A100868 -id:A100868 - cp's OEIS frontend

A050946 "Stirling-Bernoulli transform" of Fibonacci numbers.

0, 1, 1, 7, 13, 151, 421, 6847, 25453, 532231, 2473141, 63206287, 352444093, 10645162711, 69251478661, 2413453999327, 17943523153933, 708721089607591, 5927841361456981, 261679010699505967, 2431910546406522973, 118654880542567722871, 1212989379862721528101

Offset: 0

N. J. A. Sloane, Jan 02 2000

nonn

From Paul Curtz, Oct 11 2013: (Start)
Differences table:
     0,    1,    1,    7,   13,  151,  421, 6847, ...
     1,    0,    6,    6,  138,  270, 6426, ...
    -1,    6,    0,  132,  132, 6156, ...
     7,   -6,  132,    0, 6024, ...
   -13,  138, -132, 6024, ...
   151, -270, 6156, ...
  -421, 6426, ...
  6847, ... .
a(n) is an autosequence of first kind: the inverse binomial transform is the sequence signed, the main diagonal is A000004=0's.
The "Stirling-Bernoulli transform" applied to an autosequence of first kind is an autosequence of first kind.
Now consider the Akiyama-Tanigawa transform or algorithm applied to A000045(n):
     0,   1,   1,   2,   3,   5, 8, ...
    -1,   0,  -3,  -4, -10, -18, ...    = -A006490
    -1,   6,   3,  24,  40, ...
    -7,   6, -63, -64, ...
   -13, 138,   3, ...
  -151, 270, ...
  -421, ... .
Hence -a(n). The Akiyama-Tanigawa algorithm applied to an autosequence of first kind is an autosequence of first kind.
a(n+5) - a(n+1) = 150, 420, 6840, ... is divisible by 30.
For an autosequence of the second kind, the inverse binomial transform is the sequence signed with the main diagonal double of the first upper diagonal.
The Akiyama-Tanigawa algorithm applied to an autosequence leads to an autosequence of the same kind. Example: the A-T algorithm applied to the autosequence of second kind 1/n leads to the autosequence of the second kind A164555(n)/A027642(n).
Note that a2(n) = 2*a1(n+1) - a1(n) applied to the autosequence of the first kind a1(n) is a corresponding autosequence of the second kind. (End)

Alois P. Heinz, Table of n, a(n) for n = 0..447
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7

Cf. A000045, A051782, A105796, A163626.

with(combinat):
a:= n-> add((-1)^(k+1) *k! *stirling2(n+1, k+1)*fibonacci(k), k=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 17 2013

CoefficientList[Series[E^x*(1-E^x)/(1-3*E^x+E^(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Aug 13 2013 *)
t[0, k_] := Fibonacci[k]; t[n_, k_] := t[n, k] = (k+1)*(t[n-1, k] - t[n-1, k+1]); a[n_] := t[n, 0] // Abs; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Oct 22 2013, after Paul Curtz *)

{a(n)=polcoeff(sum(m=0, n, fibonacci(m)*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */

O.g.f.: Sum_{n>=1} Fibonacci(n) * n! * x^n / Product_{k=1..n} (1+k*x). - Paul D. Hanna, Jul 20 2011
A100872(n)=a(2*n) and A100868(n)=a(2*n-1).
From Paul Barry, Apr 20 2005: (Start)
E.g.f.: exp(x)*(1-exp(x))/(1-3*exp(x)+exp((2*x))).
a(n) = Sum_{k=0..n} (-1)^(n-k)*S2(n, k)*k!*Fibonacci(k). [corrected by Ilya Gutkovskiy, Apr 04 2019] (End)
a(n) ~ c * n! / (log((3+sqrt(5))/2))^(n+1), where c = 1/sqrt(5) if n is even and c = 1 if n is odd. - Vaclav Kotesovec, Aug 13 2013
a(n) = -1 * Sum_{k = 0..n} A163626(n,k)*A000045(k). - Philippe Deléham, May 29 2015

A100872 a(n) = (1/sqrt(5)) * Sum_{k>0} k^(2n)/phi^(2k) where phi = (1+sqrt(5))/2 = A001622.

Original entry on oeis.org

1, 13, 421, 25453, 2473141, 352444093, 69251478661, 17943523153933, 5927841361456981, 2431910546406522973, 1212989379862721528101, 722875495525684291639213, 507275965883448333971692021, 414031618935013558427928710653, 388884101194230308462039862028741

Offset: 1

Benoit Cloitre, Jan 08 2005

A bisection of "Stirling-Bernoulli transform" of Fibonacci numbers.

Alois P. Heinz, Table of n, a(n) for n = 1..223

Cf. A001622, A050946, A100868, A241171.

