cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-17 of 17 results.

A341895 Indices of triangular numbers that are ten times other triangular numbers.

Original entry on oeis.org

0, 4, 20, 39, 175, 779, 1500, 6664, 29600, 56979, 253075, 1124039, 2163720, 9610204, 42683900, 82164399, 364934695, 1620864179, 3120083460, 13857908224, 61550154920, 118481007099, 526235577835, 2337285022799, 4499158186320, 19983094049524, 88755280711460, 170849530073079, 758831338304095, 3370363382012699
Offset: 1

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Author

Vladimir Pletser, Feb 23 2021

Keywords

Comments

Second member of the Diophantine pair (b(n), a(n)) that satisfies a(n)^2 + a(n) = 10*(b(n)^2 + b(n)) or T(a(n)) = 10*T(b(n)) where T(x) is the triangular number of x. The T(b)'s are in A068085 and the b's are in A341893.
Can be defined for negative n by setting a(-n) = -a(n+1) - 1 for all n in Z.

Examples

			a(2) = 4 is a term because its triangular number, T(a(2)) = 4*5 / 2 = 10 is ten times a triangular number.
a(4) = 38*a(1) - a(-2) + 18 = 38*0 - (-21) + 18 = 39, etc.
		

Crossrefs

Programs

  • Maple
    f := gfun:-rectoproc({a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)]) ; #
  • Mathematica
    Rest@ CoefficientList[Series[x^2*(4 + 16*x + 19*x^2 - 16*x^3 - 4*x^4 - x^5)/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 30}], x] (* Michael De Vlieger, May 19 2022 *)

Formula

a(n) = 38*a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = -21, a(-1) = -5, a(0) = -1, a(1) = 0, a(2) = 4, a(3) = 20.
G.f.: x^2*(4 + 16*x + 19*x^2 - 16*x^3 - 4*x^4 - x^5)/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7). - Stefano Spezia, Feb 24 2021
a(n) = (A198943(n) + 1)/2 - 1. - Hugo Pfoertner, Feb 26 2021

A222390 Nonnegative integers m such that 10*m*(m+1)+1 is a square.

Original entry on oeis.org

0, 3, 15, 132, 588, 5031, 22347, 191064, 848616, 7255419, 32225079, 275514876, 1223704404, 10462309887, 46468542291, 397292260848, 1764580902672, 15086643602355, 67007605759263, 572895164628660, 2544524437949340, 21754929612286743, 96624921036315675
Offset: 1

Views

Author

Bruno Berselli, Feb 18 2013

Keywords

Comments

a(n+1)/a(n) tends alternately to (7+2*sqrt(10))/3 and (13+4*sqrt(10))/3; a(n+2)/a(n) tends to A176398^2.
Subsequence of A014601.

Crossrefs

Cf. nonnegative integers m such that k*m*(m+1)+1 is a square: A001652 (k=2), A001921 (k=3), A001477 (k=4), A053606 (k=5), A105038 (k=6), A105040 (k=7), A053141 (k=8), this sequence (k=10), A105838 (k=11), A061278 (k=12), A104240 (k=13); A105063 (k=17), A222393 (k=18), A101180 (k=19), A077259 (k=20) [incomplete list].
Cf. A221875.

Programs

  • Magma
    m:=22; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(3*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2))));
    
  • Magma
    I:=[0,3,15,132,588]; [n le 5 select I[n] else Self(n-1) +38*Self(n-2)-38*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013
    
  • Mathematica
    LinearRecurrence[{1, 38, -38, -1, 1}, {0, 3, 15, 132, 588}, 23]
    CoefficientList[Series[3 x (1 + 4 x + x^2)/((1 - x) (1 - 6 x - x^2) (1 + 6 x - x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)
  • Maxima
    makelist(expand(-1/2+((5+(-1)^n*sqrt(10))*(3-sqrt(10))^(2*floor(n/2))+(5-(-1)^n*sqrt(10))*(3+sqrt(10))^(2*floor(n/2)))/20), n, 1, 23);
    
