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a(n) = Sum_{k=1..n} d(k)*d(n+1-k), where d(k) is number of positive divisors of k.
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%I #69 Sep 19 2024 09:52:06

%S 1,4,8,14,20,28,37,44,58,64,80,86,108,108,136,134,169,160,198,192,236,

%T 216,276,246,310,288,348,310,400,344,433,396,474,408,544,450,564,512,

%U 614,522,688,560,716,638,756,636,860,676,859,772,926,758,1016,804,1032

%N a(n) = Sum_{k=1..n} d(k)*d(n+1-k), where d(k) is number of positive divisors of k.

%C a(n) is the number of ordered ways to express n+1 as a*b+c*d with 1 <= a,b,c,d <= n. - _David W. Wilson_, Jun 16 2003

%C tau(n) (A000005) convolved with itself, treating this result as a sequence whose offset is 2. - _Graeme McRae_, Jun 06 2006

%C Convolution of A341062 and nonzero terms of A006218. - _Omar E. Pol_, Feb 16 2021

%H Charles R Greathouse IV, <a href="/A055507/b055507.txt">Table of n, a(n) for n = 1..10000</a>

%H George E. Andrews, <a href="http://dx.doi.org/10.1007/BF01608779">Stacked lattice boxes</a>, Ann. Comb. 3 (1999), 115-130. See D_{0,0}.

%H A. E. Ingham, <a href="https://doi.org/10.1112/jlms/s1-2.3.202">Some asymptotic formulae in the theory of numbers</a>, Journal of the London Mathematical Society, Vol. s1-2, No. 3 (1927), pp. 202-208.

%H Yoichi Motohashi,<a href="http://www.numdam.org/item?id=ASENS_1994_4_27_5_529_0"> The binary additive divisor problem</a>, Annales scientifiques de l'École Normale Supérieure, Sér. 4, 27 no. 5 (1994), p. 529-572.

%H E. C. Titchmarsh, <a href="https://doi.org/10.1093/qmath/os-13.1.129">Some problems in the analytic theory of numbers</a>, The Quarterly Journal of Mathematics 1 (1942): 129-152.

%F G.f.: Sum_{i >= 1, j >= 1} x^(i+j-1)/(1-x^i)/(1-x^j). - _Vladeta Jovovic_, Nov 11 2001

%F Working with an offset of 2, it appears that the o.g.f is equal to the Lambert series sum {n >= 2} A072031(n-1)*x^n/(1 - x^n). - _Peter Bala_, Dec 09 2014

%F a(n) = A212151(n+2) - A212151(n+1). - _Ridouane Oudra_, Sep 12 2020

%e a(4) = d(1)*d(4) + d(2)*d(3) + d(3)*d(2) + d(4)*d(1) = 1*3 +2*2 +2*2 +3*1 = 14.

%e 3 = 1*1+2*1 in 4 ways, so a(2)=4; 4 = 1*1+1*3 (4 ways) = 2*1+2*1 (4 ways), so a(3)=8; 5 = 4*1+1*1 (4 ways) = 2*2+1*1 (2 ways) + 3*1+2*1 (8 ways), so a(4) = 14. - _N. J. A. Sloane_, Jul 07 2012

%p with(numtheory); A055507:=n->add(tau(j)*tau(n+1-j),j=1..n);

%t Table[Sum[DivisorSigma[0, k]*DivisorSigma[0, n + 1 - k], {k, 1, n}], {n, 1, 100}] (* _Vaclav Kotesovec_, Aug 08 2022 *)

%o (PARI) a(n)=sum(k=1,n,numdiv(k)*numdiv(n+1-k)) \\ _Charles R Greathouse IV_, Oct 17 2012

%o (Python)

%o from sympy import divisor_count

%o def A055507(n): return (sum(divisor_count(i+1)*divisor_count(n-i) for i in range(n>>1))<<1)+(divisor_count(n+1>>1)**2 if n&1 else 0) # _Chai Wah Wu_, Jul 26 2024

%Y Cf. A000005, A000385, A072031, A212151.

%Y Cf. A006218, A341062.

%K nonn

%O 1,2

%A _Leroy Quet_, Jun 29 2000

%E More terms from _James A. Sellers_, Jul 04 2000

%E Definition clarified by _N. J. A. Sloane_, Jul 07 2012