A080806 -id:A080806 - cp's OEIS frontend

A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

0, 1, 3, 14, 34, 143, 341, 1420, 3380, 14061, 33463, 139194, 331254, 1377883, 3279081, 13639640, 32459560, 135018521, 321316523, 1336545574, 3180705674, 13230437223, 31485740221, 130967826660, 311676696540, 1296447829381, 3085281225183, 12833510467154

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

Also nonnegative m such that 24*m^2 + 24*m + 1 is a square. - Gerald McGarvey, Apr 02 2005

			a(3) = 2*3 - 1 + 9 = 14, a(4) = 2*14 - 3 + 9 = 34, etc.
G.f. = x + 3*x^2 + 14*x^3 + 34*x^4 + 143*x^5 + 341*x^6 + 1420*x^7 + 3380*x^8 + ... - _Michael Somos_, Jul 15 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 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 (1,10,-10,-1,1).

The k values are in A077291
Cf. A077289, A077290, A077291.
Cf. A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))); // G. C. Greubel, Jul 15 2018

f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 100)]); - Vladimir Pletser, Jul 24 2020

CoefficientList[Series[x*(1 + x)^2/((1 - x)*(1 - 10 x^2 + x^4)), {x, 0, 40}],x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 3, 14, 34}, 50] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ x (1 + x)^2 / ((1 - x) (1 - 10 x^2 + x^4)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))) \\ G. C. Greubel, Jul 15 2018

Let b(n) be A072256. Then a(2*n+2) = 2*a(2*n+1) - a(2*n) + b(n+1), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + b(n+2), with a(0)=0, a(1)=1.
G.f.: x*(1+x)^2/((1-x)*(1-10*x^2+x^4)).
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(n) = 10*a(n-2) - a(n-4) + 4, n > 4. - Vladimir Pletser, Feb 29 2020
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
2*a(n) + 1 = A080806(n+1). - R. J. Mathar, Oct 01 2021

A077291 Second member of Diophantine pair (m,k) that satisfies 6*(m^2 + m) = k^2 + k: a(n) = k.

Original entry on oeis.org

0, 3, 8, 35, 84, 351, 836, 3479, 8280, 34443, 81968, 340955, 811404, 3375111, 8032076, 33410159, 79509360, 330726483, 787061528, 3273854675, 7791105924, 32407820271, 77123997716, 320804348039, 763448871240, 3175635660123, 7557364714688, 31435552253195

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

The corresponding m are given in A077288.

Numbers x such that (2*x^2 + 2*x + 3)/3 = y^2. The corresponding y are given by A080806. - Klaus Purath, Jul 30 2025

			b(3)=630 so a(3) = (-1 + sqrt(8*630 + 1))/2 = (-1 + sqrt(5041))/2 = (71 - 1)/2 = 35.

Colin Barker, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 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 (1,10,-10,-1,1).

Cf. A077288, A077289, A077290, A080806.

f := gfun:-rectoproc({a(-2) = -4, a(-1) = -1, a(0) = 0, a(1) = 3, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 26 2020

LinearRecurrence[{1,10,-10,-1,1},{0,3,8,35,84},30] (* Harvey P. Dale, Oct 11 2019 *)

concat(0, Vec(x*(x^3+3*x^2-5*x-3)/((x-1)*(x^4-10*x^2+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Let b(n) be A077290. Then a(n) = (-1 + sqrt(8*b(n) + 1))/2.
G.f.: x*(x^3+3*x^2-5*x-3) / ((x-1)*(x^4-10*x^2+1)). - Colin Barker, Mar 09 2012
From Vladimir Pletser, Jul 26 2020: (Start)
a(n) = 10*a(n-2) - a(n-4) + 4 with a(-2)=-4, a(-1)=-1, a(0)=0, a(1)=3.
Can be defined for negative n by setting a(-n) = - a(n-1) - 1 for all n in Z.
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). (End)

A237610 Positive integers k such that x^2 - 10xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

