A096952 -id:A096952 - cp's OEIS frontend

A096951 Sum of odd powers of 2 and of 3 divided by 5.

1, 7, 55, 463, 4039, 35839, 320503, 2876335, 25854247, 232557151, 2092490071, 18830313487, 169464432775, 1525146340543, 13726182847159, 123535108753519, 1111813831298023, 10006315891747615, 90056808665990167, 810511140554958031, 7294599715238808391

Offset: 0

Wolfdieter Lang, Jul 16 2004

Sequence appears in A096952 (upper bounds for Lagrange remainder in Taylor expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2)).
Divisibility of 2^(2*n+1) + 3^(2*n+1) by 5 is proved by induction.
The sequence a(n+1), with g.f. (7-36*x)/(1-13*x+36*x^2) and formula (27*9^n + 8*4^n)/5, is the Hankel transform of C(n) + 6*C(n+1), where C(n) is A000108(n). - Paul Barry, Dec 06 2006

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,-36).

Cf. A074614 (sum of even powers of 2 and of 3), A007689 (sum of powers of 2 and powers of 3).

Cf. A000108, A015441, A096952, A138233.

[(2^(2*n+1) + 3^(2*n+1))/5: n in [0..30]]; // Vincenzo Librandi, May 31 2011

LinearRecurrence[{13, -36},{1, 7},19] (* Ray Chandler, Jul 14 2017 *)

a(n) = (2^(2*n+1) + 3^(2*n+1))/5.
G.f.: (1-6*x)/((1-4*x)*(1-9*x)).
From Reinhard Zumkeller, Mar 07 2008: (Start)
a(n+1) = 4*a(n) + 3^(2*n+1), a(0) = 1.
a(n) = A138233(n)/5. (End)
From Elmo R. Oliveira, Aug 02 2025: (Start)
E.g.f.: exp(4*x)*(2 + 3*exp(5*x))/5.
a(n) = 13*a(n-1) - 36*a(n-2).
a(n) = A015441(2*n+1). (End)

A096953 Denominators of upper bounds for Lagrange-remainder in Taylor's expansion of log((1+x)/(1-x)) multiplied by 6/5.

Original entry on oeis.org

1, 108, 1296, 326592, 15116544, 665127936, 28298170368, 235092492288, 47958868426752, 1929639176699904, 10968475320188928, 3027299188372144128, 4738381338321616896, 4605706660848611622912, 178087324219479649419264, 6853291511342734094893056

Offset: 0

Wolfdieter Lang, Jul 16 2004

An upper bound for the Lagrange-remainder in the expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2), is R(2*n):=(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1).

			n=4: R(2*4)=(5/6)* A096952(4)/a(4) = (5/6)*4039/15116544 = 20195/90699264 = 0.0002226589..., therefore log(2)-2*sum(((1/3)^(2*k-1))/(2*k-1),k=1..4) < 0.0002226589... In fact, the partial sum is 0.0000124233...

M. Barner and F. Flohr, Analysis I, de Gruyter, 5te Auflage, 2000; p. 293.

W. Lang, More comments.

Numerators are given in A096952.

[Denominator((6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1)): n in [0..20]]; // Vincenzo Librandi, Jul 06 2015

vector(30, n, n--; denominator((6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1))) \\ Michel Marcus, Jul 06 2015

a(n)=denominator(A(n)), where A(n):=(6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1) = A096951(n)/((2*n+1)*6^(2*n)).

cp's OEIS Frontend

A096951 Sum of odd powers of 2 and of 3 divided by 5.

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