A000967 Sum of Fermat coefficients.

1, 2, 4, 8, 18, 40, 91, 210, 492, 1165, 2786, 6710, 16267, 39650, 97108, 238824, 589521, 1459960, 3626213, 9030450, 22542396, 56393792, 141358274, 354975429, 892893120, 2249412290, 5674891000, 14335757256, 36259245522, 91815545800

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn

Examples

			n...Sum_{c=1..n} (n:c).....a(n)
--------------------------------
.1........1.................1
.2........2.................2
.3........4.................4
.4........8+1/3.............8
		

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000 (corrected by Sean A. Irvine, April 17, 2019)
• R. P. Loh, A. G. Shannon, A. F. Horadam, Divisibility Criteria and Sequence Generators Associated with Fermat Coefficients, Preprint, 1980.
• P. A. Piza, Fermat coefficients, Math. Mag., 27 (1954), 141-146.

Crossrefs

Cf. A143858, A258708.

Programs

Haskell

import Data.Function (on)
a000967 n = round $ sum $
            zipWith ((/) `on` fromIntegral) (a258993_row n) [1, 3 ..]
-- Reinhard Zumkeller, Jun 22 2015

Magma

[Round((&+[Binomial(n+k,n-k)/(2*k+1): k in [0..n-1]])): n in [1..35]]; // G. C. Greubel, Apr 16 2019

Maple

FermatCoeff:=(n,c)->binomial(2*n-c,c-1)/c:seq(round(add(FermatCoeff(n,c),c=1..n)),n=1..40); # Pab Ter, Oct 13 2005

Mathematica

Table[Round[Sum[Binomial[n+k, n-k]/(2*k+1), {k, 0, n-1}]], {n,1,35}] (* G. C. Greubel, Apr 16 2019 *)

PARI

{a(n) = round(sum(k=0,n-1, binomial(n+k,n-k)/(2*k+1)))}; \\ G. C. Greubel, Apr 16 2019

Sage

[round(sum(binomial(n+k,n-k)/(2*k+1) for k in (0..n-1))) for n in (1..35)] # G. C. Greubel, Apr 16 2019

Formula

Following Piza's definition for the Fermat coefficients: (n:c)=binomial(2n-c, c-1)/c, a(n)= Round( sum_ {c=1..n} (n:c) ). - Pab Ter (pabrlos2(AT)yahoo.com), Oct 13 2005

a(n) = rounded(sum(A258993(n,k)/(2*k+1)): k = 0..n-1). - Reinhard Zumkeller, Jun 22 2015

Extensions

More terms from Pab Ter (pabrlos2(AT)yahoo.com), Oct 13 2005