A106525 -id:A106525 - cp's OEIS frontend

A276600 Values of m such that m^2 + 6 is a triangular number (A000217).

0, 2, 3, 7, 15, 20, 42, 88, 117, 245, 513, 682, 1428, 2990, 3975, 8323, 17427, 23168, 48510, 101572, 135033, 282737, 592005, 787030, 1647912, 3450458, 4587147, 9604735, 20110743, 26735852, 55980498, 117214000, 155827965, 326278253, 683173257, 908231938

Offset: 1

Colin Barker, Sep 07 2016

2*a(n+2) gives the y members of all positive solutions (x(n), y(n)), proper and improper, of the Pell equation x^2 - 2*y^2 = 7^2, n >= 0. The corresponding x members are x(n) = A106525(n). - Wolfdieter Lang, Sep 29 2016

			7 is in the sequence because 7^2 + 6 = 55, which is a triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Cf. A000217, A230044.
Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.
Cf. A275794, A275796, A106525.

I:=[0,2,3,7,15,20]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..41]]; // G. C. Greubel, Sep 15 2021

LinearRecurrence[{0,0,6,0,0,-1}, {0,2,3,7,15,20}, 41] (* G. C. Greubel, Sep 15 2021 *)

concat(0, Vec(x^2*(2+3*x+7*x^2+3*x^3+2*x^4)/(1-6*x^3+x^6) + O(x^40)))

def A276600_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x^2*(2+3*x+7*x^2+3*x^3+2*x^4)/(1-6*x^3+x^6) ).list()
a=A276600_list(41); a[1:] # G. C. Greubel, Sep 15 2021

a(n) = 6*a(n-3) - a(n-6) for n>6.
G.f.: x^2*(2 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/(1 - 6*x^3 + x^6).
From Wolfdieter Lang, Sep 29 2016: (Start)
Trisection:
a(2+3*n) = 15*S(n-1,6) - 2*S(n-2,6) = A275794(n),
a(3+3*n) = 20*S(n-1,6) - 3*S(n-2,6) = A275796(n),
a(4+3*n) = 7*(6*S(n-1,6) - S(n-2,6)) = 7*A001109(n+1) for n >= 0, with the Chebyshev polynomials S(n, 6) = A001109(n+1), n >= -1, with S(-2, 6) = -1.
(End)

A275793 The x members of the positive proper solutions (x = x1(n), y = y1(n)) of the first class for the Pell equation x^2 - 2*y^2 = +7^2.

Original entry on oeis.org

9, 43, 249, 1451, 8457, 49291, 287289, 1674443, 9759369, 56881771, 331531257, 1932305771, 11262303369, 65641514443, 382586783289, 2229879185291, 12996688328457, 75750250785451, 441504816384249, 2573278647520043, 14998167068736009, 87415723764896011

Offset: 0

Wolfdieter Lang, Sep 27 2016

This gives the (increasingly sorted) positive x members of the first class of the proper solutions (x1(n), y1(n)) to the Pell equation x^2 - 2*y^2 = +7^2. For the y1(n) solutions see 2*A275794(n). The solutions for the second class (x2(n), y2(n)) are found in A275795(n) and 2*A275796(n).
All solutions, including the improper ones, are given in A106525(n) and 2*A276600(n+2).
See also the comments on A263012 which apply here mutatis mutandis.
This is for the Pell equations x^2 - 2*y^2 = z^2, besides z^2 = 1 the first instance with proper solutions. For z^2 > 1 there seem to be always two classes of such solutions. For z^2 = 1 there is only one class of proper solutions. These z^2 values seem to appear for z from A058529 (prime factors are +1 or -1 (mod 8)).

			The first positive proper fundamental solution (x = x1(n), y = y1(n)) of x^2 - 2*y^2  = 49 are [9, 4], [43, 30], [249, 176], [1451, 1026], [8457, 5980], [49291, 34854], [287289, 203144], [1674443, 1184010], ...
The first positive proper fundamental solution of the second class (x = x2(n), y = y2(n)) are [11, 6], [57, 40], [331, 234], [1929, 1364], [11243, 7950], [65529, 46336], [381931, 270066], [2226057, 1574060], ...

T. Nagell, Introduction to Number Theory, Wiley, 1951, Theorem 109, pp. 207-208.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A001109, A058529, A106525.

Cf. A263012, A275794, A275795, A275796, 2*A276600.