FullSimplify[Table[PolyLog[-2k, GoldenRatio^(-2)]/Sqrt[5], {k, 1, 10}]] (* Vladimir Reshetnikov, Feb 16 2011 *)
T[n_, k_] /; 1 <= k <= n := T[n, k] = k(2k-1) T[n-1, k-1] + k^2 T[n-1, k]; T[, 1] = 1; T[, ] = 0; a[n] := Sum[2^(k-1) T[n, k], {k, 1, n}]; Array[a, 15] (* Jean-François Alcover, Jul 03 2019 *)

a(n)=round(1/sqrt(5)*sum(k=1,500,k^(2*n)/((1+sqrt(5))/2)^(2*k)))

a(n) = A050946(2*n).
From Peter Bala, Aug 20 2014: (Start)
E.g.f.: -1/2 + (1/2)*exp(z)/(3*exp(z) - exp(2*z) - 1) = z^2/2! + 13*z^4/4! + 421*z^6/6! + ....
a(n) = Sum_{k = 1..n} 2^(k-1)*A241171(n,k).
a(n) = Sum_{1 <= j <= k <= n} (-1)^(k-j)*binomial(2*k,k+j)*j^(2*n).
(End)

A303675 Triangle read by rows: coefficients in the sum of odd powers as expressed by Faulhaber's theorem, T(n, k) for n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 6, 1, 120, 30, 1, 5040, 1680, 126, 1, 362880, 151200, 17640, 510, 1, 39916800, 19958400, 3160080, 168960, 2046, 1, 6227020800, 3632428800, 726485760, 57657600, 1561560, 8190, 1, 1307674368000, 871782912000, 210680870400, 22313491200, 988107120, 14217840, 32766, 1

Offset: 1

Kolosov Petro, May 08 2018

T(n,k) are the coefficients in an identity due to Faulhaber: Sum_{j=0..n} j^(2*m-1) = Sum_{k=1..m} T(m,k) binomial(n+k, 2*k). See the Knuth reference, page 10.

More explicitly, Faulhaber's theorem asserts that, given integers n >= 0, m >= 1 and odd, Sum_{k=1..n} k^m = Sum_{k=1..(m+1)/2} C(n+k,n-k)*[(1/k)*Sum_{j=0..k-1} (-1)^j*C(2*k,j)*(k-j)^(m+1)]. The coefficients T(m, k) are indicated by square brackets. Sums similar to this inner part are A304330, A304334, A304336; however, these triangles are (0,0)-based and lead to equivalent but slightly more systematic representations. - Peter Luschny, May 12 2018

			The triangle begins (see the Knuth reference p. 10):
         1;
         6,          1;
       120,         30,         1;
      5040,       1680,       126,        1;
    362880,     151200,     17640,      510,       1;
  39916800,   19958400,   3160080,   168960,    2046,    1;
6227020800, 3632428800, 726485760, 57657600, 1561560, 8190, 1;
.
Let S(n, m) = Sum_{j=1..n} j^m. Faulhaber's formula gives for m = 7 (m odd!):
F(n, 7) = 5040*C(n+4, 8) + 1680*C(n+3, 6) + 126*C(n+2, 4) + C(n+1, 2).
Faulhaber's theorem asserts that for all n >= 1 S(n, 7) = F(n, 7).
If n = 43 the common value is 1600620805036.

John H. Conway and Richard Guy, The Book of Numbers, Springer (1996), p. 107.

Donald E. Knuth, Johann Faulhaber and Sums of Powers, arXiv:9207222 [math.CA], 1992.
Petro Kolosov, Polynomial identities involving central factorial numbers, GitHub, 2024. See pp. 3, 6.

First column is a bisection of A000142, second column is a bisection of A001720.
Row sums give A100868.
Cf. A008955, A008957, A036969 and A304330, A304334, A304336.

T := proc(n,k) local m; m := n-k;
2*(2*m+1)!*add((-1)^(j+m)*(j+1)^(2*n)/((j+m+2)!*(m-j)!), j=0..m) end:
seq(seq(T(n, k), k=1..n), n=1..8); # Peter Luschny, May 09 2018

(* After Peter Luschny's above formula. *)
T[n_, k_] := (1/(n-k+1))*Sum[(-1)^j*Binomial[2*(n-k+1), j]*((n-k+1) - j)^(2*n), {j, 0, n-k+1}]; Column[Table[T[n, k], {n, 1, 10}, {k, 1, n}], Center]

def A303675(n, k): return factorial(2*(n-k)+1)*A008957(n, k)
for n in (1..7): print([A303675(n, k) for k in (1..n)]) # Peter Luschny, May 10 2018

T(n, k) = (2*(n-k)+1)!*A008957(n, k), n >= 1, 1 <= k <= n.

T(n, k) = (1/m)*Sum_{j=0..m} (-1)^j*binomial(2*m,j)*(m-j)^(2*n) where m = n-k+1. - Peter Luschny, May 09 2018

New name by Peter Luschny, May 10 2018

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A050946 "Stirling-Bernoulli transform" of Fibonacci numbers.

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A100872 a(n) = (1/sqrt(5)) * Sum_{k>0} k^(2n)/phi^(2k) where phi = (1+sqrt(5))/2 = A001622.

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A303675 Triangle read by rows: coefficients in the sum of odd powers as expressed by Faulhaber's theorem, T(n, k) for n >= 1, 1 <= k <= n.

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