  • PARI
    x='x+O('x^30); concat([0], Vec(3*x*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2)))) \\ G. C. Greubel, Jul 15 2018

Formula

G.f.: 3*x*(1+4*x+x^2)/((1-x)*(1-6*x-x^2)*(1+6*x-x^2)).
a(n) = a(-n+1) = a(n-1)+38*a(n-2)-38*a(n-3)-a(n-4)+a(n-5).
a(n) = -1/2+((5+t*(-1)^n)*(3-t)^(2*floor(n/2))+(5-t*(-1)^n)*(3+t)^(2*floor(n/2)))/20, where t=sqrt(10).
2*a(n)+1 = A221875(n).

A341893 Indices of triangular numbers that are one-tenth of other triangular numbers.

Original entry on oeis.org

0, 1, 6, 12, 55, 246, 474, 2107, 9360, 18018, 80029, 355452, 684228, 3039013, 13497834, 25982664, 115402483, 512562258, 986657022, 4382255359, 19463867988, 37466984190, 166410301177, 739114421304, 1422758742216, 6319209189385, 28066884141582, 54027365220036, 239963538895471, 1065802482958830, 2051617119619170
Offset: 1

Views

Author

Vladimir Pletser, Feb 23 2021

Keywords

Comments

The indices of triangular numbers that are one-tenth of other triangular numbers [t of T(t) such that T(t)=T(u)/10].
First member of the Diophantine pair (t, u) that satisfies 10*(t^2 + t) = u^2 + u; a(n) = t.
The T(t)'s are in A068085 and the u's are in A341895.
Also, nonnegative t such that 40*t^2 + 40*t + 1 is a square.
Can be defined for negative n by setting a(n) = a(-1-n) for all n in Z.

Examples

			a(4) = 12 is a term because its triangular number, (12*13) / 2 = 78 is one-tenth of 780, the triangular number of 39.
a(4) = 38 a(1) - a(-2) +18 = 0 - 6 +18 = 12 ;
a(5) = 38 a(2) - a(-1) + 18 = 38*1 - 1 +18 = 55.
		

Crossrefs

Programs

  • Maple
    f := gfun:-rectoproc({a(-3) = 6, a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(2) = 6, a(n) = 38*a(n-3)-a(n-6)+18}, a(n), remember); map(f, [`$`(0 .. 1000)])[] ;
  • Mathematica
    Rest@ CoefficientList[Series[(x^2*(1 + 5*x + 6*x^2 + 5*x^3 + x^4))/(1 - x - 38*x^3 + 38*x^4 + x^6 - x^7), {x, 0, 31}], x] (* Michael De Vlieger, May 19 2022 *)

Formula

a(n) = (-1 + sqrt(8*b(n) + 1))/2 where b(n) = A068085(n).
a(n) = 38 a(n-3) - a(n-6) + 18 for n > 3, with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
a(n) = a(n-1) + 38*(a(n-3) - a(n-4)) - (a(n-6) - a(n-7)) for n >= 4 with a(-2) = 6, a(-1) = 1, a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 6.
G.f.: x^2*(1 + 4*x+x^2)*(1+x+x^2)/ ((1-x)*(1-38*x^3+x^6)). - Stefano Spezia, Feb 24 2021
a(n) = A180003(n) - 1. - Hugo Pfoertner, Feb 28 2021

A222393 Nonnegative integers m such that 18*m*(m+1)+1 is a square.

Original entry on oeis.org

0, 4, 12, 152, 424, 5180, 14420, 175984, 489872, 5978292, 16641244, 203085960, 565312440, 6898944364, 19203981732, 234361022432, 652370066464, 7961375818340, 22161378278060, 270452416801144, 752834491387592, 9187420795420572, 25574211328900084
Offset: 1

Views

Author

Bruno Berselli, Feb 19 2013

Keywords

Comments

a(n+2)/a(n) tends to A156164.
a(n) is congruent to {0,2,4} (mod 5, 6 and 10).