I:=[9,43]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2): n in [1..31]]; // G. C. Greubel, Sep 15 2021

RecurrenceTable[{a[n]== 6a[n-1] -a[n-2], a[-1]==11, a[0]==9}, a, {n,0,25}] (* Michael De Vlieger, Sep 28 2016 *)
Table[9*Fibonacci[2*n+1, 2] - Fibonacci[2*n, 2], {n,0,30}] (* G. C. Greubel, Sep 15 2021 *)

a(n) = round((((3-2*sqrt(2))^n*(-8+9*sqrt(2))+(3+2*sqrt(2))^n*(8+9*sqrt(2)))) / (2*sqrt(2))) \\ Colin Barker, Sep 28 2016

Vec((9-11*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Oct 02 2016

def P(n): return lucas_number1(n, 2, -1);
[9*P(2*n+1) - P(2*n) for n in (0..30)] # G. C. Greubel, Sep 15 2021

a(n) = 43*S(n-1, 6) - 9*S(n-2, 6), with the Chebyshev polynomials S(n, 6) = A001109(n+1), n >= -1, with S(-2, 6) = -1.
O.g.f: (9 - 11*x)/(1 - 6*x + x^2).
a(n) = 6*a(n-1) - a(n-2) for n >= 1, with a(-1) = 11 and a(0) = 9.
a(n) = (((3-2*sqrt(2))^n*(-8+9*sqrt(2))+(3+2*sqrt(2))^n*(8+9*sqrt(2)))) / (2*sqrt(2)). - Colin Barker, Sep 28 2016
a(n) = 9*A000129(2*n+1) - A000129(2*n). - G. C. Greubel, Sep 15 2021

A106526 Values of y in x^2 - 49 = 2*y^2.

Original entry on oeis.org

4, 6, 14, 30, 40, 84, 176, 234, 490, 1026, 1364, 2856, 5980, 7950, 16646, 34854, 46336, 97020, 203144, 270066, 565474, 1184010, 1574060, 3295824, 6900916, 9174294, 19209470, 40221486, 53471704, 111960996, 234428000, 311655930, 652556506

Offset: 1

Andras Erszegi (erszegi.andras(AT)chello.hu), May 07 2005

nonn

The expression 2*n^2 + c with c = 49 yields more squares than any other value of c in the range 1 < c < 100 and n < 5*10^4. - K. D. Bajpai, Nov 04 2013

			a(12) = 2856; as 12 mod 3 = 0, a(12) = 14*A001109(12/3) = 204*14 = 2856; also 2*2856^2 = 4039^2 - 49, i.e., A106525(12)^2 - 49;
a(13) = 5980; as 13 mod 3 = 1, a(13) = A001109(4+2) - A001109(4+1) + A001109(4) + A001109(4-1) = 6930 - 1189 + 204 + 35 = 5980; also 2*5980^2 = 8457^2 - 49, i.e., A106525(13)^2 - 49;
a(14) = 7950; as 14 mod 3 = 2, a(14) = A001109(4+2) + A001109(4+1) - A001109(4) + A001109(4-1) = 6930 + 1189 - 204 + 35 = 7950; also 2*7950^2 = 11243^2 - 49, i.e., A106525(14)^2 - 49.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).

Cf. A001109, A106525, A276600.

I:=[4,6,14,30,40,84]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..41]]; // G. C. Greubel, Aug 12 2021

LinearRecurrence[{0,0,6,0,0,-1}, {4,6,14,30,40,84}, 40] (* T. D. Noe, Nov 04 2013 *)

def A106526_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (2*x)*(2 +3*x +7*x^2 +3*x^3 +2*x^4)/(1 -6*x^3 +x^6) ).list()
a=A106526_list(41); a[1:] # G. C. Greubel, Aug 12 2021

a(n) = 6*a(n-3) - a(n-6), with initial terms 4, 6, 14, 30, 40, 84. - T. D. Noe, Nov 04 2013
From G. C. Greubel, Aug 12 2021: (Start)
a(n) = 2*A276600(n+1).
G.f.: (2*x)*(2 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/(1 - 6*x^3 + x^6). (End)

A333641 11-gonal (or hendecagonal) square numbers.

Original entry on oeis.org

0, 1, 196, 29241, 1755625, 261468900, 38941102225, 2337990844401, 348201795147556, 51858411008887561, 3113535139359330841, 463705205422871375236, 69060571958250748760481, 4146338334574433921200225, 617522713934165528806340100, 91968930524758079223806760025

Offset: 1

Bernard Schott, Mar 31 2020

The 11-gonal square numbers correspond to the nonnegative integer solutions of the Diophantine equation k*(9*k-7)/2 = m^2, equivalent to (18*k-7)^2 - 72*m^2 = 49. Substituting x = 18*k-7 and y = m gives the Pell equation x^2-72*y^2 = 49. The integer solutions (x,y) = (-7,0), (11,1), (119,14), (1451,171), (11243,1325), ... correspond to the following solutions (k,m) = (0,0), (1,1), (7,14), (81,171), (625,1325), ...

