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A123612 Antidiagonal sums of triangle A123610. 3
1, 1, 2, 3, 5, 8, 17, 31, 68, 145, 325, 728, 1685, 3891, 9140, 21565, 51311, 122666, 295037, 712477, 1728262, 4207027, 10276693, 25178708, 61866141, 152397945, 376309596, 931239093, 2309219447, 5737078442, 14278587533, 35595622719 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

The g.f. was suggested by P. D. Hanna. It can be proved either by letting y=x in the bivariate g.f. for sequence A123610 or by using the formula of A. Howroyd (below) for this sequence and the l.g.f. for sequence A167539. The second proof proceeds as follows: Sum_{n>=1} a(n)*x^n = Sum_{n>=1} (1/n)*Sum_{d|n} phi(n/d)*g(d), where g(d) = A167539(d). Then Sum_{n>=1} a(n)*x^n = Sum_{m>=1} (phi(m)/m)*Sum_{d>=1} g(d)*(x^m)^d/d = Sum_{m>=1} (phi(m)/m)*G(x^m), where G(x) = l.g.f. of sequence g(n) = A167539(n). - Petros Hadjicostas, Oct 25 2017

LINKS

Andrew Howroyd, Table of n, a(n) for n = 0..200

FORMULA

a(n) = (1/n) * Sum_{d | n} phi(n/d) * A167539(d) for n>0. - Andrew Howroyd, Apr 02 2017

G.f.: 1-Sum_{n>=1} (phi(n)/n)*f(x^n), where f(x) = log((1-x-x^2+sqrt((1+x+x^2)*(1-3*x+x^2))/2) = -log((1-x-x^2-sqrt((1+x+x^2)*(1-3*x+x^2))/(2*x^3)). - Petros Hadjicostas, Oct 25 2017

MATHEMATICA

Total /@ Table[Function[m, If[k == 0, 1, 1/m DivisorSum[m, If[GCD[k, #] == #, EulerPhi[#] Binomial[m/#, k/#]^2, 0] &]]][n - k + 1], {n, -1, 30}, {k, 0, Ceiling[n/2]}] (* Michael De Vlieger, Apr 03 2017, after Jean-Fran├žois Alcover at A123610 *)

PROG

(PARI) {a(n)=sum(k=0, n\2, if(k==0, 1, (1/(n-k))*sumdiv(n-k, d, if(gcd(k, d)==d, eulerphi(d)*binomial((n-k)/d, k/d)^2, 0))))}

CROSSREFS

Cf. A123610 (triangle), A123611 (row sums); central terms: A123617, A123618, A167539.

Sequence in context: A093000 A122630 A108054 * A077177 A303874 A145793

Adjacent sequences:  A123609 A123610 A123611 * A123613 A123614 A123615

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Oct 03 2006

STATUS

approved

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Last modified August 20 18:56 EDT 2019. Contains 326154 sequences. (Running on oeis4.)