A139011 -id:A139011 - cp's OEIS frontend

A201730 Triangle T(n,k), read by rows, given by (2,1/2,3/2,0,0,0,0,0,0,0,...) DELTA (0,1/2,-1/2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

1, 2, 0, 5, 1, 0, 14, 6, 0, 0, 41, 26, 1, 0, 0, 122, 100, 10, 0, 0, 0, 365, 363, 63, 1, 0, 0, 0, 1094, 1274, 322, 14, 0, 0, 0, 0, 3281, 4372, 1462, 116, 1, 0, 0, 0, 0, 9842, 14760, 6156, 744, 18, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Dec 04 2011

Riordan array ((1-2x)/(1-4x+3x^2),x^2/(1-4x+3x^2)).

A007318*A201701 as lower triangular matrices.

			Triangle begins:
1
2, 0
5, 1, 0
14, 6, 0, 0
41, 26, 1, 0, 0
122, 100, 10, 0, 0, 0
365, 363, 63, 1, 0, 0, 0

Cf. A007051 (1st column), A261064 (2nd column).

A201730 := proc(n,k)
    (1-2*x)/(1-4*x+(3-y)*x^2) ;
    coeftayl(%,y=0,k) ;
    coeftayl(%,x=0,n) ;
end proc:
seq(seq(A201730(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Dec 06 2011

m = 13;
(* DELTA is defined in A084938 *)
DELTA[Join[{2, 1/2, 3/2}, Table[0, {m}]], Join[{0, 1/2, -1/2}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)

G.f.: (1-2x)/(1-4x+(3-y)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A139011(n), A000079(n), A007051(n), A006012(n), A001075(n), A081294(n), A001077(n), A084059(n), A108851(n), A084128(n), A081340(n), A084132(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k, k>+0} T(n+k,k) = A081704(n) .
T(n,k) = 3*T(n-1,k)+ Sum_{j>0} T(n-1-j,k-1).
T(n,k) = 4*T(n-1,k)+ T(n-2,k-1) - 3*T(n-2,k) with T(0,0)=1, T(1,0)= 2, T(1,1) = 0 and T(n,k) = 0 if k<0 or if n

A316657 For any n >= 0 with base-5 expansion Sum_{k=0..w} d_k * 5^k, let f(n) = Sum_{k=0..w} [d_k > 0] * (2 + i)^k * i^(d_k - 1) (where [] is an Iverson bracket and i denotes the imaginary unit); a(n) equals the real part of f(n).

Original entry on oeis.org

0, 1, 0, -1, 0, 2, 3, 2, 1, 2, -1, 0, -1, -2, -1, -2, -1, -2, -3, -2, 1, 2, 1, 0, 1, 3, 4, 3, 2, 3, 5, 6, 5, 4, 5, 2, 3, 2, 1, 2, 1, 2, 1, 0, 1, 4, 5, 4, 3, 4, -4, -3, -4, -5, -4, -2, -1, -2, -3, -2, -5, -4, -5, -6, -5, -6, -5, -6, -7, -6, -3, -2, -3, -4, -3

Offset: 0

Rémy Sigrist, Jul 09 2018

See A316658 for the imaginary part of f.
See A316707 for the square of the modulus of f.
The function f has nice fractal features (see scatterplot in Links section).
It appears that f defines a bijection from the nonnegative integers to the Gaussian integers.

Rémy Sigrist, Table of n, a(n) for n = 0..15624
Rémy Sigrist, Colored scatterplot of (a(n), A316658(n)) for n=0..5^8-1 (where the hue is function of n)
Wikipedia, Gaussian integer

Cf. A139011, A316658, A316707.

a[n_] := Module[{d, z}, d = IntegerDigits[n, 5] // Reverse; z = Sum[ If[d[[i]]>0, (2+I)^(i-1)*I^(d[[i]]-1), 0], {i, 1, Length[d]}]; Re[z]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 06 2021, after PARI code *)

a(n) = my (d=Vecrev(digits(n, 5)), z=sum(i=1, #d, if (d[i], (2+I)^(i-1) * I^(d[i]-1), 0))); real(z)

a(5^n) = A139011(n) for any n >= 0.

a(3 * 5^n) = -A139011(n) for any n >= 0.