8, 15, 20, 23, 24, 32, 47, 60, 71, 72, 80, 87, 92, 95, 96, 116, 128, 135, 152, 159, 167, 180, 188, 191, 200, 207, 212, 215, 216, 239, 240, 263, 276, 284, 288, 303, 311, 320, 335, 344, 348, 359, 368, 375, 380, 383, 384, 392, 404, 423, 431, 447, 456, 464, 479

Offset: 1

Colin Barker, Feb 10 2014

nonn

			15 is in the sequence because x^2 - 10xy + y^2 + 15 = 0 has integer solutions, for example (x, y) = (2, 19).

Cf. A072256 (k = 8), A129445 (k = 15), A080806 (k = 20), A074061 (k = 23), A001079 (k = 24).

Cf. A031363, A084917, A237351, A237599, A237606, A237609.

is(n)=m=bnfisintnorm(bnfinit(x^2-10*x+1),-n);#m>0&&denominator(polcoeff(m[1],1))==1 \\ Ralf Stephan, Feb 11 2014

A080811 a(1) = 8, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

Original entry on oeis.org

8, 81, 68921, 671898241, 67499118303941862584001, 6576499147308118090309591239435044518621593475579845001

Offset: 1

Amarnath Murthy, Mar 22 2003

Cf. A080804, A080805, A080806, A080807, A080808, A080809, A080810, A080812.

buffedStr := proc(n,candid) local f ; if length(n) = 0 then RETURN(true) ; fi ; f := SearchText(substring(n,1),candid) ; if f = 0 then RETURN(false) ; else if buffedStr(substring(n,2..-1),substring(candid,f+1..-1)) = true then RETURN(true) ; else RETURN(false) ; fi ; fi ; end: A080811 := proc(preva,n) local i,tst ; i := 1 ; while true do tst := i^n ; if buffedStr(convert(preva,string),convert(tst,string)) = true then RETURN(tst) ; fi ; i := i+1 ; od: end: an :=8 ; for n from 2 to 15 do an := A080811(an,n) ; end ; # R. J. Mathar, Sep 20 2006

2 more terms from R. J. Mathar, Sep 20 2006

One more term. Sean A. Irvine, Aug 31 2009

A080809 a(1) = 6; thereafter, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

Original entry on oeis.org

6, 16, 216, 2085136, 62019685191371643, 82626753081964483505319130781618465733184

Offset: 1

Amarnath Murthy, Mar 22 2003

Cf. A080804, A080805, A080806, A080807, A080808, A080810, A080811, A080812.

buffedStr := proc(n,candid) local f ; if length(n) = 0 then RETURN(true) ; fi ; f := SearchText(substring(n,1),candid) ; if f = 0 then RETURN(false) ; else if buffedStr(substring(n,2..-1),substring(candid,f+1..-1)) = true then RETURN(true) ; else RETURN(false) ; fi ; fi ; end: A080809 := proc(preva,n) local i,tst ; i := 1 ; while true do tst := i^n; if buffedStr(convert(preva,string),convert(tst,string)) = true then RETURN(tst) ; fi ; i := i+1 ; od: end: an :=6 ; for n from 2 to 15 do an := A080809(an,n) ; end ; # R. J. Mathar, Sep 20 2006

a(n)=A080514(n), n>1. - R. J. Mathar, Sep 18 2008

More terms from R. J. Mathar, Sep 20 2006

Copied a term from A080514. - Sean A. Irvine, Sep 01 2009

A080810 a(1) = 7, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

Original entry on oeis.org

7, 576, 17576, 1475789056, 1420778345789277056227207, 149247077443060991182553045537892735703095362592472033442634721

Offset: 1

Amarnath Murthy, Mar 22 2003

			24^2=576. 26^3=17576. 196^4=1475789056. 67687^5=1420778345789277056227207.