Crossrefs

Cf. nonnegative integers n such that k*n*(n+1)+1 is a square: A001652 (k=2), A001921 (k=3), A001477 (k=4), A053606 (k=5), A105038 (k=6), A105040 (k=7), A053141 (k=8), A222390 (k=10), A105838 (k=11), A061278 (k=12), A104240 (k=13); A105063 (k=17), this sequence (k=18), A101180 (k=19), A077259 (k=20) [incomplete list].

Programs

  • Magma
    m:=22; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(4*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2))));
    
  • Magma
    I:=[0,4,12,152,424]; [n le 5 select I[n] else Self(n-1)+34*Self(n-2)-34*Self(n-3)-Self(n-4)+Self(n-5): n in [1..25]]; // Vincenzo Librandi, Aug 18 2013
    
  • Mathematica
    LinearRecurrence[{1, 34, -34, -1, 1}, {0, 4, 12, 152, 424}, 23]
    CoefficientList[Series[4 x (1 + x)^2 / ((1 - x) (1 - 6 x + x^2) (1 + 6 x + x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 18 2013 *)
  • Maxima
    makelist(expand(-1/2+((3+sqrt(2)*(-1)^n)*(3-2*sqrt(2))^(2*floor(n/2))+(3-sqrt(2)*(-1)^n)*(3+2*sqrt(2))^(2*floor(n/2)))/12), n, 1, 23);
    
  • PARI
    x='x+O('x^30); concat([0], Vec(4*x*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)))) \\ G. C. Greubel, Jul 15 2018

Formula

G.f.: 4*x*(1+x)^2/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)).
a(n) = a(-n+1) = a(n-1)+34*a(n-2)-34*a(n-3)-a(n-4)+a(n-5).
a(n) = -1/2+((3+t*(-1)^n)*(3-2*t)^(2*floor(n/2))+(3-t*(-1)^n)*(3+2*t)^(2*floor(n/2)))/12, where t=sqrt(2).

A257939 x-values in the solutions to x^2 + x = 5*y^2 + y.

Original entry on oeis.org

0, 2, 116, 798, 37512, 257114, 12078908, 82790070, 3889371024, 26658145586, 1252365390980, 8583840088782, 403257766524696, 2763969850442378, 129847748455561292, 889989708002357094, 41810571744924211488, 286573922006908542050, 13462874254117140538004
Offset: 1

Views

Author

Arkadiusz Wesolowski, May 13 2015

Keywords

Crossrefs

Subsequence of A077259.
Cf. A257940.

Programs

  • Magma
    I:=[0, 2, 116, 798, 37512]; [n le 5 select I[n] else Self(n-1)+322*Self(n-2)-322*Self(n-3)-Self(n-4)+Self(n-5): n in [1..19]];
    
  • Mathematica
    LinearRecurrence[{1, 322, -322, -1, 1}, {0, 2, 116, 798, 37512}, 30] (* Vincenzo Librandi, May 15 2015 *)
  • PARI
    concat(0, Vec(-2*(3*x^3+19*x^2+57*x+1)/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)) + O(x^100))) \\ Colin Barker, May 14 2015

Formula

a(1) = 0, a(2) = 2, a(3) = 116, a(4) = 798, a(5) = 37512; for n > 5, a(n) = a(n-1) + 322*a(n-2) - 322*a(n-3) - a(n-4) + a(n-5).
a(n) = 322*a(n-2) - a(n-4) + 160.
a(n) = 161*a(n-2) + 360*A257940(n-2) + 116.
G.f.: -2*x^2*(3*x^3+19*x^2+57*x+1) / ((x-1)*(x^2-18*x+1)*(x^2+18*x+1)). - Colin Barker, May 14 2015

A341894 For square n > 0, a(n) = 0; for nonsquare n > 0, a(n) is the rank r such that t(r) + t(r-1) = u(r) - u(r-1) - 1, where u(r) and t(r) are indices of some triangular numbers in the Diophantine relation T(u(r)) = n*T(t(r)).

