sqrt(4/3))^5 - (1-sqrt(4/3))^5)/(4*sqrt(4/3)) + 2/2.
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 last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows:
last(2n+1) = (e(2n+1)^2 + k^2*e(2n)^2 + k*(k-1)*e(2n+1)*e(n))/k and, for n > 0, last(2n) = (k*(u(2n)^2 + u(2n-1)^2 + (k-1)*u(2n)*u(2n-1)))/(k+1); e.g., a(3) = 1322 = (40^2 + 2^2*9^2 + 2*1*40*9)/2 and a(4) = 13082 = (2*(109^2 + 49^2 + 1*109*49))/3.
If b(0)=1, b(1)=4k+2 and, for n > 1, b(n) = (4k+2)*b(n-1) - b(n-2), and last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple as defined above, then Sum_{i=0..n} (k*last(i) - k(k-1)/2) = k(k+1)/2*b(n); e.g., if n=3, then 1 + 2 + 13 + 14 + 133 + 134 = 297 = 3*99.
If first(n) is the first term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} first(n+1)/first(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

			a(3) = 134 since 108^2 + 109^2 + 110^2 = 133^2 + 134^2.

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.

Paolo Xausa, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A001653, A157089, A157093, A157097.

LinearRecurrence[{11, -11, 1}, {2, 14, 134}, 25] (* Paolo Xausa, May 29 2025 *)

for(y=0,oo,yy=(y+2)^2+(y+1)^2;for(x=sqrtint(yy\3),ceil(sqrt(yy/3)),x^2+(x-1)^2+(x-2)^2==yy&&print1(y+2,", "))) \\ For illustrative purpose. - M. F. Hasler, Oct 04 2014

For n > 1, a(n) = 10*a(n-1) - a(n-2) - 4; e.g., 1322 = 10*134 - 14 - 4.
Limit_{n->oo} a(n+1)/a(n) = 2*(1+sqrt(3/2))^2 = 5 + 2*sqrt(6).
G.f.: 2*(1-4*x+x^2)/((1-x)*(x^2-10*x+1)). - Colin Barker, Jan 01 2012
a(n) = 2*A253175(n+1). - R. J. Mathar, Feb 07 2022

Edited by M. F. Hasler, Oct 04 2014

A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 6, 55, 540, 5341, 52866, 523315, 5180280, 51279481, 507614526, 5024865775, 49741043220, 492385566421, 4874114620986, 48248760643435, 477613491813360, 4727886157490161, 46801248083088246, 463284594673392295, 4586044698650834700, 45397162391834954701

Offset: 1

Colin Barker, Jan 02 2015

Also positive integers x in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0, the corresponding values of y being A054318.
Also indices of centered hexagonal numbers (A003215) which are also hexagonal numbers (A000384).
Also indices of terms in sequence A193218 which are the square root of a sum of 5th powers (A000539). - Daniel Poveda Parrilla, Jun 10 2017

			6 is in the sequence because the 6th centered square number is 61, which is also the 5th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A000539, A001844, A003215, A054318, A193218, A253175.

LinearRecurrence[{11, -11, 1}, {1, 6, 55}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(x*(5*x-1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: x*(5*x-1) / ((x-1)*(x^2-10*x+1)).
a(n) = sqrt((-2-(5-2*sqrt(6))^n-(5+2*sqrt(6))^n)*(2-(5-2*sqrt(6))^(1+n)-(5+2*sqrt(6))^(1+n)))/(4*sqrt(2)). - Gerry Martens, Jun 04 2015
2*a(n) = 1+A054320(n-1). - R. J. Mathar, Feb 07 2022

A006244 Hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 91, 8911, 873181, 85562821, 8384283271, 821574197731, 80505887094361, 7888755361049641, 773017519495770451, 75747828155224454551, 7422514141692500775541, 727330638057709851548461, 71270980015513872950973631, 6983828710882301839343867371, 684343942686450066382748028721

Offset: 1

N. J. A. Sloane, Jeffrey Shallit

Equivalently, triangular hex numbers.

			a(1)=91 because 91 is the sixth centered hexagonal number and the seventh hexagonal number.

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 19.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Jon E. Schoenfield, Table of n, a(n) for n = 1..500
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
S. C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011, pp. 339-350.
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A001570, A001921, A000384, A003215, A253175.

CP := n -> 1+1/2*6*(n^2-n): N:=10: u:=5: v:=1: x:=6: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+24*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007
A006244:=-(1-8*z+z**2)/(z-1)/(z**2-98*z+1); # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
a := n -> (Matrix([[91,1,1]]). Matrix([[99,1,0],[ -99,0,1],[1,0,0]])^n)[1,3]; seq (a(n), n=1..20); # Alois P. Heinz, Aug 14 2008

CoefficientList[Series[(1 - 8*x + x^2)/(1 - 99*x + 99*x^2 - x^3), {x, 0, 20}], x] (* Jean-François Alcover, Feb 26 2015 *)

Vec(-x*(x^2-8*x+1)/((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Jan 08 2015

From Richard Choulet, Sep 19 2007: (Start)
We must solve 2*r^2-r=3*p^2-3*p+1, which gives X^2=6*Y^2+3 with X=4*r-1 and Y=2*p-1. We obtain at the same time the following sequences:
X is given by 3, 27, 267, ... sequence for which a(n+2)=10*a(n+1)-a(n) and a(n+1)=5*a(n)+2*(6a(n)^2-18)^0.5
Y is given by 1, 11, 109, ... sequence for which a(n+2)=10*a(n+1)-a(n) and a(n+1)=5*a(n)+2*(6a(n)^2+3)^0.5
p is given by 1, 6, 55, 540, ... sequence for which a(n+2)=10*a(n+1)-a(n)-4 and a(n+1)=5*a(n)-2+(24*a(n)^2-24*a(n)+9)^0.5
r is given by 1, 7, 67, 661, ... sequence for which a(n+2)=10*a(n+1)-a(n)-2 and a(n+1)=5*a(n)-1+(24*a(n)^2-12*a(n)-3)^0.5
a(n+2) = 98*a(n+1)-a(n)-6, a(n+1)=49*a(n)-3+5*(96*a(n)^2-12*a(n)-3)^0.5.
G.f.: z*(1-8*z+z^2)/((1-z)*(1-98*z+z^2)). (End)
Define x(n) + y(n)*sqrt(24) = (6+sqrt(24))*(5+sqrt(24))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+6*(s(n)^2-s(n))). - Steven Schlicker, Apr 24 2007
a(n) = (A007667(n+1)-1)/4. - Ralf Stephan, Mar 03 2004
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3). - Colin Barker, Jan 08 2015

Edited by N. J. A. Sloane, Sep 25 2007
More terms from Alois P. Heinz, Aug 14 2008
More terms from Jon E. Schoenfield, Dec 26 2008

cp's OEIS Frontend

A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

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A157085 Consider all Consecutive Integer Pythagorean quintuples (X, X+1, X+2, Z-1, Z) ordered by increasing Z; sequence gives Z values.

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A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

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A006244 Hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

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