Cf. A080804, A080805, A080806, A080807, A080808, A080809, A080811, A080812.

buffedStr := proc(n,candid) local f ; if length(n) = 0 then RETURN(true) ; fi ; f := SearchText(substring(n,1),candid) ; if f = 0 then RETURN(false) ; else if buffedStr(substring(n,2..-1),substring(candid,f+1..-1)) = true then RETURN(true) ; else RETURN(false) ; fi ; fi ; end: A080810 := proc(preva,n) local i,tst ; i := 1 ; while true do tst := i^n ; if buffedStr(convert(preva,string),convert(tst,string)) = true then RETURN(tst) ; fi ; i := i+1 ; od: end: an :=7 ; for n from 2 to 15 do an := A080810(an,n) ; end ; # R. J. Mathar, Sep 20 2006

Corrected and extended by R. J. Mathar, Sep 20 2006

One more term. Sean A. Irvine, Sep 01 2009

A080807 a(1) = 3, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

Original entry on oeis.org

3, 36, 32768, 20632736881, 2096638274713626813358181376, 2095266319949834829710838437415365780027681350359080101817236461769

Offset: 1

Amarnath Murthy, Mar 22 2003

Next term is > 3*10^43. - David Wasserman, May 14 2004

Cf. A080804, A080805, A080806, A080808, A080809, A080810, A080811, A080812.

One more term from Sean A. Irvine, Sep 01 2009

A082651 Positive integer values of n such that 5n^2+11 is a square.

Original entry on oeis.org

1, 7, 25, 127, 449, 2279, 8057, 40895, 144577, 733831, 2594329, 13168063, 46553345, 236291303, 835365881, 4240075391, 14990032513, 76085065735, 268985219353, 1365291107839, 4826743915841, 24499154875367, 86612405265785, 439619496648767, 1554196550868289

Offset: 1

John W. Layman, May 16 2003

The corresponding sequence for which 5n^2+4 is a square is A001906 (a bisection of the Fibonacci sequence).

			25 is a term of the sequence since 5*25^2 + 11 = 56^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,18,0,-1).

Cf. A001906, A080806.

LinearRecurrence[{0, 18, 0, -1}, {1, 7, 25, 127}, 50] (* Paolo Xausa, Mar 18 2024 *)

Vec(x*(1+7*x+7*x^2+x^3)/(1-18*x^2+x^4) + O(x^100)) \\ Colin Barker, Nov 06 2014

a(n) = 18*a(n-2) - a(n-4).

G.f.: x*(1+7*x+7*x^2+x^3)/(1-18*x^2+x^4). - Colin Barker, Jun 14 2012

More terms from Colin Barker, Nov 06 2014

A256545 Composite numbers k such that k*phi(k) is in A002378.

Original entry on oeis.org

6, 30, 434, 510, 616, 912, 1640, 2989, 3003, 5934, 7280, 8600, 10726, 12700, 13825, 14288, 18699, 19389, 54153, 59394, 59906, 70563, 72816, 116052, 117964, 121954, 131070, 134212, 140752, 177000, 206514, 210728, 274023, 319522, 418610, 437736, 456666

Offset: 1

Robert Israel, Apr 01 2015

nonn

Composite k such that 4*A002618(k)+1 is a square.
For all primes p, 4*A002618(p) + 1 = (2*p-1)^2.
The only semiprime < 10^7 in the sequence is 6.
k = 2*p with p prime is in the sequence if 2*p-1 is in A001653. However, the only such p < 10^3000 is 3.
Similarly, k = 3*p with p prime is in the sequence if 2*p-1 is in A080806. However, the only such p < 10^3000 is 2.

			a(1) = 6 is in the sequence because 6*phi(6) = 12 = 4*3.

