The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A080239 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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A080239

Antidiagonal sums of triangle A035317.
(history; published version)

#77 by Joerg Arndt at Fri Aug 23 02:09:04 EDT 2024

STATUS

reviewed

approved

#76 by Stefano Spezia at Fri Aug 23 02:00:45 EDT 2024

STATUS

proposed

reviewed

#75 by Jason Yuen at Thu Aug 22 22:17:45 EDT 2024

STATUS

editing

proposed

#74 by Jason Yuen at Thu Aug 22 22:17:40 EDT 2024

FORMULA

a(n) = Sum_{j=0..floor(n/2)} Sum_{k=0..floor((n-j)/2)} binomial(n-j-2k, j-2k} ) for n>=0.

STATUS

approved

editing

#73 by Charles R Greathouse IV at Thu Sep 08 08:45:09 EDT 2022

PROG

(MAGMAMagma) I:=[1, 1, 2, 3, 6, 9]; [n le 6 select I[n] else Self(n-1)+Self(n-2)+Self(n-4)-Self(n-5)-Self(n-6): n in [1..50]]; // Vincenzo Librandi, Jun 07 2015

Discussion

Thu Sep 08

08:45

OEIS Server: https://oeis.org/edit/global/2944

#72 by Susanna Cuyler at Sat Jul 13 21:19:56 EDT 2019

STATUS

proposed

approved

#71 by G. C. Greubel at Sat Jul 13 17:08:48 EDT 2019

STATUS

editing

proposed

#70 by G. C. Greubel at Sat Jul 13 17:08:22 EDT 2019

MATHEMATICA

(*f[n] is the Fibonacci sequence and a[n] is the sequence of A080239*) f[n_] := f[n] = f[n - 1] + f[n - 2]; f[1] = 1; f[2] = 1; a[n_] := Which[n == 1, 1, Mod[n, 4] == 2, f[(n + 2)/2]^2, Mod[n, 4] == 3, (f[(n + 5)/2]^2 - 2f[(n + 1)/2]^2 - 1)/3, Mod[n, 4] == 0, (f[(n + 4)/2]^2 + f[n/2]^2 - 1)/3, Mod[n, 4] == 1, (2f[(n + 3)/2]^2 - f[(n - 1)/2]^2 + 1)/3] (* Hiroshi Matsui and Ryohei Miyadera, Aug 08 2006 *)

f=Fibonacci; a2[n_] := Block[{m, s}, s = Mod[n, 4]; m = (n - s)/4;

Which[n == 1, 1, n == 2, 1, n == 3, 2, s == 0, 3 + Sum[f[4 i], {i, 2, m}], s == 1, 1 + Sum[f[4 i 4i+ 1], {i, 1, m}], s == 2, 1 + Sum[f[4 i 4i+ 2], {i, 1, m}], s == 3, 2 + Sum[f[4 i 4i+ 3], {i, 1, m}]]]; Table[a2[n], {n, 1, 3340}] (* Ryohei Miyadera, Apr 11 2014, minor update by Jean-François Alcover, Apr 29 2014 *)

LinearRecurrence[{1, 1, 0, 1, -1, -1}, {1, 1, 2, 3, 6, 9}, 5041] (* Vincenzo Librandi, Jun 07 2015 *)

PROG

(PARI) vector(40, n, f=fibonacci; sum(k=0, ((n-1)\4), f(n-4*k))) \\ G. C. Greubel, Jul 13 2019

(Sage) [sum(fibonacci(n-4*k) for k in (0..floor((n-1)/4))) for n in (1..40)] # G. C. Greubel, Jul 13 2019

(GAP) List([1..40], n-> Sum([0..Int((n-1)/4)], k-> Fibonacci(n-4*k) )); # G. C. Greubel, Jul 13 2019

CROSSREFS

Cf. A035317, A000045, A004695, A026636, A026647, A004695A035317.

STATUS

approved

editing

#69 by Alois P. Heinz at Mon Nov 23 01:21:12 EST 2015

STATUS

proposed

approved

#68 by Jon E. Schoenfield at Mon Nov 23 01:19:02 EST 2015

STATUS

editing

proposed