			1755625 is a term because 625*(9*625-7)/2 = 1325^2 = 1755625; that means that 1755625 is the 625th 11-gonal number and the square of 1325.

Index entries for linear recurrences with constant coefficients, signature (1,0,1331714,-1331714,0,-1,1).

Intersection of A000290 (squares) and A051682 (11-gonals).
Cf. A106525.
Cf. A001110 (square triangulars), A036353 (square pentagonals), A046177 (square hexagonals), A036354 (square heptagonals), A036428 (square octagonals), A036411 (square 9-gonals), A188896 (only {0,1} are square 10-gonals), this sequence (square 11-gonals), A342709 (square 12-gonals).

for k from 0 to 8000000 do
d:= k*(9*k-7)/2;
if issqr(d) then print(k,sqrt(d),d); else fi; od:

Last /@ Solve[(18*x - 7)^2 - 72*y^2 == 49 && x >= 0 && y >= 0 && y < 10^16, {x, y}, Integers] /. Rule -> (#2^2 &) (* Amiram Eldar, Mar 31 2020 *)

concat(0, Vec(-x*(1 + 195*x + 29045*x^2 + 394670*x^3 + 29045*x^4 + 195*x^5 + x^6)/(-1 + x + 1331714*x^3 - 1331714*x^4 - x^6 + x^7) + O(x^20))) \\ Jinyuan Wang, Mar 31 2020

a(n) = k*(9*k-7)/2 for n > 1, where k = (A106525(4*n-6) + 7)/18. - Jinyuan Wang, Mar 31 2020

More terms from Amiram Eldar, Mar 31 2020

A359438 For n >= 0, let S be the sequence of numbers m such that (m^2 - 2n^2 + 1)/2 is a square. Then a(n) is the number k such that S(j) = 6S(j-k) - S(j-2k) for all j for which S(j-2k) is defined.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 4, 2, 4, 2, 4, 2, 2, 4, 4, 2, 2, 4, 2, 2, 4, 4, 2, 3, 4, 4, 4, 4, 2, 4, 2, 8, 2, 2, 4, 2, 2, 2, 6, 2, 2, 4, 4, 2, 2, 4, 2, 4, 8, 4, 2, 4, 6, 2, 4, 4, 2, 2, 2, 8, 4, 4, 4, 2, 4, 4, 4, 2, 4, 4, 2, 4, 4, 4, 2, 2, 8, 4, 4, 2, 4

Offset: 0

Jon E. Schoenfield, Dec 31 2022

nonn

			For n = 0, {S(j)} = A002315 (the NSW numbers), which satisfies S(j) = 6*S(j-1) - S(j-2), so a(0) = 1.
For n = 1, {S(j)} = A001541, which also satisfies S(j) = 6*S(j-1) - S(j-2), so a(1) = 1.
For n = 2, {S(j)} = A077443, which satisfies S(j) = 6*S(j-2) - S(j-4), so a(2) = 2.
For n = 5, {S(j)} = A106525, which satisfies S(j) = 6*S(j-3) - S(j-6), so a(5) = 3.

Cf. A000005, A002315, A001541, A077443, A077242, A106525.

a(0) = 1; for n >= 1, a(n) = A000005(2*n^2 - 1).

cp's OEIS Frontend

A276600 Values of m such that m^2 + 6 is a triangular number (A000217).

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A275793 The x members of the positive proper solutions (x = x1(n), y = y1(n)) of the first class for the Pell equation x^2 - 2*y^2 = +7^2.

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A106526 Values of y in x^2 - 49 = 2*y^2.

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A333641 11-gonal (or hendecagonal) square numbers.

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A359438 For n >= 0, let S be the sequence of numbers m such that (m^2 - 2*n^2 + 1)/2 is a square. Then a(n) is the number k such that S(j) = 6*S(j-k) - S(j-2k) for all j for which S(j-2k) is defined.

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A359438 For n >= 0, let S be the sequence of numbers m such that (m^2 - 2n^2 + 1)/2 is a square. Then a(n) is the number k such that S(j) = 6S(j-k) - S(j-2k) for all j for which S(j-2k) is defined.