Amiram Eldar, Table of n, a(n) for n = 1..258 (terms below 10^11; terms 1..62 from Robert Israel)

Cf. A000010, A001653, A002378, A002618, A080806.

select(n -> not isprime(n) and issqr(1+4*n*numtheory:-phi(n)), [$1..10^6]);

Select[Range[10^6],!PrimeQ[#]&&IntegerQ[Sqrt[4*#*EulerPhi[#]+1]]&] (* Ivan N. Ianakiev, Apr 02 2015 *)

lista(nn) = {forcomposite (n=1, nn, if (ispolygonal(n*eulerphi(n)/2, 3), print1(n ", ")););} \\ Michel Marcus, Apr 02 2015

A216073 The list of the a(n)-values in the common solutions to k+1=b^2 and 6*k+1=a^2.

Original entry on oeis.org

1, 7, 17, 71, 169, 703, 1673, 6959, 16561, 68887, 163937, 681911, 1622809, 6750223, 16064153, 66820319, 159018721, 661452967, 1574123057, 6547709351, 15582211849, 64815640543, 154247995433, 641608696079, 1526897742481, 6351271320247, 15114729429377, 62871104506391

Offset: 1

Paul Weisenhorn, Sep 01 2012

The equations are equivalent to the Pell equation a^2 - 6*b^2 = -5 with the 2 fundamental solutions (1;1) and (7;3) and the solution (5;2) for the unit form.
The associated b(n) are in A080806.
A181442(n) = (A080806(n) + 1)/2.
A180483(n) = (a(n) + 5)/2.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Cf. A080806, A180483, A181442.

a(1)=1: a(2)=7: a(3)=17: a(4)=71:
for n from 5 to 20 do
  a(n)=10*a(n-2)-a(n-4):
  printf("%9d%20d\n",n,a(n)):
end do:

LinearRecurrence[{0,10,0,-1}, {1,7,17,71}, 50] (* G. C. Greubel, Feb 22 2017 *)

a(n) = if(n<1, 0, if(n<5, [1,7,17, 71][n], 10*a(n-2)-a(n-4) ) );
/* Joerg Arndt, Sep 03 2012 */

def b(n): return (1/2)*(1+(-1)^n)*chebyshev_U(n//2, 5)
def A216073(n): return b(n) +7*b(n-1) +7*b(n-2) +b(n-3)
[A216073(n) for n in (0..50)] # G. C. Greubel, Apr 28 2022

a(n) = 10*a(n-2) - a(n-4).
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+7*x+7*x^2+x^3)/(1-10*x^2+x^4).
a(2*n+1) = ((1+r)*(5+2*r)^n + (1-r)*(5-2*r)^n)/2  where r=sqrt(6) and 0<=n.
a(2*n+2) = ((7+3*r)*(5+2*r)^n + (7-3*r)*(5-2*r)^n)/2  where r=sqrt(6) and 0<=n.
a(n) = -((5-2*r)^(1/4)*((2*r+5)^((-1)^n/4+n/2)*(-1)^n - r*(2*r+5)^((-1)^n/4+n/2)) + (2*r+5)^(1/4)*((5-2*r)^((-1)^n/4+n/2)*(-1)^n + (5-2*r)^((-1)^n/4+n/2)*r))/(2*(5-2*r)^(1/4)*(2*r+5)^(1/4)) with r=sqrt(6) and 1<=n. - Alexander R. Povolotsky, Sep 01 2012
a(n) = b(n) +7*b(n-1) +7*b(n-2) +b(n-3), where b(n) = (1/2)*(1 +(-1)^n)* ChebyshevU(n/2, 5). - G. C. Greubel, Apr 28 2022

cp's OEIS Frontend

A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

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A077291 Second member of Diophantine pair (m,k) that satisfies 6*(m^2 + m) = k^2 + k: a(n) = k.

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A237610 Positive integers k such that x^2 - 10xy + y^2 + k = 0 has integer solutions.

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A080811 a(1) = 8, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

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A080809 a(1) = 6; thereafter, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

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A080810 a(1) = 7, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

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A080807 a(1) = 3, a(n)= smallest n-th power obtained by inserting digits anywhere in a(n-1).

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Extensions

A082651 Positive integer values of n such that 5n^2+11 is a square.

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