A007610 -id:A007610 - cp's OEIS frontend

A090251 a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29.

2, 29, 839, 24302, 703919, 20389349, 590587202, 17106639509, 495501958559, 14352450158702, 415725552643799, 12041688576511469, 348793243166188802, 10102962363242963789, 292637115290879761079, 8476373381072270107502

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004

a(n+1)/a(n) converges to ((29+sqrt(837))/2) =28.9654761... Lim a(n)/a(n+1) as n approaches infinity = 0.0345238... =2/(29+sqrt(837)) =(29-sqrt(837))/2. Lim a(n+1)/a(n) as n approaches infinity = 28.9654761... = (29+sqrt(837))/2=2/(29-sqrt(837)). Lim a(n)/a(n+1) = 29 - Lim a(n+1)/a(n).
A Chebyshev T-sequence with a Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 93*(3*b)^2 =+4 with companion sequence b(n)=A097782(n+1), n>=0.

			a(4) =703919 = 29a(3) - a(2) = 29*24302 - 839= ((29+sqrt(837))/2)^4 + ((29-sqrt(837))/2)^4 = 703918.99999857 + 0.00000142 =703919.
(x,y) = (2;0), (29;1), (839;29), (24302,840), ..., give the
nonnegative integer solutions to x^2 - 93*(3*y)^2 =+4.

O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (29, -1).

Cf. A083148, A007610.
a(n)=sqrt(4 + 93*(3*A097782(n-1))^2), n>=1.
Cf. A077428, A078355 (Pell +4 equations).
Cf. A090248 for 2*T(n, 27/2).

a[0] = 2; a[1] = 29; a[n_] := 29a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{29,-1},{2,29},30] (* Harvey P. Dale, May 28 2013 *)

[lucas_number2(n,29,1) for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008

a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29. a(n) = ((29+sqrt(837))/2)^n + ((29-sqrt(837))/2)^n, (a(n))^2 =a(2n)+2.
a(n) = S(n, 29) - S(n-2, 29) = 2*T(n, 29/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 27)=A097781(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (29+3*sqrt(93))/2 and am := (29-3*sqrt(93))/2.
G.f.: (2-29*x)/(1-29*x+x^2).

More terms from Robert G. Wilson v, Jan 30 2004

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A068066 The sum or half the sum of n consecutive primes starting at a(n) is prime.

Original entry on oeis.org

2, 2, 5, 2, 5, 2, 17, 83, 3, 3, 5, 2, 29, 2, 3, 11, 3, 3, 11, 23, 7, 5, 7, 11, 5, 3, 7, 3, 13, 3, 13, 5, 7, 17, 5, 3, 5, 73, 13, 5, 7, 5, 7, 5, 7, 29, 7, 53, 11, 29, 17, 31, 3, 23, 3, 47, 97, 5, 29, 2, 3, 37, 13, 2, 3, 17, 19, 5, 19, 71, 3, 47, 5, 19, 3, 59, 23, 89, 7, 19, 11, 37, 53, 3

Offset: 1

Robert G. Wilson v, Feb 16 2002

nonn

This eliminates the impossibles out of A007610 and the 'or' in the title is the exclusive or.

			a(3) = 5 because 5+7+11 = prime 23 and a(10) = 3 because 3+5+7+11+13+17+19+23+29+31 = 158 and half that or 79 is a prime.

Cf. A007610.

Do[k = n; a = Table[Prime[i], {i, 1, n} ]; While[ !PrimeQ[Plus @@ a] && !PrimeQ[Plus @@ a/2], k++; a = Drop[a, 1]; a = Append[a, Prime[k]]]; Print[a[[1]]], {n, 1, 100} ]

cp's OEIS Frontend

A090251 a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29.

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