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A123612
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Antidiagonal sums of triangle A123610.
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3
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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
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MATHEMATICA
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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-François Alcover at A123610 *)
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PROG
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(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))))}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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