A099456 Expansion of 1/(1 - 4x + 5x^2).

Original entry on oeis.org

1, 4, 11, 24, 41, 44, -29, -336, -1199, -3116, -6469, -10296, -8839, 16124, 108691, 354144, 873121, 1721764, 2521451, 1476984, -6699319, -34182196, -103232189, -242017776, -451910159, -597551756, -130656229, 2465133864

Offset: 0

Paul Barry, Oct 16 2004

Associated to the knot 9_44 by the modified Chebyshev transform A(x) -> (1/(1+x^2)^2)A(x/(1+x^2)). See A099457 and A099458.

Imaginary part of (2+i)^n. - Gary W. Adamson, Apr 05 2008; Franklin T. Adams-Watters, Jan 06 2009

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Mei Bai, Wenchang Chu, and Dongwei Guo, Reciprocal Formulae among Pell and Lucas Polynomials, Mathematics, 10, 2691, (2022). See p. 5.
Beata Bajorska-Harapińska, Barbara Smoleń, and Roman Wituła, On Quaternion Equivalents for Quasi-Fibonacci Numbers, Shortly Quaternaccis, Advances in Applied Clifford Algebras (2019) Vol. 29, 54.
Dror Bar-Natan, The Rolfsen Knot Table
Index entries for linear recurrences with constant coefficients, signature (4,-5).

Cf. A139011, A090131 (inv. bin. transf.)

seq(((2+I)^(n+1) - (2-I)^(n+1))/(2*I),n=0..30);  # James R. Buddenhagen, Dec 29 2017

CoefficientList[Series[1/(1-4*x+5*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 09 2013 *)
Table[((1+2*I)*(2-I)^n + (1-2*I)*(2+I)^n)/2,{n,0,20}] (* Vaclav Kotesovec, Oct 09 2013 *)

[lucas_number1(n,4,5) for n in range(1, 29)] # Zerinvary Lajos, Apr 22 2009

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-5)^k*4^(n-2k).
E.g.f. (with offset 1): exp(2*x)*sin(x). - Zerinvary Lajos, Apr 06 2009 [corrected by Joerg Arndt, Apr 24 2011]
a(n) = 4*a(n-1) - 5*a(n-2), a(0)=1, a(1)=4. - Vincenzo Librandi, Mar 22 2011
From Paul Curtz, Apr 24 2011: (Start)
a(n) - a(n-4) = 40 * A118444(n);
a(n) - a(n-2) = 10 * A139011(n). (End)
a(n) = ((1+2*i)*(2-i)^n + (1-2*i)*(2+i)^n)/2. - Vaclav Kotesovec, Oct 09 2013
a(n) = ((2+i)^(n+1) - (2-i)^(n+1))/(2*i).
Lim sup n->infinity |a(n)|/5^((n+1)/2) = 1. - Vaclav Kotesovec, Oct 09 2013
a(n) = Sum_{k=0..n} (-1)^k*2^(n-2*k)*binomial(n+1,2*k+1). - Gerry Martens, Sep 18 2022
E.g.f.: exp(2*x)*(cos(x) + 2*sin(x)). - Stefano Spezia, Jul 24 2025

A176594 a(n) = 5^(2^n).

Original entry on oeis.org

5, 25, 625, 390625, 152587890625, 23283064365386962890625, 542101086242752217003726400434970855712890625, 293873587705571876992184134305561419454666389193021880377187926569604314863681793212890625

Offset: 0

Vincenzo Librandi, Apr 21 2010

nonn

Also the hypotenuse of primitive Pythagorean triangles obtained by repeated application of basic formula c(n)=p(n)^2+q(n)^2 starting p(0)=2, q(0)=1, see A100686, A098122. Example: a(2)=25 since starting (2,1) gives Pythagorean triple (3,4,5) using (3,4) as new generators gives triple (7,24,25) hypotenuse 25=a(2). - Carmine Suriano, Feb 04 2011

Paolo Xausa, Table of n, a(n) for n = 0..10

Cf. A001146, A011764, A078886, A098122, A100686, A165423.

