A214838 -id:A214838

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A077241

Combined Diophantine Chebyshev sequences A054488 and A077413.

+10
14

1, 2, 8, 13, 47, 76, 274, 443, 1597, 2582, 9308, 15049, 54251, 87712, 316198, 511223, 1842937, 2979626, 10741424, 17366533, 62605607, 101219572, 364892218, 589950899, 2126747701, 3438485822, 12395593988, 20040964033, 72246816227, 116807298376

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

-8*a(n)^2 + b(n)^2 = 17, with the companion sequence b(n)= A077242(n).

The number a > 0 belongs to the sequence A077241, if a^2 belongs to the sequence A034856. - Alzhekeyev Ascar M, Apr 27 2012

Numbers k such that k^2 + 2 is a triangular number (see A214838). - Alex Ratushnyak, Mar 07 2013

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Index entries for sequences related to Chebyshev polynomials.

Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

FORMULA

a(2k) = A054488(k) and a(2k+1)= A077413(k) for k>=0.

G.f.: (1+x)*(1+x+x^2)/(1-6*x^2+x^4).

a(n) = (-1)^n*((4-5*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor((n+1)/2))+(4+5*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor((n+1)/2)))/8. [Bruno Berselli, Mar 10 2013]

EXAMPLE

8*a(2)^2 + 17 = 8*8^2+17 = 529 = 23^2 = A077242(2)^2.

MATHEMATICA

LinearRecurrence[{0, 6, 0, -1}, {1, 2, 8, 13}, 30] (* Bruno Berselli, Mar 10 2013 *)

CoefficientList[Series[(1 + x) (1 + x + x^2)/(1 - 6 x^2 + x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 18 2014 *)

PROG

(Maxima) makelist(expand((-1)^n*((4-5*sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor((n+1)/2))+(4+5*sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor((n+1)/2)))/8), n, 0, 30); /* Bruno Berselli, Mar 10 2013 */

(Magma) I:=[1, 2, 8, 13]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Feb 18 2014

KEYWORD

nonn,easy

AUTHOR

Wolfdieter Lang, Nov 08 2002

STATUS

approved

A328791

Triangular numbers of the form k^2 + 3.

+10
2

3, 28, 903, 30628, 1040403, 35343028, 1200622503, 40785822028, 1385517326403, 47066803275628, 1598885794044903, 54315050194251028, 1845112820810490003, 62679520857362409028, 2129258596329511416903, 72332112754346025765628, 2457162575051435364614403

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

There exist triangular numbers of the form k^2 + j for j=0 (A001110), j=1 (A164055), j=2 (A214838), and j=3 (this sequence), but not for j=4,7,8,13,16,18,... (A328792).

LINKS

Table of n, a(n) for n=1..17.

FORMULA

a(1) = 3, a(2) = 28; for n > 2, a(n) = 34*a(n-1) - a(n-2) - 46.

PROG

(Python)

limit = 10**7 # rough limit for k

A000217 = set(k*(k+1)//2 for k in range(14*limit//10))

A117950 = set(k**2 + 3 for k in range(limit))

print(sorted(A000217 & A117950)) # Michael S. Branicky, Mar 28 2021

CROSSREFS

Cf. A001110, A164055, A214838, A328792.

Intersection of A000217 and A117950.

Cf. A276598 (the k's).

KEYWORD

nonn

AUTHOR

Jon E. Schoenfield, Oct 27 2019

STATUS

approved

A328792

Numbers that are not the difference between any triangular number and the largest square that does not exceed it.

+10
2

4, 7, 8, 13, 16, 18, 22, 23, 25, 26, 31, 33, 34, 37, 38, 40, 43, 47, 48, 49, 52, 58, 59, 60, 61, 63, 64, 67, 68, 70, 73, 76, 79, 81, 83, 85, 86, 88, 92, 93, 94, 97, 98, 99, 102, 103, 106, 108, 112, 113, 114, 115, 118, 121, 123, 124, 125, 130, 133, 134, 138

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

LINKS

Table of n, a(n) for n=1..61.

EXAMPLE

For any triangular number t, let f(t) = t - floor(sqrt(t))^2.

0 is not a term: for each term t in A001110, f(t) = 0.

1 is not a term: for each term t > 1 in A164055, f(t) = 1.

2 is not a term: for each term t in A214838, f(t) = 2.

3 is not a term: for each term t > 3 in A328791, f(t) = 3.

4 is a term, however: there exists no triangular number t such that f(t) = 4.

CROSSREFS

The complement of A230044.

Cf. A001110, A064784, A164055, A214838, A328791.

KEYWORD

nonn

AUTHOR

Jon E. Schoenfield, Oct 27 2019

STATUS

approved

A305292

Numbers k such that k-1 is a square and k+1 is a triangular number.

+10
1

2, 5, 65, 170, 2210, 5777, 75077, 196250, 2550410, 6666725, 86638865, 226472402, 2943171002, 7693394945, 99981175205, 261348955730, 3396416785970, 8878171099877, 115378189547777, 301596468440090, 3919462027838450, 10245401755863185, 133146330756959525

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

It is easy to prove that there are no numbers k such that k-1 is a triangular number and k+1 is a square.

LINKS

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

FORMULA

G.f.: x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((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) = 35*a(n-2) - 35*a(n-4) + a(n-6) = 34*a(n-2) - a(n-4) + 2.

a(n) = (-2 + (13*sqrt(2) + 7*(-1)^n)*(1 + sqrt(2))^(2*n+1) - (13*sqrt(2) - 7* (-1)^n)*(1 - sqrt(2))^(2*n+1))/32.

a(n) = A214838(n) - 1.

a(n) = A077241(n-1)^2 + 1.

MATHEMATICA

LinearRecurrence[{1, 34, -34, -1, 1}, {2, 5, 65, 170, 2210}, 25]

PROG

(PARI) Vec(x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30)) \\ Colin Barker, Jun 14 2018

CROSSREFS

Cf. A000045, A000290, A077241, A214838.

KEYWORD

nonn,easy

AUTHOR

Bruno Berselli, Jun 11 2018

STATUS

approved

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