Original entry on oeis.org

0, 1, 1, 0, 2, 2, 2, 2, 0, 3, 2, 2, 4, 2, 2, 0, 2, 2, 3, 2, 4, 4, 2, 2, 0, 3, 2, 4, 4, 2, 4, 2, 2, 2, 2, 0, 2, 2, 2, 4, 4, 2, 4, 2, 4, 6, 2, 2, 0, 3, 3, 4, 4, 2, 4, 2, 4, 4, 2, 2, 8, 2, 2, 0, 2, 4, 4, 2, 4, 4, 4, 2, 6, 2, 2, 6, 4, 4, 2, 2, 0, 3, 2, 2, 8, 4, 2, 4, 4, 2, 6, 4, 4, 4, 2, 4, 4, 2, 2, 0, 2, 2, 4, 2, 4, 8, 2, 2, 8, 2
Offset: 1

Views

Author

Vladimir Pletser, Mar 06 2021

Keywords

Comments

Let t(i) and u(j) be the indices of triangular numbers that satisfy the Diophantine relation T(u(j)) = n*T(t(i)) for some integers i and j. The number of solutions (t(i), u(j)) of T(u(j)) = n*T(t(i)) is 0 or 1 for square n, and an infinity for nonsquare n.
For square n, a(n) is arbitrarily set to 0.
For nonsquare n, a(n) is the index r in the sequence of t(i) and u(j) such that t(r) + t(r-1) = u(r) - u(r-1) - 1.
Alternatively, for nonsquare n, a(n) is the index r such that the ratio t(i)/t(i-r) is decreasing monotonically without jumps for increasing values of i.
Alternatively, for n > 4, a(n) is the index r such that the ratio t(r)/t(r-1) varies between (s+1)/(s-1) and (s+2)/s, with s = [sqrt(n)], where [x] = floor(x).
Alternatively, for nonsquare n, a(n) is the number of fundamental solutions (X_f, Y_f) of the generalized Pell equation X^2 - n*Y^2 = 1 - n providing odd solutions, i.e., with X_f odd and Y_f odd (or Y_f even if y_f is odd, where y_f is the fundamental solution of the associated simple Pell equation x^2 - n*y^2 = 1).

Examples

			The following table gives the first values of nonsquare n and a(n) and the sequences yielding the values of t, u, T(t) and T(u) such that T(u) = n*T(t).
n       2       3       5       6       7       8      10
a(n)    1       1       2       2       2       2       3
t    A053141 A061278 A077259 A077288 A077398 A336623  A341893*
u    A001652 A001571 A077262 A077291 A077401 A336625* A341895*
T(t) A075528 A076139 A077260 A077289 A077399 A336624  A068085*
T(u) A029549 A076140 A077261 A077290 A077400 A336626*   -
With a(n) = r, the definition t(r) + t(r-1) = u(r) - u(r-1) - 1 yields:
- For n = 2, a(n) = 1: A053141(1) + A053141(0) = A001652(1) - A001652(0) - 1, i.e., 2 + 0 = 3 + 0 - 1 = 2.
- For n = 5, a(n) = 2: A077259(2) + A077259(1) = A077262(2) - A077262(1) - 1, i.e., 6 + 2 = 14 - 5 - 1 = 8.
- For n = 10, a(n) = 3: A341893(3+1*) + A341893(2+1*) = A341895(3+1*) - A341895(2+1*) - 1, i.e., 12 + 6 = 39 - 20 - 1 = 18.
Note that for those sequences marked with an *, the first term 0 appears for n = 1, contrary to all the other sequences, where the first term 0 appears for n = 0; the numbering must therefore be adapted and 1 must be added to compensate for this shift in indices.
The monotonic decrease of t(i)/t(i-r) can be seen also as:
- For n = 2, a(n) = 1: for 1 <= i <= 6, A053141(i)/A053141(i-1) decreases monotonically from 7 to 5.829.
- For n = 5, a(n) = 2: for 3 <= i <= 8, A077259(i)/A077259(i-2) decreases monotonically from 22 to 17.948, while A077259(i)/A077259(i-1) takes values alternatively varying between 3 and 2.618 and between 7.333 and 6.855.
- For n = 10, a(n) = 3: for 4 <= i <= 10, A341893(i)/A341893(i-3) decreases monotonically from 55 to 38, while A077259(i) / A077259(i-1) takes values alternatively varying between 6 and 4.44 and between 2 and 1.925.
For n > 4, the relation (s+1)/(s-1) <=  t(r)/t(r-1) <= (s+2)/s, with s = [sqrt(n)], yields:
- For n = 5, a(n) = 2: A077259(2)/A077259(1) = 6/2 = 3, and s = [sqrt(5)] = 2, (s+1)/(s-1) = 3 and (s+2)/s = 2.
- For n = 10, a(n) = 3: A077259(3+1*)/A077259(2+1*) = 12/6 = 2, and s = [sqrt(10)] = 3, (s+1)/(s-1) = 2 and (s+2)/s = 5/3 = 1.667.
Finally, the number of fundamental solutions of the generalized Pell equation is as follows.
- For n = 2, X^2 - 2*Y^2 = -1 has a single fundamental solution, (X_f, Y_f) = (1, 1), and the rank a(n) is 1.
- For n = 5, X^2 - 5*Y^2 = -4 has two fundamental solutions, (X_f, Y_f) = (1, 1) and (-1, 1), and the rank a(n) is 2.
- For n = 10, X^2 - 10*Y^2 = -9 has three fundamental solutions, (X_f, Y_f) = (1, 1), (-1, 1), and (9, 3), and the rank a(n) is 3.
		