Cf. A185457, A120905, A139011. - Carmine Suriano, Feb 04 2011

NestList[#^2 &, 5, 8] (* Paolo Xausa, Jul 13 2025 *)

a(n) = 5^(2^n); \\ Michel Marcus, Jan 26 2016

a(n) = A165423(n+3).
a(n+1) = a(n)^2 with a(0)=5.
a(n-1) = (Im((2+i)^(2^n))^2 + Re((2+i)^(2^n))^2)^(1/2). - Carmine Suriano, Feb 04 2011
Sum_{n>=0} 1/a(n) = A078886. - Amiram Eldar, Nov 09 2020
Product_{n>=0} (1 + 1/a(n)) = 5/4. - Amiram Eldar, Jan 29 2021

Offset corrected by R. J. Mathar, Jun 18 2010

A230710 Values of x such that x^2 + y^2 = 5^n with x and y coprime and 0 < x < y.

Original entry on oeis.org

1, 3, 2, 7, 38, 44, 29, 336, 718, 237, 2642, 10296, 8839, 16124, 108691, 164833, 24478, 922077, 2521451, 1476984, 6699319, 34182196, 35553398, 32125393, 306268562, 597551756, 130656229, 2465133864, 8701963882, 6890111163, 15949374758, 98248054847, 135250416961

Offset: 1

Colin Barker, Oct 28 2013

nonn

The corresponding y-values are in A230711.
For all non-coprime solutions (x,y) to the equation x^2 + y^2 = p^n, x and y are both divisible by the prime p.
Using de Moivre's Theorem (in essence), define (c,d)*(e,f) as (ce-df,cf+de). Then a(n) = min{|u(n)|, |v(n)|}, where (u(n),v(n)) = (2,1)^n = (2,1)*(2,1)^[n-1]. Proof: It can be readily seen that u^2(n) + v^2(n) = 5^n. To show that u(n) and v(n) are relatively prime, assume that x,y are relatively prime. Then (2,1)*(x,y) = (2x-y, x+2y). If a prime p were to divide both of 2x-y and x+2y, then p would divide 5y, so p=5. Now suppose x == 2 (mod 5) and y == 1 (mod 5). It can be seen that 2x-y == -2 (mod 5) and x+2y == -1 (mod 5). The reverse also holds. Because u(1)=2 and v(1)=1, the result follows inductively. - Richard Peterson, May 21 2021

			a(4)=7 because 7^2 + 24^2 = 625 = 5^4.

Zak Seidov, Table of n, a(n) for n = 1..200
Chris Busenhart, Lorenz Halbeisen, Norbert Hungerbühler, and Oliver Riesen, On primitive solutions of the Diophantine equation x^2+ y^2= M, Eidgenössische Technische Hochschule (ETH Zürich, Switzerland, 2020).

Cf. A000351, A230711.
Cf. A188948, A230622, A230644.
Cf. A139011.

Table[Select[PowersRepresentations[5^n, 2, 2], CoprimeQ[#[[1]], #[[2]]] &][[1,1]], {n, 33}] (* T. D. Noe, Nov 04 2013 *)

Typo in data fixed by Colin Barker, Nov 02 2013

A180957 Generalized Narayana triangle for (-1)^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, -2, -5, -2, 1, 1, -5, -15, -15, -5, 1, 1, -9, -30, -41, -30, -9, 1, 1, -14, -49, -77, -77, -49, -14, 1, 1, -20, -70, -112, -125, -112, -70, -20, 1, 1, -27, -90, -126, -117, -117, -126, -90, -27, 1, 1, -35, -105, -90, 45, 131, 45, -90, -105, -35, 1

Offset: 0

Paul Barry, Sep 28 2010

			Triangle begins
  1;
  1,   1;
  1,   1,    1;
  1,   0,    0,    1;
  1,  -2,   -5,   -2,    1;
  1,  -5,  -15,  -15,   -5,    1;
  1,  -9,  -30,  -41,  -30,   -9,    1;
  1, -14,  -49,  -77,  -77,  -49,  -14,   1;
  1, -20,  -70, -112, -125, -112,  -70, -20,    1;
  1, -27,  -90, -126, -117, -117, -126, -90,  -27,   1;
  1, -35, -105,  -90,   45,  131,   45, -90, -105, -35, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A056241, A139011.

Variant: A061176.

