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1, 4, 7, 14, 15, 28, 29, 40, 43, 64, 53, 86, 79, 90, 99, 134, 109, 160, 133, 160, 171, 216, 167, 232, 223, 244, 233, 310, 233, 344, 297, 328, 339, 366, 315, 450, 403, 422, 387, 522, 401, 560, 473, 490, 539, 636, 485, 648, 569, 634, 609, 758, 603, 726, 655
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OFFSET
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1,2
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COMMENTS
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Let A(x) = sum {n >= 1} x^n/(1 - x^n) be the Lambert series generating function of the divisor function tau(n) - see A000005. Then it appears that A(x)^2 is equal to the Lambert series sum {n >= 2} a(n-1)*x^n/(1 - x^n) (checked up to x^1000 using Luschny's formula for a(n)).
This conjecture is equivalent to the following identity for the divisor function: for n >= 2 there holds sum {k = 1..n} tau(k)*tau(n-k) = sum {d divides n} ( sum {k = 1..d} tau(k*(d - k)) ), where we take tau(0) = 0. (End)
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LINKS
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MAPLE
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A072031 := n -> add(numtheory[tau](j*(n+1-j)), j=1..n);
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MATHEMATICA
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T[n_, k_] := T[n, k] = Which[n < 1 || k < 1, 0, n == k, 1, n < k, T[k, n], True, 1 + T[k, n - k]];
a[n_] := Sum[T[n - k + 1, k], {k, 1, n}];
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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