References

  • J. S. Chahal and H. D'Souza, "Some remarks on triangular numbers", in A.D. Pollington and W. Mean, eds., Number Theory with an Emphasis on the Markov Spectrum, Lecture Notes in Pure Math, Dekker, New York, 1993, 61-67.

Crossrefs

A351354 Numbers k such that the k-th centered 40-gonal numbers (A195317) is a square.

Original entry on oeis.org

1, 3, 7, 45, 117, 799, 2091, 14329, 37513, 257115, 673135, 4613733, 12078909, 82790071, 216747219, 1485607537, 3889371025, 26658145587, 69791931223, 478361013021, 1252365390981, 8583840088783, 22472785106427, 154030760585065, 403257766524697, 2763969850442379
Offset: 1

Views

Author

Lamine Ngom, Feb 08 2022

Keywords

Comments

Corresponding square roots are listed in A351353.
3 and 7 are the unique primes in this sequence, a(2*n+1) and a(2*n) always sharing common factors that are closely linked to Fibonacci (A000045) and Lucas (A000032) numbers (detailed in formula section).
In addition, the ratio a(2*n+1)/a(2*n) converges to 2.618033988 ... = golden ratio squared: A104457.

Examples

			45 is in the sequence because the 45th centered 40-gonal number is 39601, which is a square: 199^2 = A000032(11)^2.
799 is in the sequence because the 799th centered 40-gonal number is 12752041, which is a square: 3571^2 = A000032(17)^2.
		

Crossrefs

Programs

  • Maple
    a[1] := 1: a[2] := 3: a[3] := 7: a[4] := 45: a[5] := 117:
    for n from 6 to 30 do a[n] := a[n - 1] + 18*a[n - 2] - 18*a[n - 3] - a[n - 4] + a[n - 5]: od:
    seq(a[n], n = 1 .. 30);
  • Mathematica
    LinearRecurrence[{1, 18, -18, -1, 1}, {1, 3, 7, 45, 117}, 30] (* Amiram Eldar, Feb 08 2022 *)

Formula

a(n) = A077259(n-1) + 1.
a(1)=1, a(2)=3, a(3)=7, a(4)=45, a(5)=117 and a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
gcd(a(2*n+1), a(2*n)) = A000045(n)*(A000032(2*n) - 1)/2, if n is odd.
gcd(a(2*n+1), a(2*n)) = A000032(n)*(A000032(2*n) - 1)/2, if n is even.
A195317(a(n)) = A000032(A007310(n))^2 = A351353(n)^2.
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