A157809 -id:A157809 - cp's OEIS frontend

A215149 a(n) = n * (1 + 2^(n-1)).

0, 2, 6, 15, 36, 85, 198, 455, 1032, 2313, 5130, 11275, 24588, 53261, 114702, 245775, 524304, 1114129, 2359314, 4980755, 10485780, 22020117, 46137366, 96469015, 201326616, 419430425, 872415258, 1811939355, 3758096412, 7784628253, 16106127390, 33285996575, 68719476768, 141733920801, 292057776162

Offset: 0

Paul Curtz, Aug 04 2012

Related to Bernoulli numbers.

Essentially the same as A135854.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).

Cf. A001477, A001787, A001792, A027641, A027642, A094373, A132045, A135854, A157809, A164555, A164558.

[n*(1 + 2^(n-1)): n in [0..40]]; // G. C. Greubel, Apr 19 2018

Table[n(1+2^(n-1)),{n,0,40}] (* or *) LinearRecurrence[{6,-13,12,-4},{0,2,6,15}, 40] (* Harvey P. Dale, Oct 18 2013 *)

a(n) = n*(1+2^(n-1)) \\ Michel Marcus, Mar 10 2013

def A215149(n): return n*(pow(2,n)+2)//2
print([A215149(n) for n in range(41)]) # G. C. Greubel, Jan 18 2025

a(n) = (A157809(n) - A164555(n)) / A027642(n).
a(n) = n (the nonnegative integers A001477(n)) + n*2^(n-1) (their binomial transform A001787(n)).
a(n+1) - a(n) = 2,4,9,21,... = A001792(n) + 1.
a(n+1) - 2*a(n) = 2 before A132045(n+1).
a(n) is the binomial transform of b(n) = 0,2,2,3,4,5,... = A001477(n) with 2 instead of 1. b(n) = (A164558(n) - A027641(n))/A027642(n)?
G.f.: x*(2-6*x+5*x^2) / ( (1-x)^2*(1-2*x)^2 ). - R. J. Mathar, Aug 06 2012
E.g.f.: x*exp(x)*(1 + exp(x)). - G. C. Greubel, Jan 18 2025
a(n) = n * A094373(n). - Alois P. Heinz, Jan 18 2025

A256595 Triangle A074909(n) with 0's as second column.

Original entry on oeis.org

1, 1, 0, 1, 0, 3, 1, 0, 6, 4, 1, 0, 10, 10, 5, 1, 0, 15, 20, 15, 6, 1, 0, 21, 35, 35, 21, 7, 1, 0, 28, 56, 70, 56, 28, 8, 1, 0, 36, 84, 126, 126, 84, 36, 9, 1, 0, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 55, 165, 330, 462, 462, 330, 165, 55, 11

Offset: 0

Paul Curtz, Apr 03 2015

For Bernoulli numbers, B(1) excluded.
B(n) is calculated via
B(0) = 1;
B(0) + 0 = 1;
B(0) + 0 + 3*B(2) = 3/2;
B(0) + 0 + 6*B(2) + 4*B(3) = 2;
etc.
The diagonal is A026741(n+1)/A040001(n).
Row sums: 1, 1, 4, 11, 26, 57, ..., essentially Euler numbers A000295. See A130103, A008292 and A173018.
There is an infinitude of Bernoulli number sequences. They are of the form
B(n,q) = 1, q, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, ... .
Chronologically, the first, and the most regular, is, for q=1/2, A164555(n)/A027642(n), from Jacob Bernoulli (1654-1705), published in Ars Conjectandi in 1713 and(?) Seko Kowa (1642-1708) in 1712. See A159688. The second is, for q=-1/2, B(n,-1/2) = A027641(n)/A027642(n), from B(n,1/2) via Pascal's triangle. We could choose Be(n,q) instead of B(n,q) to avoid confusion with Sloane's B(n,p) for A027641(n)/A027642(n) (p=-1), A164555(n)/A027642(n) (p=1), A164558(n)/A027642(n) (p=2), A157809(n)/A027642(n) (p=3), ..., successive binomial transforms of the previous sequence.
This motivates the proposal of the (independent of q) sequence Bernoulli(n+2):
B(n+2) = 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, ... and its inverse binomial transform. See A190339.

			1,
1, 0,
1, 0,  3,
1, 0,  6, 4,
1, 0, 10, 10,  5,
1, 0, 15, 20, 15, 6,
1, 0, 21, 35, 35, 21, 7,
etc.

Jacob Bernoulli, Ars Conjectandi (1713).

Wikipedia, Seki Takakazu (also known as Seki Kowa).

Cf. A007318, A074909, A026741, A004001, A000295, A130103, A008292, A173018, A027641/A027642, A164555/A027642, A176327/A176289.

T[, 0] = 1; T[, 1] = 0; T[n_, k_] := Binomial[n+1, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 11 2016 *)

A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

Original entry on oeis.org

1, 1, 3, 1, 2, 13, 0, 1, 5, 3, -1, -1, 2, 29, 119, 0, -1, -1, 1, 31, 5, 1, 1, -1, -8, -1, 43, 253, 0, 1, 1, 4, -4, -1, 41, 7, -1, -1, -1, 4, 8, 4, -1, 29, 239, 0, -1, -1, -8, -4, 4, 8, 1, 31, 9, 5, 5, 7, -4, -116, -32, -116, -4, 7, 71, 665, 0

Offset: 0

Paul Curtz, Feb 03 2015

The difference table of Bernoulli(n,2) or B(n,2) = A164558(n)/A027642(n) is defined by placing the fractions in the upper row and calculating further rows as the differences of their preceding row:
1,       3/2, 13/6,     3, 119/30, ...
1/2,     2/3,  5/6, 29/30, ...
1/6,     1/6, 2/15, ...
0,     -1/30, ...
-1/30, ...
etc.
The first column is A164555(n)/A027642(n).
In particular, the sums of the antidiagonals
1                              = 1
1/2 + 3/2                      = 2
1/6 + 2/3 + 13/6               = 3
0   + 1/6 +  5/6 + 3           = 4
etc. are the positive natural numbers. (This is rewritten for Bernoulli(n,3) in A157809).
We also have for Bernoulli(.,2)
B(0,2) = 1
B(0,2) + 2*B(1,2) = 4
B(0,2) + 3*B(1,2) + 3*B(2,2) = 12
B(0,2) + 4*B(1,2) + 6*B(2,2) + 4*B(3,2) = 32
etc. with right hand sides provided by A001787.
More generally sum_{s=0..t-1} binomial(t,s)*Bernoulli(s,q) gives A027471(t) for q=3, A002697 for q=4 etc, by reading A104002 downwards the q-th column.

Cf. A027641, A027642, A074909, A085737, A085738, A104002, A157809, A157920, A157930, A157945, A157946, A157965, A164555, A164558, A190339, A158302, A181131 (numerators and denominators of the main diagonal).

nmax = 11; A164558 = Table[BernoulliB[n,2], {n, 0, nmax}]; D164558 = Table[ Differences[A164558, n], {n, 0, nmax}]; Table[ D164558[[n-k+1, k+1]] // Numerator, {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 04 2015 *)

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A215149 a(n) = n * (1 + 2^(n-1)).

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A256595 Triangle A074909(n) with 0's as second column.

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A254630 Ascending antidiagonal numerators of the table of repeated differences of A164558(n)/A027642(n).

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