A180957:= func< n,k | (&+[ (-1)^(k-j)*Binomial(n, j)*Binomial(n-j, 2*(k-j)) : j in [0..n]]) >;
[A180957(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Apr 06 2021

T[n_, k_]:= Sum[(-1)^(k-j)*Binomial[n, j]*Binomial[n-j, 2*(k-j)], {j,0,n}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 06 2021 *)

def A180957(n,k): return sum( (-1)^(k+j)*binomial(n,j)*binomial(n-j, 2*(k-j)) for j in (0..n))
flatten([[A180957(n,k) for k in (0..n)] for n in [0..15]]) # G. C. Greubel, Apr 06 2021

G.f.: 1/(1 -x -x*y + x/(1 -x -x*y)) = (1 -x*(1+y))/(1 -2*x*(1+y) +x^2*(1 +3*y +y^2)).
E.g.f.: exp((1+y)*x) * cos(sqrt(y)*x).
T(n, k) = Sum_{j=0..n} (-1)^(k-j)*binomial(n,j)*binomial(n-j, 2*(k-j)).
Sum_{k=0..n} T(n, k) = A139011(n) (row sums).
Sum_{k=0..floor(n/2)} T(n-k, k) = A180958(n) (diagonal sums).

A292495 Triangle read by rows: T(n,k) = (-2)T(n-1,k-1) + T(n,k-1) with T(2m,0) = 0 and T(2*m+1,0) = (-1)^m.

Original entry on oeis.org

0, 1, 1, 0, -2, -4, -1, -1, 3, 11, 0, 2, 4, -2, -24, 1, 1, -3, -11, -7, 41, 0, -2, -4, 2, 24, 38, -44, -1, -1, 3, 11, 7, -41, -117, -29, 0, 2, 4, -2, -24, -38, 44, 278, 336, 1, 1, -3, -11, -7, 41, 117, 29, -527, -1199, 0, -2, -4, 2, 24, 38, -44, -278, -336, 718

Offset: 0

Seiichi Manyama, Sep 22 2017

			First few rows are:
   0;
   1,  1;
   0, -2, -4;
  -1, -1,  3,  11;
   0,  2,  4,  -2, -24;
   1,  1, -3, -11,  -7,  41;
   0, -2, -4,   2,  24,  38,  -44;
  -1, -1,  3,  11,   7, -41, -117, -29;
   0,  2,  4,  -2, -24, -38,   44, 278, 336.

Seiichi Manyama, Rows n = 0..139, flattened

The diagonal of the triangle is related to A099456.
The next diagonal of the triangle is related to A139011.
Cf. A006495, A006496.
T(n,k) = b*T(n-1,k-1) + T(n,k-1): A292789 (b=-3), this sequence (b=-2), A117918 and A228405 (b=1), A227418 (b=2), A292466 (b=4).

T(n+1,n)^2 + T(n,n)^2 = 5^n.

Showing 1-7 of 7 results.

cp's OEIS Frontend

A201730 Triangle T(n,k), read by rows, given by (2,1/2,3/2,0,0,0,0,0,0,0,...) DELTA (0,1/2,-1/2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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Maple

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A316657 For any n >= 0 with base-5 expansion Sum_{k=0..w} d_k * 5^k, let f(n) = Sum_{k=0..w} [d_k > 0] * (2 + i)^k * i^(d_k - 1) (where [] is an Iverson bracket and i denotes the imaginary unit); a(n) equals the real part of f(n).

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Mathematica

PARI

Formula

A099456 Expansion of 1/(1 - 4*x + 5*x^2).

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Maple

Mathematica

Sage

Formula

A176594 a(n) = 5^(2^n).

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Mathematica

PARI

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A230710 Values of x such that x^2 + y^2 = 5^n with x and y coprime and 0 < x < y.

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Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A180957 Generalized Narayana triangle for (-1)^n.

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Magma

Mathematica

Sage

Formula

A292495 Triangle read by rows: T(n,k) = (-2)*T(n-1,k-1) + T(n,k-1) with T(2*m,0) = 0 and T(2*m+1,0) = (-1)^m.

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Formula

A099456 Expansion of 1/(1 - 4x + 5x^2).

A292495 Triangle read by rows: T(n,k) = (-2)T(n-1,k-1) + T(n,k-1) with T(2m,0) = 0 and T(2*m+1,0) = (-1)^m.